1.4.1 Vertical asymptotes

Achievable AP Calculus AB

1. Limits

1.4. Limits and infinity

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Vertical asymptotes

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What you’ll learn

How to evaluate limits approaching vertical asymptotes (infinite limits)

An infinite limit happens when a function grows without bound (toward $+ \infty$ or $- \infty$ ) as the input $x$ approaches a finite number.

We express this as

$x \to a lim f (x) = + \infty$

$or$

$x \to a lim f (x) = - \infty$

Graphically, an infinite limit indicates a vertical asymptote at $x = a$ , and $f (a)$ is not defined. These limits most commonly appear when evaluating one-sided limits of rational functions, though they can also occur with trigonometric and logarithmic functions.

Constant over zero:

If direct substitution gives a non-zero constant in the numerator and $0$ in the denominator, the expression is not indeterminate. Instead, it suggests a vertical asymptote at $x = a$ .

Examples

Evaluate

$x \to 2^{-} lim \frac{1}{x - 2}$

(spoiler)

Direct substitution gives $\frac{1}{0}$ , which suggests a vertical asymptote at $x = 2$ .

The simplest way to evaluate the left-hand limit of this rational expression is to plug in values close to but less than $2$ .

Using values like $1.9, 1.99, 1.999,$ etc. the denominator $(x - 2)$ stays negative but gets closer to $0$ .

As a result, $\frac{1}{x - 2}$ becomes a large negative number. This behavior can be confirmed with either its graph or a table:

$x$	$\frac{1}{x - 2}$
$1.9$	$- 10$
$1.99$	$- 100$
$1.999$	$- 1000$
$1.9999$	$- 10000$

A similar analysis can be made for trigonometric functions.

$x \to \frac{π}{2}^{+} lim \frac{x}{cos ( x )} =$

(spoiler)

Direct substitution results in $\frac{π /2}{0}$ , which suggests a vertical asymptote.

As $x$ approaches $\frac{π}{2}$ from the right, $cos (x)$ is negative and approaching $0$ . Since the denominator becomes smaller in magnitude ( $- 0.1, - 0.01, - 0.001,$ etc.), the fraction as a whole becomes increasingly negative.

Therefore

$x \to \frac{π}{2}^{+} lim \frac{x}{cos ( x )} = - \infty$

Asymptotes vs. holes

Limits can also be used to distinguish between vertical asymptotes and holes, also called removable discontinuities.

Vertical asymptote:

If a rational function has a vertical asymptote at $x = a$ , direct substitution results in

$\frac{c}{0} with c \neq = 0$

Removable discontinuity:

A rational function has a hole at $x = a$ if $f (a)$ is undefined but $x \to a lim f (x)$ exists and is a finite value.

This often occurs when direct substitution gives $\frac{0}{0}$ , and a common factor cancels from the numerator and denominator.

A result of $\frac{0}{0}$ alone is not sufficient to conclude a hole - factoring is required to confirm whether the discontinuity is removable or an asymptote.

Example

Find the vertical asymptote(s) of the graphs of

a) $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

b) $y = \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4}$

Solutions

a) $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

(spoiler)

A rational function is undefined when its denominator is $0$ :

$x^{2} - 3 x + 2 = 0 (x - 1) (x - 2) = 0 x = 1, 2$

Next, check what happens at each value.

At $x = 1$ :

Substituting into the original function gives $\frac{0}{0}$ .

Compute the limit by factoring and canceling:

$x \to 1 lim \frac{x - 1}{x ^{2} - 3 x + 2} = x \to 1 lim \frac{( x - 1 )}{( x - 1 ) ( x - 2 )} = x \to 1 lim \frac{1}{x - 2} = - 1$

Although $f (1)$ is still undefined, the function approaches $- 1$ as $x$ approaches $1$ . Therefore the graph has a hole at $(1, - 1)$ , not an asymptote.

