Calculus Help
2.2 The Limit of a Function
1. Determine the following limits.
f (x) 4
4
2
−5
5
−2
f (x)
f (x)
Name:
The Limit of a Function
Section:
For use with OpenStax Calculus, free at https://openstax.org/details/books/calculus-volume-1
16
2
lim
x→4
2
f (x) =
x
4
lim
x→−3
f (x) =
x
lim
4
2
3
−2
2
1
x→∞
x
f (x) =
lim
x→3
lim
x→2
f (x) = lim
x→0
f (x) = lim
x→2
f (x) = lim
x→0
f (x) =
f (x) =
2. Determine the following limits.
(a) lim
−1 x5 − x1/3 = (b) lim 3x3−1
= (c) lim 7x2−35 =
x→−8 3
x→ √3 1/3 x
x→0
x−5
(d) lim
x→−11
(x+1)(x+11)
x+11
= (e) lim
x→2
x2+4x−12
x−2
= (f) lim
x→0+
ln xx+2 ln x2
3. Sketch and carefully label a graph that has all of the following limits.
2
lim
=
x→1
f (x) = 2 lim
x→−4
f (x) = 2 lim
x→0
f (x) = 1
lim
x 5
→
exist
f (x) does not
4. Use the graph to determine the following:
f (x)
(a) lim
x→1
(b) lim
f (x) =
f (x) =
4
2
−
1
1
2
3
−2
x 2+
→
x
(c) lim
x→0
f (x) =
(d) lim
x→1+
f (x) =
(e) lim
x→1−
f (x) =
(f) lim
x→−1
f (x) =
5. Determine the following limits.
x
(a) lim
x→0
ex−1
xe2x−xex
(b) lim
sin(2x)
x→π
x→0
sin x
(c) lim sin 1
6. Determine the following limits.
(a) lim 3
x→7 x−7
(b) lim 5
x→2 (2−x)2
(c) lim x
x→0 sin x
2.3 The Limit Laws
Limit Laws
|
Limit of the Identity Function |
lim x = x→a |
|
Limit of the Constant Function |
For a constant c, lim c = x→a |
|
Sum Law of Limits |
lim( f (x) + g(x)) = x→a |
|
Difference Law of Limits |
lim( f (x) − g(x)) = x→a |
|
Constant Multiple Law for Limits |
lim(c · f (x)) = x→a |
|
Product Law for Limits |
lim( f (x) · g(x)) = x→a |
|
Quotient Law for Limits |
lim f (x) = x→a g(x) |
|
Power Law for Limits |
lim( f (x))n = x→a |
|
Squeeze Theorem |
Given functions f , g, and h such that ≤ ≤ , if = = L, then lim g(x) = x→a |
|
Special Limits |
lim ex−1 = lim sin x = lim cos x−1 = x→0 x x→0 x x→0 x |
1. Find the following limits.
Name:
The Limit Laws
Section:
For use with OpenStax Calculus, free at https://openstax.org/details/books/calculus-volume-1
19
(a) lim
x2+5x
(b) lim x−1
lim (3+h)2−9
x→−1
x4+2
x→1 x2−1 (c)
h→0 h
(d) lim
t→0
√t2+9 3
t2
−
(e) lim x
x→0
| |
2. Find the following limits:
(a) lim (7+h)2−49
(b) lim (−3+h)2−(−3)2
(c) lim (4+h)2+2−(42+2)
h→0 h
h→0 h
h→0 h
(d) lim (8+h)7−(8)7
3 2 3 2
h→0
h (e) lim (−1+h) −(−1+h) +17−((−1 −(−1) +17))
