MAT 170 -Precalculus - Introduction to Limits and Continuity
1. Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
1. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
2. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
3. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
4. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
5. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
6. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
7. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
8. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
9. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
10. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
11. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
12. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
13. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
14. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
15. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
16. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
17. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
18. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
19. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
20. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
21. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
22. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
23. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
24. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
25. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
26. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
27. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
28. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
29. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
30. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
31. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
32. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
33. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
34. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
35. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
36. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
37. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
38. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
39. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
40. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
41. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
42. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
43. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
44. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
45. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
46. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
47. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
48. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
49. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
50. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
51. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
52. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
53. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
54. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
55. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
56. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
57. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
58. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
59. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
60. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
61. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
62. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
63. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
64. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
65. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
66. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
67. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
68. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
69. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
70. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
71. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
72. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
73. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
74. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
75. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
76. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
77. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
78. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
79. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
80. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
81. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
82. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
83. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
84. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
85. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
86. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
87. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
88. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
89. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
90. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
91. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
92. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
93. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
94. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
95. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
96. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using L'Hôpital's
Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
97. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
98. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
99. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
100. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
101. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
102. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
103. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
104. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
105. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
106. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
107. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
108. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
109. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
110. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
111. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
112. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
113. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
114. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
115. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
116. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
117. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
118. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
119. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
120. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
121. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
122. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
123. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
124. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
125. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
126. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
127. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
128. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
129. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
130. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
131. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
132. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
133. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
134. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
135. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
136. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
137. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
138. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
139. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
140. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
141. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
142. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
143. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
144. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
145. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
146. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
147. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
148. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
149. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
150. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
151. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
152. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
153. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
154. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
155. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
156. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
157. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
158. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
159. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
160. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
161. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
162. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
163. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
164. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
165. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
166. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
167. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
168. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
169. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
170. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
171. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
172. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
173. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
174. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
175. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
176. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
177. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
178. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
179. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
180. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
181. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
182. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
183. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
184. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
185. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
186. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
187. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
188. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
189. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
190. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
191. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
192. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
193. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
194. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
195. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
196. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
197. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
198. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
199. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
200. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
201. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
202. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
203. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
204. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
205. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
206. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
207. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
208. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
209. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
210. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
211. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
212. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
213. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
214. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
215. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
216. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
217. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
218. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
219. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
220. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
221. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
222. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
223. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
224. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
225. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
226. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
227. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
228. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
229. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
230. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
231. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
232. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
233. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
234. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
235. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
236. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
237. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
238. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
239. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
240. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
241. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
242. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
243. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
244. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
245. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
246. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
247. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
248. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
249. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
250. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
251. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
252. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
253. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
254. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
255. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
256. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
257. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
258. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
259. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
260. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
261. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
262. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
263. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
264. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
265. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
266. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
267. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
268. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
269. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
270. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
271. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
272. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
273. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
274. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
275. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
276. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
277. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
278. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
279. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
280. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
281. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
282. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
283. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
284. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
285. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
286. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
287. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
288. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
289. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
290. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
291. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
292. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
293. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
294. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
295. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
296. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
297. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
298. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
299. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
300. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
301. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
302. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
303. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
304. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
305. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
306. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
307. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
308. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
309. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
310. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
311. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
312. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
313. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
314. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
315. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
316. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
317. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
318. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
319. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
320. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
321. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
322. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
323. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
324. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
325. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
326. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
327. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
328. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
329. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
330. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
331. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
332. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
333. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
334. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
335. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
336. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
337. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
338. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
339. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
340. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
341. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
342. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
343. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
344. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
345. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
346. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
347. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
348. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
349. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
350. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
351. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
352. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
353. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
354. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
355. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
356. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
357. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
358. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
359. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
360. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
361. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
362. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
363. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
364. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
365. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
366. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
367. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
368. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
369. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
370. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
371. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
372. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
373. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
374. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
375. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
376. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
377. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
378. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
379. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
380. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
381. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
382. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
383. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
384. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
385. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
386. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
387. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
388. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
389. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
390. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
391. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
392. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
393. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
394. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
395. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
396. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
397. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
398. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
399. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
400. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
401. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
402. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
403. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
404. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
405. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
406. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
407. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
408. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
409. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
410. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
411. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
412. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
413. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
414. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
415. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
416. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
417. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
418. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
419. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
420. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
421. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
422. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
423. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
424. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
425. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
426. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
427. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
428. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
429. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
430. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
431. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
432. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
433. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
434. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
435. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
436. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
437. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
438. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
439. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
440. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
441. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
442. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
443. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
444. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
445. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
446. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
447. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
448. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
449. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
450. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
451. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
452. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
453. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
454. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
455. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
456. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
457. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
458. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
459. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
460. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
461. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
462. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
463. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
464. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
465. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
466. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
467. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
468. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
469. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
470. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
471. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
472. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
473. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
474. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
475. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
476. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
477. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
478. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
479. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
480. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
481. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
482. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
483. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
484. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
485. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
486. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
487. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
488. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
489. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
490. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
491. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
492. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
493. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
494. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
495. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
496. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
497. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
498. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
499. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
500. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
501. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
502. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
503. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
504. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
505. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
506. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
507. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
508. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
509. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
510. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
511. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
512. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
513. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
514. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
515. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
516. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
517. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
518. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
519. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
520. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
521. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
522. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
523. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
524. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
525. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
526. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
527. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
528. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
529. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
530. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
531. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
532. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
533. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
534. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
535. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
536. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
537. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
538. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
539. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
540. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
541. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
542. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
543. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
544. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
Understanding Limits
• Definition of a Limit: The limit of f(x)f(x)f(x) as xxx approaches a value aaa, denoted as
limx→af(x)\lim_{x \to a} f(x)limx→af(x), is the value that f(x)f(x)f(x) approaches as xxx gets
close to aaa.
o Example: For f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, find limx→1f(x)\lim_{x \to 1}
f(x)limx→1f(x).