At $x = 2$ :

Direct substitution gives $\frac{1}{0}$ , which indicates a vertical asymptote.

In fact, after factoring and simplifying the function, its graph behaves like the simplified expression $\frac{1}{x - 2}$ , but with the hole at $(1, - 1)$ .

b) $y = \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4}$

(spoiler)

The function is undefined when its denominator is $0$ :

$x^{2} - 4 x + 4 = 0 (x - 2)^{2} = 0 x = 2$

Direct substitution of $x = 2$ results in $\frac{0}{0}$ . However, when the limit is evaluated after factoring and canceling:

$x \to 2 lim \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4} = x \to 2 lim \frac{( x - 2 ) ( x - 5 )}{( x - 2 ) ( x - 2 )} = x \to 2 lim \frac{x - 5}{x - 2} = \frac{- 3}{0}$

Since the limit does not exist (the simplified version is still undefined when $x = 2$ ), there is instead a vertical asymptote.

Challenge problem

If the graph of

$y = \frac{x - 1}{3 x ^{2} + b x + 2}$

has a removable discontinuity at $x = 1$ and only one vertical asymptote, find the value of $b$ .

(spoiler)

A hole at $x = 1$ means that direct substitution of $x = 1$ results in $\frac{0}{0}$ .

In other words, the denominator $3 x^{2} + b x + 2$ equals $0$ when $x = 1$ . Solving for $b$ ,

$3 (1)^{2} + b (1) + 2 = 0 3 + b + 2 = 0 b = - 5$

To confirm,

$x \to 1 lim \frac{x - 1}{3 x ^{2} - 5 x + 2} = x \to 1 lim \frac{x - 1}{( x - 1 ) ( 3 x - 2 )} = x \to 1 lim \frac{1}{3 x - 2} = 1$

The hole occurs at $(1, 1)$ , and by the simplified version $y = \frac{1}{3 x - 2}$ , the only vertical asymptote is $x = \frac{2}{3}$ .

Infinite limits

Function grows without bound as x approaches a finite value → vertical asymptote at x = a
“Constant over zero” (c/0, c ≠ 0): not indeterminate, indicates vertical asymptote
Evaluate one-sided limits by testing values close to a from the appropriate direction

Vertical asymptotes vs. removable discontinuities

Vertical asymptote: direct substitution gives c/0 (c ≠ 0) → limit does not exist
Removable discontinuity (hole): direct substitution gives 0/0, but a common factor cancels and the limit exists as a finite value
0/0 alone is insufficient — must factor to determine which type of discontinuity

Identifying asymptotes in rational functions

Set denominator = 0 to find candidate x-values
Test each candidate: factor and simplify to check if discontinuity is a hole or asymptote
After canceling, if the simplified expression still gives c/0, a vertical asymptote remains

Vertical asymptotes

What you’ll learn

How to evaluate limits approaching vertical asymptotes (infinite limits)

An infinite limit happens when a function grows without bound (toward $+ \infty$ or $- \infty$ ) as the input $x$ approaches a finite number.

We express this as

$x \to a lim f (x) = + \infty$

$or$

$x \to a lim f (x) = - \infty$

Constant over zero:

If direct substitution gives a non-zero constant in the numerator and $0$ in the denominator, the expression is not indeterminate. Instead, it suggests a vertical asymptote at $x = a$ .

Examples

Evaluate

$x \to 2^{-} lim \frac{1}{x - 2}$

(spoiler)

Direct substitution gives $\frac{1}{0}$ , which suggests a vertical asymptote at $x = 2$ .

The simplest way to evaluate the left-hand limit of this rational expression is to plug in values close to but less than $2$ .

Using values like $1.9, 1.99, 1.999,$ etc. the denominator $(x - 2)$ stays negative but gets closer to $0$ .

