CSUN MATH 150A: Comprehensive Study Guide on Asymptotes
(Calculus I)
A CSUN Students Insight on Mastering Asymptotes
Welcome, fellow Math 150A student! Asymptotes are one of the most crucial
concepts youll encounter in Calculus I, as they connect limits (Chapter 2) with
curve sketching (Chapter 4). Dont just memorize the rules; understand why these
lines exist. An asymptote is a limiting behavior—its a line that a function approaches
but never truly touches as its input (x) or output (y) heads towards infinity.
Mastering this topic means mastering limits involving infinity, which is fundamental
to all future calculus courses. Focus on the connection between the degree of the
polynomials in a rational function and its horizontal asymptote—thats where the
limit at infinity comes into play.
I. Foundational Concepts and Definitions
An asymptote is a straight line such that the distance between the curve and the
line approaches zero as one or both of the
x
or
y
coordinates tend to infinity. In
Math 150A, we classify them into three main types:
1. Vertical Asymptotes (VA)
2. Horizontal Asymptotes (HA)
3. Slant (or Oblique) Asymptotes (SA)
A. Vertical Asymptotes (VA)
A vertical line
x=a
is a Vertical Asymptote of the function
f(x)
if at least one of the
following limit statements is true:
lim
x→ a−
f(x)=∞or lim
x → a−
f(x)=− ∞ lim
x →a+¿f(x)=∞or lim
x→a+¿f(x)=−∞ ¿¿¿ ¿
Conceptual Understanding: VAs occur where the functions denominator is
zero and the numerator is non-zero. This creates an "infinite discontinuity" or
a blow-up in the graph.
Procedure for Finding VAs (for Rational Functions
f(x)= N(x)
D(x)
):
a. Factor both the numerator
N(x)
and the denominator
D(x)
completely.
b. Cancel any common factors. (If a factor cancels, it indicates a hole or
removable discontinuity, not a VA.)
c. Set the remaining, simplified denominator equal to zero and solve
for
x
. These
x
-values are the locations of the Vertical Asymptotes.
B. Horizontal Asymptotes (HA)
A horizontal line
y=L
is a Horizontal Asymptote of the function
f(x)
if:
lim
x→ ∞
f(x)=Lor lim
x→ −∞
f(x)=L
Conceptual Understanding: HAs describe the end behavior of the function
—what
y
value the graph settles on as
x
gets extremely large (positive or
negative). A function can have at most two HAs (one for
x → ∞
and one for
x → − ∞
, though rational functions usually only have one).
Procedure for Finding HAs (for Rational Functions
f(x)= anxn+…
bmxm+…
):
Let
n
be the degree of the numerator and
m
be the degree of the
denominator.
a. Case 1: (Degree of Numerator < Degree of Denominator)
n<m
The HA is always
y=0
(the
x
-axis).
Why? The denominator grows much faster than the numerator,
forcing the fraction to approach zero.
b. Case 2: (Degree of Numerator = Degree of Denominator)
n=m
The HA is
y=an
bm
, the ratio of the leading coefficients.
Why? As
x → ± ∞
, the lower-order terms become insignificant,
and the function behaves like
f(x)≈anxn
bmxm=an
bm
.
c. Case 3: (Degree of Numerator > Degree of Denominator)
n>m
There is NO Horizontal Asymptote. The function grows
without bound. (It may have a Slant Asymptote if
n=m+1
).
C. Slant (or Oblique) Asymptotes (SA)
A non-horizontal, non-vertical line
y=mx+b
is a Slant Asymptote of the function
f(x)
if:
lim
x→ ∞
[f(x)−(mx+b)]=0 or lim
x → −∞
[f(x)−(mx +b)]=0
Conceptual Understanding: A SA exists only when the numerators degree is
exactly one greater than the denominators degree (
n=m+1
). It describes
the linear function that the graph resembles for large absolute values of
x
.
Procedure for Finding SAs (for Rational Functions):
a. Perform Polynomial Long Division (or Synthetic Division, if
applicable) of the numerator
N(x)
by the denominator
D(x)
.
b. The division results in a quotient
Q(x)
and a remainder
R(x)
:
f(x)=Q(x)+ R(x)
D(x)
c. Since
deg (N)=deg (D)+1
, the quotient
Q(x)
will be a linear function,
Q(x)=mx+b
.
d. As
x → ± ∞
, the remainder term
R(x)
D(x)
approaches 0 (since
deg (R)<deg(D)
).
e. Therefore, the Slant Asymptote is the linear quotient:
y=mx+b
.