▪ Solution: Substitute x=1x = 1x=1: f(1)=2(1)+3=5f(1) = 2(1) + 3 =
5f(1)=2(1)+3=5.
• Left-Hand and Right-Hand Limits:
o The left-hand limit of f(x)f(x)f(x) as xxx approaches aaa (denoted
limx→a−f(x)\lim_{x \to a^-} f(x)limx→a−f(x)) is the value f(x)f(x)f(x) approaches
from values less than aaa.
o The right-hand limit (denoted limx→a+f(x)\lim_{x \to a^+} f(x)limx→a+f(x)) is the
value approached from values greater than aaa.
o Example: Find limx→3−\lim_{x \to 3^-}limx→3− and limx→3+\lim_{x \to
3^+}limx→3+ for f(x)={x+2,x<3x2−1,x≥3f(x) = \begin{cases} x + 2, & x < 3 \\ x^2 - 1, &
x \geq 3 \end{cases}f(x)={x+2,x2−1,x<3x≥3.
• Limit Laws:
o Sum Law: limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)\lim_{x \to a} [f(x) + g(x)]
= \lim_{x \to a} f(x) + \lim_{x \to a} g(x)limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
o Product Law: limx→a[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x))\lim_{x \to a} [f(x)
\cdot g(x)] = (\lim_{x \to a} f(x)) \cdot (\lim_{x \to a} g(x))limx→a
[f(x)
⋅
g(x)]=(limx→af(x))
⋅
(limx→ag(x)).
o Quotient Law: limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a}
\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}limx→ag(x)f(x)
=limx→ag(x)limx→af(x), provided limx→ag(x)≠0\lim_{x \to a} g(x) \neq 0limx→a
g(x) =0.
• Techniques for Finding Limits:
o Direct Substitution: Plug in the value of aaa directly if f(x)f(x)f(x) is continuous at
aaa.
o Factoring: Factor expressions where possible to simplify, especially in cases with
indeterminate forms like 0/00/00/0.
▪ Example: limx→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4.
Factor the numerator: (x−2)(x+2)x−2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x +
2x−2(x−2)(x+2)=x+2. Substitute x=2x = 2x=2: limit is 444.
2. Indeterminate Forms and L'Hôpital's Rule
• Indeterminate Forms: Limits that yield forms like 0/00/00/0 or ∞/∞\infty / \infty∞/∞
after substitution.
• L'Hôpital's Rule: For limits that result in 0/00/00/0 or ∞/∞\infty / \infty∞/∞, take the
derivative of the numerator and denominator separately:
limx→af(x)g(x)=limx→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}
\frac{f'(x)}{g'(x)}x→alimg(x)f(x)=x→alimg′(x)f′(x)
o Example: limx→0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}limx→0xsin(x). Taking
derivatives gives limx→0cos(x)1=1\lim_{x \to 0} \frac{\cos(x)}{1} = 1limx→0
1cos(x)=1.
3. Continuity
• Definition of Continuity at a Point: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if:
1. f(a)f(a)f(a) is defined.
2. limx→af(x)\lim_{x \to a} f(x)limx→af(x) exists.
3. limx→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→af(x)=f(a).
o Example: Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1 is
continuous at x=1x = 1x=1.
▪ Factor f(x)=x+1f(x) = x + 1f(x)=x+1 for x≠1x \neq 1x =1. Substitute x=1x =
1x=1: f(1)=2f(1) = 2f(1)=2, so f(x)f(x)f(x) is continuous for all x≠1x \neq 1x =1.
• Types of Discontinuities:
o Removable Discontinuity: A "hole" in the graph, typically due to a factor canceling in
the numerator and denominator.
o Jump Discontinuity: When the left-hand and right-hand limits do not match.
o Infinite Discontinuity: Vertical asymptote at a certain point where the function goes
to infinity.
4. Infinite Limits and Asymptotes
• Vertical Asymptotes:
o If limx→af(x)=±∞\lim_{x \to a} f(x) = \pm \inftylimx→af(x)=±∞, then x=ax = ax=a is
a vertical asymptote of f(x)f(x)f(x).
o Example: For f(x)=1x−2f(x) = \frac{1}{x - 2}f(x)=x−21, as x→2x \to 2x→2, f(x)→∞f(x)
\to \inftyf(x)→∞, so there is a vertical asymptote at x=2x = 2x=2.
• Horizontal Asymptotes:
o If limx→∞f(x)=L\lim_{x \to \infty} f(x) = Llimx→∞f(x)=L, then y=Ly = Ly=L is a
horizontal asymptote.
o Example: For f(x)=3xx+1f(x) = \frac{3x}{x + 1}f(x)=x+13x, find the limit as x→∞x \to
\inftyx→∞: limx→∞3xx+1=3\lim_{x \to \infty} \frac{3x}{x + 1} = 3limx→∞x+13x=3,
so y=3y = 3y=3 is a horizontal asymptote.
5. Practice Problems
545. Evaluate limx→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4x−4x2−16.
546. Determine if f(x)={x2,x<12x+1,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1
\end{cases}f(x)={x2,2x+1,x<1x≥1 is continuous at x=1x = 1x=1.
547. Find the horizontal asymptote of f(x)=2x2+3x2+1f(x) = \frac{2x^2 + 3}{x^2 +
1}f(x)=x2+12x2+3.
548. Evaluate limx→0tan(x)x\lim_{x \to 0} \frac{\tan(x)}{x}limx→0xtan(x) using
L'Hôpital's Rule.