As a result, $\frac{1}{x - 2}$ becomes a large negative number. This behavior can be confirmed with either its graph or a table:

$x$	$\frac{1}{x - 2}$
$1.9$	$- 10$
$1.99$	$- 100$
$1.999$	$- 1000$
$1.9999$	$- 10000$

A similar analysis can be made for trigonometric functions.

$x \to \frac{π}{2}^{+} lim \frac{x}{cos ( x )} =$

(spoiler)

Direct substitution results in $\frac{π /2}{0}$ , which suggests a vertical asymptote.

Therefore

$x \to \frac{π}{2}^{+} lim \frac{x}{cos ( x )} = - \infty$

Asymptotes vs. holes

Limits can also be used to distinguish between vertical asymptotes and holes, also called removable discontinuities.

Vertical asymptote:

If a rational function has a vertical asymptote at $x = a$ , direct substitution results in

$\frac{c}{0} with c \neq = 0$

Removable discontinuity:

A rational function has a hole at $x = a$ if $f (a)$ is undefined but $x \to a lim f (x)$ exists and is a finite value.

This often occurs when direct substitution gives $\frac{0}{0}$ , and a common factor cancels from the numerator and denominator.

A result of $\frac{0}{0}$ alone is not sufficient to conclude a hole - factoring is required to confirm whether the discontinuity is removable or an asymptote.

Example

Find the vertical asymptote(s) of the graphs of

a) $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

b) $y = \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4}$

Solutions

a) $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

(spoiler)

A rational function is undefined when its denominator is $0$ :

$x^{2} - 3 x + 2 = 0 (x - 1) (x - 2) = 0 x = 1, 2$

Next, check what happens at each value.

At $x = 1$ :

Substituting into the original function gives $\frac{0}{0}$ .

Compute the limit by factoring and canceling:

$x \to 1 lim \frac{x - 1}{x ^{2} - 3 x + 2} = x \to 1 lim \frac{( x - 1 )}{( x - 1 ) ( x - 2 )} = x \to 1 lim \frac{1}{x - 2} = - 1$

Although $f (1)$ is still undefined, the function approaches $- 1$ as $x$ approaches $1$ . Therefore the graph has a hole at $(1, - 1)$ , not an asymptote.

At $x = 2$ :

Direct substitution gives $\frac{1}{0}$ , which indicates a vertical asymptote.

In fact, after factoring and simplifying the function, its graph behaves like the simplified expression $\frac{1}{x - 2}$ , but with the hole at $(1, - 1)$ .

b) $y = \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4}$

(spoiler)

The function is undefined when its denominator is $0$ :

$x^{2} - 4 x + 4 = 0 (x - 2)^{2} = 0 x = 2$

Direct substitution of $x = 2$ results in $\frac{0}{0}$ . However, when the limit is evaluated after factoring and canceling:

$x \to 2 lim \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4} = x \to 2 lim \frac{( x - 2 ) ( x - 5 )}{( x - 2 ) ( x - 2 )} = x \to 2 lim \frac{x - 5}{x - 2} = \frac{- 3}{0}$

Since the limit does not exist (the simplified version is still undefined when $x = 2$ ), there is instead a vertical asymptote.

Challenge problem

If the graph of

$y = \frac{x - 1}{3 x ^{2} + b x + 2}$

has a removable discontinuity at $x = 1$ and only one vertical asymptote, find the value of $b$ .

(spoiler)

A hole at $x = 1$ means that direct substitution of $x = 1$ results in $\frac{0}{0}$ .

In other words, the denominator $3 x^{2} + b x + 2$ equals $0$ when $x = 1$ . Solving for $b$ ,

$3 (1)^{2} + b (1) + 2 = 0 3 + b + 2 = 0 b = - 5$

To confirm,

$x \to 1 lim \frac{x - 1}{3 x ^{2} - 5 x + 2} = x \to 1 lim \frac{x - 1}{( x - 1 ) ( 3 x - 2 )} = x \to 1 lim \frac{1}{3 x - 2} = 1$

The hole occurs at $(1, 1)$ , and by the simplified version $y = \frac{1}{3 x - 2}$ , the only vertical asymptote is $x = \frac{2}{3}$ .