II. Advanced Concepts and Limiting Behavior
A. Rigorous Limit Definitions
The core of asymptotes lies in limits involving infinity.
Limit at Infinity (for HA): The formal definition of
lim
x→ ∞
f(x)=L
is that for
every number
ϵ>0
, there exists a corresponding number
N
such that if
x>N
,
then
¿f(x)− L∨¿ϵ
.
Infinite Limit (for VA): The formal definition of
lim
x→ a
f(x)=∞
is that for every
positive number
M
, there exists a corresponding number
δ>0
such that if
0<¿x −a∨¿δ
, then
f(x)>M
.
B. The Dominance Principle for Limits at Infinity
When evaluating
lim
x→ ± ∞
f(x)
for a rational function, the highest degree terms
dominate the behavior. All other terms become negligible. This principle justifies the
three HA rules:
lim
x →∞
anxn+…
bmxm+…=
lim
x →∞
anxn
bmxm
Example of Dominance Principle:
Evaluate
lim
x →∞
3x2+5x − 1
2x2+4
.
lim
x →∞
3x2+5x− 1
2x2+4=
lim
x→ ∞
3x2
x2+5x
x2−1
x2
2x2
x2+4
x2
=
lim
x → ∞
3+5
x−1
x2
2+4
x2
=3+0−0
2+0=3
2
This confirms the HA rule:
y=3/2
(ratio of leading coefficients).
C. Asymptotes in Non-Rational Functions
Asymptotes are not exclusive to rational functions.
Functions with
√
x2
: When dealing with functions involving square roots, be
careful with
x → − ∞
. Remember that
√
x2=¿x∨¿
.
oIf
x → ∞
, then
¿x∨¿x
.
oIf
x → − ∞
, then
¿x∨¿− x
.
Exponential and Logarithmic Functions:
o
f(x)=ex
: HA at
y=0
as
x → − ∞
.
o
f(x)=ln(x)
: VA at
x=0
(since the domain requires
x>0
).
Trigonometric Functions:
o
f(x)=tan(x)
: VA at
x=π
2+nπ
for any integer
n
.
III. Solved Examples and Practice Problems
Example Set 1: Vertical Asymptotes and Holes
Problem 1: Find the asymptotes and any holes for the function
f(x)= x2− x −2
x2−4
.
1. Factor:
f(x)=(x − 2)(x+1)
(x −2)(x+2)
2. Identify Hole: The factor
(x − 2)
cancels.
oA hole exists at
x=2
. To find the
y
-coordinate of the hole, use the
simplified function:
g(x)= x+1
x+2
.
o
g(2)= 2+1
2+2=3
4
. Hole at
(2,3/4)
.
3. Identify VA: Set the remaining denominator to zero:
x+2=0⇒x=−2
.
oVertical Asymptote (VA):
x=−2
.
4. Confirm VA with Limit:
lim
x→ −2+¿f(x)= lim
x→ −2+¿x+1
x+2¿
¿ ¿
¿
As
x → −2+¿ ¿
,
x+1→ −1
(negative) and
x+2→0+¿ ¿
(small positive).
lim
x→ −2+¿f(x)=negative
positive, small =−∞ ¿
¿
(This confirms
x=−2
is a VA).
Example Set 2: Horizontal Asymptotes
Problem 2a (Case
n<m
): Find the HA for
f(x)= 5x2+1
3x3−2x+4
.
Degree of Numerator (
n
): 2
Degree of Denominator (
m
): 3
Since
n<m
(2 < 3), the HA is
y=0
.
Limit Confirmation:
lim
x → ∞
5x2+1
3x3−2x+4=
lim
x →∞
5/x+1/x3
3−2/x2+4/x3=0+0
3−0+0=0
Horizontal Asymptote (HA):
y=0
.
Problem 2b (Case
n=m
): Find the HA for
g(x)= 6x4− x2
2x4+7x3
.
Degree of Numerator (
n
): 4
Degree of Denominator (
m
): 4
Since
n=m
(4 = 4), the HA is the ratio of the leading coefficients:
y=6
2=3
.
Limit Confirmation:
lim
x →− ∞
6x4− x2
2x4+7x3=
lim
x →− ∞
6−1/x2
2+7/x=6−0
2+0=3
Horizontal Asymptote (HA):
y=3
.