h→0 h
3. Find the following limits:
x2−6x+9
c2−c
(c) lim
√ 1− p √
(a) lim
c→1
x→3
√x−3 (b) lim
√c−1
p→1
3−p− 2
4. Use the squeeze theorem to determine the following limits.
(a) lim θ2 cos 1 (b) lim ex−1
θ→0
θ
x→0
x
5. Consider the function: f (x) = −2x3 − 7x2 + 1. Determine the following limits.
(a) lim f (h)− f (0) (b) lim f (x+h)− f (x)
h→0 h h→0 h
6. Find the following limits:
(a) lim
θ→π
2
sin θ
1 − 1
(c) lim
tan θ−tan a
tan θ (b) lim
h→0
a+h a
h
θ→π 1+tan θ tan a
2.4 Continuity
|
Composite Functions Theorem |
If f (x) is continuous at L and lim g(x) = L, then: x→a lim f (g(x)) = = x→a |
|
Intermediate Value Theorem |
For any closed, bounded interval [a, b], if z is a real number between and , then there exists a number c in [a, b] such that f (c) = |
1. Evaluate lim ln sin x 2. Evaluate lim sin ex−1
x→0
x
x→0
x
3. Determine whether each of the following functions is continuous over its domain. If it is not, state where it is discontinuous.
Name:
Continuity
Section:
For use with OpenStax Calculus, free at https://openstax.org/details/books/calculus-volume-1
22
f (x) 2
1
f (x) 4
2
1
x
1 x −2
f (x) 4
2
1
x
−2
4. Determine whether each of the following functions is continuous over its domain. If it is not, state where it is discontinuous.
(a) f (x) = 1
x2−1
(b) f (x) = 4
+
x2 1
(c) f (x) = |x−2|
x−2
5. For each of the following, determine whether the function is continuous at the given point. If it is not, state what type of discontinuity it is.
(a) f (x) = 2x2−5x+3 at x = 1
x−1
(a) f (y) = sin(πy) ta y = 1
tan(πy)
(b) f (x) =
x2 − ex ; x < 0
x − 1 ; x ≥ 0
at x = 0
(c) h(θ) = sin θ−cos θ at θ = π
tan θ
6. For each of the following functions and intervals, determine whether or not the Intermediate Value Theorem applies and whether or not given value of f can be found on the interval.
(a) f (x) = 2x|x−1| , f (x) = 0 on [0, 2]
3x−3
(b) f (x) = x3+x2−6x , f (x) = 2 on [−4, 0]
x−2
(c) f (x) = x3+x2−6x , f (x) = −2 on [−4, 0]
x−2
7. Use the Intermediate Value Theorem to determine whether the equation 2x = x3 has a solution over either of the intervals [1.25, 1.375] or [1.375, 1.5]. Explain your answer for each interval.
8. Suppose that y = f (x) is defined for all x. Sketch a graph that meets the conditions listed for each of the following.
(a) f is discontinuous at x = 1 (b) f is only discontinuous at (c) f is discontinous at x = 0,
=
with lim
x→−1
f (x) = −1 and
x = 2 and lim 1
x→0 2
with lim
x→2+
f (x) = 2 and
lim
x→2
f (x) = 4
lim
x→2−
f (x) = −1
A.2 Limits Practice I
Evaluate the following limits.
3x5− 1 x−120
1. lim
x→2
2
x−1
2. lim log
(x + 243) 3. lim sin x + π
Name:
Limits Practice I
Section:
For use with OpenStax Calculus, free at https://openstax.org/details/books/calculus-volume-1
91
4. lim 7x5−2x3+x
x→0
3
x→0
6
5. lim
4x2−6x−28
6. lim
3x3+5x2−6x−10
x→0 x
x→−2
x+2
x→−5/3
5 +3
7. lim sin(x+π)
x2+6x+9
√27+h− √27
x→0
x 8. lim
x→0
√x+3 9. lim h
→
h 0
Sketch a graph that satisfies the given criteria.
10.
x lim
f (x) = −2, lim
f (x) = 2,
11.
x lim
f (x) = ∞, lim
f (x) = −∞,
→lim−∞
x
→∞
x→−4+
f (x) = ∞,
x
lim f (x) = 5
x→2
→lim−∞
x→−5+
f (x) = −
x
10, lim
→∞
x→1
f (x) = 5