Achievable AP Calculus AB

1. Limits

1.4. Limits and infinity

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Vertical asymptotes

5 min read

Font

Discuss

Feedback

What you’ll learn

How to evaluate limits approaching vertical asymptotes (infinite limits)

An infinite limit happens when a function grows without bound (toward $+ \infty$ or $- \infty$ ) as the input $x$ approaches a finite number.

We express this as

$x \to a lim f (x) = + \infty$

$or$

$x \to a lim f (x) = - \infty$

Constant over zero:

If direct substitution gives a non-zero constant in the numerator and $0$ in the denominator, the expression is not indeterminate. Instead, it suggests a vertical asymptote at $x = a$ .

Examples

Evaluate

$x \to 2^{-} lim \frac{1}{x - 2}$

(spoiler)

Direct substitution gives $\frac{1}{0}$ , which suggests a vertical asymptote at $x = 2$ .

The simplest way to evaluate the left-hand limit of this rational expression is to plug in values close to but less than $2$ .

Using values like $1.9, 1.99, 1.999,$ etc. the denominator $(x - 2)$ stays negative but gets closer to $0$ .

As a result, $\frac{1}{x - 2}$ becomes a large negative number. This behavior can be confirmed with either its graph or a table:

$x$	$\frac{1}{x - 2}$
$1.9$	$- 10$
$1.99$	$- 100$
$1.999$	$- 1000$
$1.9999$	$- 10000$

A similar analysis can be made for trigonometric functions.

$x \to \frac{π}{2}^{+} lim \frac{x}{cos ( x )} =$

(spoiler)

Direct substitution results in $\frac{π /2}{0}$ , which suggests a vertical asymptote.

Therefore

$x \to \frac{π}{2}^{+} lim \frac{x}{cos ( x )} = - \infty$

Asymptotes vs. holes

Limits can also be used to distinguish between vertical asymptotes and holes, also called removable discontinuities.

Vertical asymptote:

If a rational function has a vertical asymptote at $x = a$ , direct substitution results in

$\frac{c}{0} with c \neq = 0$

Removable discontinuity:

A rational function has a hole at $x = a$ if $f (a)$ is undefined but $x \to a lim f (x)$ exists and is a finite value.

This often occurs when direct substitution gives $\frac{0}{0}$ , and a common factor cancels from the numerator and denominator.

A result of $\frac{0}{0}$ alone is not sufficient to conclude a hole - factoring is required to confirm whether the discontinuity is removable or an asymptote.

Example

Find the vertical asymptote(s) of the graphs of

a) $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

b) $y = \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4}$

Solutions

a) $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

(spoiler)

A rational function is undefined when its denominator is $0$ :

$x^{2} - 3 x + 2 = 0 (x - 1) (x - 2) = 0 x = 1, 2$

Next, check what happens at each value.

At $x = 1$ :

Substituting into the original function gives $\frac{0}{0}$ .

Compute the limit by factoring and canceling:

$x \to 1 lim \frac{x - 1}{x ^{2} - 3 x + 2} = x \to 1 lim \frac{( x - 1 )}{( x - 1 ) ( x - 2 )} = x \to 1 lim \frac{1}{x - 2} = - 1$

Although $f (1)$ is still undefined, the function approaches $- 1$ as $x$ approaches $1$ . Therefore the graph has a hole at $(1, - 1)$ , not an asymptote.

At $x = 2$ :

Direct substitution gives $\frac{1}{0}$ , which indicates a vertical asymptote.