Example Set 3: Slant Asymptotes
Problem 3: Find the asymptotes for
h(x)= x3+x2−1
x2
.
1. Check Degrees:
deg (N)=3
and
deg (D)=2
. Since
3=2+1
, a Slant Asymptote
(SA) exists. (No HA exists).
2. Find VA: Set the denominator to zero:
x2=0⇒x=0
.
oVertical Asymptote (VA):
x=0
.
3. Perform Long Division (or simplification):
h(x)= x3
x2+x2
x2−1
x2=x+1−1
x2
oQuotient
Q(x)=x+1
.
oRemainder term
R(x)
D(x)=−1
x2
.
4. Identify SA: The Slant Asymptote is the quotient
y=x+1
.
oSlant Asymptote (SA):
y=x+1
.
5. Limit Confirmation (for SA):
lim
x→ ∞
[h(x)−(x+1)]= lim
x →∞
[
(
x+1−1
x2
)
−(x+1)
]
=lim
x → ∞
(
−1
x2
)
=0
*(This confirms
y=x+1
is the SA).
Example Set 4: Functions with Square Roots
Problem 4: Find the Horizontal Asymptotes for
k(x)=
√
4x2+3
x − 5
.
Key Insight: Remember
√
x2=¿x∨¿
. We must evaluate the limits at
∞
and
− ∞
separately.
Degree Comparison: The highest power in the numerator is
√
x2≈ x
. The
degree is
≈1
. The degree of the denominator is 1. Since
n=m
, we use the ratio
of leading coefficients, being mindful of signs.
1. Limit as
x → ∞
(
x
is positive):
√
x2=x
.
lim
x →∞
√
4x2+3
x −5=
lim
x →∞
√
x2(4+3/x2)
x(1−5/x)=
lim
x → ∞
x
√
4+3/x2
x(1−5/x)=
lim
x →∞
√
4+3/x2
1−5/x=
√
4+0
1−0=2
oHA for
x → ∞
:
y=2
. (Ratio of leading coefficients:
√
4
1=2
).
2. Limit as
x → − ∞
(
x
is negative):
√
x2=− x
.
lim
x →− ∞
√
4x2+3
x − 5=
lim
x → −∞
√
x2(4+3/x2)
x(1−5/x)=
lim
x →− ∞
¿x∨
√
4+3/x2
x(1−5/x)=
lim
x →− ∞
− x
√
4+3/x2
x(1−5/x)=
lim
x →− ∞
−
√
4+3/x2
1−5/x=−
√
4+0
1−0=−2
oHA for
x → − ∞
:
y=−2
. (Ratio of leading coefficients:
√
4
−1=−2
).
Conclusion: This function has two Horizontal Asymptotes:
y=2
and
y=−2
.
IV. Summary Checklist and Key Takeaways
Asymptote Type When does it occur?
How to Find it (Rational
f(x)=N(x)/ D(x)
)
Vertical (VA) At an
x
-value where the
function blows up to
± ∞
.
Set the simplified
denominator
D(x)
equal to
zero and solve for
x
.
Horizontal (HA) Describes the functions end
behavior as
x → ± ∞
.
Compare
deg (N)
and
deg (D)
:
n<m⇒y=0
;
n=m⇒y=an/bm
;
n>m⇒
None.
Slant (SA) Occurs when the degree of the
numerator is exactly one
greater than the degree of the
denominator (
n=m+1
).
Use Polynomial Long
Division. The SA is the
quotient
y=Q(x)
.
Hole (Removable
Discontinuity)
A factor
(x − a)
cancels out from
the numerator and
The
x
-value
a
causes both
N(x)
and
D(x)
to be zero
Asymptote Type When does it occur?
How to Find it (Rational
f(x)=N(x)/ D(x)
)
denominator. (indeterminate form
0/0
).
The Cardinal Rule of Asymptotes for Curve Sketching:
A rational function can have at most one HA or one SA, but not both.
If
n<m
or
n=m
, the function has a HA.
If
n=m+1
, the function has a SA.
If
n ≥ m+2
, the function has neither (it has a curvilinear asymptote, but we
typically do not cover this in Math 150A).
By diligently practicing the three cases for HA/SA and always factoring to check for
VAs or holes, you will master this section of Math 150A and be well-prepared for
exam questions involving limits and curve sketching. Good luck!