In fact, after factoring and simplifying the function, its graph behaves like the simplified expression $\frac{1}{x - 2}$ , but with the hole at $(1, - 1)$ .

b) $y = \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4}$

(spoiler)

The function is undefined when its denominator is $0$ :

$x^{2} - 4 x + 4 = 0 (x - 2)^{2} = 0 x = 2$

Direct substitution of $x = 2$ results in $\frac{0}{0}$ . However, when the limit is evaluated after factoring and canceling:

$x \to 2 lim \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4} = x \to 2 lim \frac{( x - 2 ) ( x - 5 )}{( x - 2 ) ( x - 2 )} = x \to 2 lim \frac{x - 5}{x - 2} = \frac{- 3}{0}$

Since the limit does not exist (the simplified version is still undefined when $x = 2$ ), there is instead a vertical asymptote.

Challenge problem

If the graph of

$y = \frac{x - 1}{3 x ^{2} + b x + 2}$

has a removable discontinuity at $x = 1$ and only one vertical asymptote, find the value of $b$ .

(spoiler)

A hole at $x = 1$ means that direct substitution of $x = 1$ results in $\frac{0}{0}$ .

In other words, the denominator $3 x^{2} + b x + 2$ equals $0$ when $x = 1$ . Solving for $b$ ,

$3 (1)^{2} + b (1) + 2 = 0 3 + b + 2 = 0 b = - 5$

To confirm,

$x \to 1 lim \frac{x - 1}{3 x ^{2} - 5 x + 2} = x \to 1 lim \frac{x - 1}{( x - 1 ) ( 3 x - 2 )} = x \to 1 lim \frac{1}{3 x - 2} = 1$

The hole occurs at $(1, 1)$ , and by the simplified version $y = \frac{1}{3 x - 2}$ , the only vertical asymptote is $x = \frac{2}{3}$ .

Infinite limits

Function grows without bound as x approaches a finite value → vertical asymptote at x = a
“Constant over zero” (c/0, c ≠ 0): not indeterminate, indicates vertical asymptote
Evaluate one-sided limits by testing values close to a from the appropriate direction

Vertical asymptotes vs. removable discontinuities

Vertical asymptote: direct substitution gives c/0 (c ≠ 0) → limit does not exist
Removable discontinuity (hole): direct substitution gives 0/0, but a common factor cancels and the limit exists as a finite value
0/0 alone is insufficient — must factor to determine which type of discontinuity

Identifying asymptotes in rational functions

Set denominator = 0 to find candidate x-values
Test each candidate: factor and simplify to check if discontinuity is a hole or asymptote
After canceling, if the simplified expression still gives c/0, a vertical asymptote remains

Vertical asymptotes

What you’ll learn

How to evaluate limits approaching vertical asymptotes (infinite limits)

An infinite limit happens when a function grows without bound (toward $+ \infty$ or $- \infty$ ) as the input $x$ approaches a finite number.

We express this as

$x \to a lim f (x) = + \infty$

$or$

$x \to a lim f (x) = - \infty$

Constant over zero:

If direct substitution gives a non-zero constant in the numerator and $0$ in the denominator, the expression is not indeterminate. Instead, it suggests a vertical asymptote at $x = a$ .

Examples

Evaluate

$x \to 2^{-} lim \frac{1}{x - 2}$

(spoiler)

Direct substitution gives $\frac{1}{0}$ , which suggests a vertical asymptote at $x = 2$ .

The simplest way to evaluate the left-hand limit of this rational expression is to plug in values close to but less than $2$ .

Using values like $1.9, 1.99, 1.999,$ etc. the denominator $(x - 2)$ stays negative but gets closer to $0$ .

As a result, $\frac{1}{x - 2}$ becomes a large negative number. This behavior can be confirmed with either its graph or a table:

$x$	$\frac{1}{x - 2}$
$1.9$	$- 10$
$1.99$	$- 100$
$1.999$	$- 1000$
$1.9999$	$- 10000$

A similar analysis can be made for trigonometric functions.

$x \to \frac{π}{2}^{+} lim \frac{x}{cos ( x )} =$

(spoiler)

Direct substitution results in $\frac{π /2}{0}$ , which suggests a vertical asymptote.

Therefore

$x \to \frac{π}{2}^{+} lim \frac{x}{cos ( x )} = - \infty$

Asymptotes vs. holes

Limits can also be used to distinguish between vertical asymptotes and holes, also called removable discontinuities.

Vertical asymptote:

If a rational function has a vertical asymptote at $x = a$ , direct substitution results in

$\frac{c}{0} with c \neq = 0$

Removable discontinuity:

A rational function has a hole at $x = a$ if $f (a)$ is undefined but $x \to a lim f (x)$ exists and is a finite value.

This often occurs when direct substitution gives $\frac{0}{0}$ , and a common factor cancels from the numerator and denominator.

A result of $\frac{0}{0}$ alone is not sufficient to conclude a hole - factoring is required to confirm whether the discontinuity is removable or an asymptote.

Example

Find the vertical asymptote(s) of the graphs of

a) $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

b) $y = \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4}$

Solutions

a) $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

(spoiler)

A rational function is undefined when its denominator is $0$ :

$x^{2} - 3 x + 2 = 0 (x - 1) (x - 2) = 0 x = 1, 2$

Next, check what happens at each value.

At $x = 1$ :

Substituting into the original function gives $\frac{0}{0}$ .

Compute the limit by factoring and canceling:

$x \to 1 lim \frac{x - 1}{x ^{2} - 3 x + 2} = x \to 1 lim \frac{( x - 1 )}{( x - 1 ) ( x - 2 )} = x \to 1 lim \frac{1}{x - 2} = - 1$

Although $f (1)$ is still undefined, the function approaches $- 1$ as $x$ approaches $1$ . Therefore the graph has a hole at $(1, - 1)$ , not an asymptote.

At $x = 2$ :

Direct substitution gives $\frac{1}{0}$ , which indicates a vertical asymptote.

In fact, after factoring and simplifying the function, its graph behaves like the simplified expression $\frac{1}{x - 2}$ , but with the hole at $(1, - 1)$ .

b) $y = \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4}$

(spoiler)

The function is undefined when its denominator is $0$ :

$x^{2} - 4 x + 4 = 0 (x - 2)^{2} = 0 x = 2$

Direct substitution of $x = 2$ results in $\frac{0}{0}$ . However, when the limit is evaluated after factoring and canceling:

$x \to 2 lim \frac{x ^{2} - 7 x + 10}{x ^{2} - 4 x + 4} = x \to 2 lim \frac{( x - 2 ) ( x - 5 )}{( x - 2 ) ( x - 2 )} = x \to 2 lim \frac{x - 5}{x - 2} = \frac{- 3}{0}$

Since the limit does not exist (the simplified version is still undefined when $x = 2$ ), there is instead a vertical asymptote.

Challenge problem

If the graph of

$y = \frac{x - 1}{3 x ^{2} + b x + 2}$

has a removable discontinuity at $x = 1$ and only one vertical asymptote, find the value of $b$ .

(spoiler)

A hole at $x = 1$ means that direct substitution of $x = 1$ results in $\frac{0}{0}$ .

In other words, the denominator $3 x^{2} + b x + 2$ equals $0$ when $x = 1$ . Solving for $b$ ,

$3 (1)^{2} + b (1) + 2 = 0 3 + b + 2 = 0 b = - 5$

To confirm,

$x \to 1 lim \frac{x - 1}{3 x ^{2} - 5 x + 2} = x \to 1 lim \frac{x - 1}{( x - 1 ) ( 3 x - 2 )} = x \to 1 lim \frac{1}{3 x - 2} = 1$

The hole occurs at $(1, 1)$ , and by the simplified version $y = \frac{1}{3 x - 2}$ , the only vertical asymptote is $x = \frac{2}{3}$ .