What you’ll learn:

How to evaluate limits approaching vertical asymptotes (infinite limits)
Evaluate limits at infinity (horizontal asymptotes)
Dominant term analysis for rational functions
Handling limits at infinity involving square roots

Infinite limits & vertical asymptotes

An infinite limit occurs when a function approaches $\pm \infty$ as the input ( $x$ ) approaches a finite value. When the limit

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

$or$

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

it typically indicates the function is getting close to a vertical asymptote. Most of the problems you’ll encounter will involve rational functions and evaluating one-sided limits, although trigonometric functions may also appear.

Sidenote

Constant over zero

If direct substitution in a limit problem results in a non-zero constant over a denominator of zero, this is not indeterminate - instead, it signals a vertical asymptote at $x = a$ .

Examples

1. Evaluate

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

Solution

Direct substitution results in $\frac{1}{0}$ , which is not an indeterminate form and suggests a vertical asymptote at $x = 2$ .

To determine if the graph increases or decreases without bound as $x \to 2$ from either side, let’s reason through what would happen when input values of $x$ are slightly smaller than $2$ (approaching from the left).

$f (x)$ would become $1$ divided by a negative number very close to 0, and overall $f (x)$ becomes a very negative number. This is confirmed by a table of values:

$x$	$f (x)$
$1.9$	$- 10$
$1.99$	$- 100$
$1.999$	$- 1000$
$1.9999$	$- 10000$

From the left:

As $x \to 2^{-}$ , $f (x)$ quickly becomes smaller and smaller. So we can estimate that

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

From the right:

On the other hand, inputting a value of $x$ that is slightly larger than $2$ results in $1$ over a very small positive number, and the overall value of $f (x)$ is overwhelmingly large. Then

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

So the overall limit does not exist.

2. Find the vertical asymptotes of $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

Solution

(spoiler)

Answer: $x = 2 only$

This function is undefined when the denominator $= 0$ .

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

However, directly substituting $x = 1$ into the function results in $\frac{0}{0}$ . This indeterminate form means $x = 1$ is not a vertical asymptote. In fact, by taking the limit

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

$= x \to 1 lim \frac{( x - 1 )}{Factor & cancel ( x - 1 ) ( x - 2 )}$

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

$= - 1$

we see that the function approaches $- 1$ as $x$ approaches $1$ . $f (1)$ is still undefined but there is a hole at $x = 1$ rather than an asymptote.

Direct substitution of $x = 2$ results in $\frac{1}{0}$ which means $x = 2$ is a vertical asymptote. In fact, the graph in this problem looks exactly like the one from the first example, $f (x) = \frac{1}{x - 2}$ , except with a hole at $x = 1$ .

Limits at infinity & horizontal asymptotes

On the other hand, limits at infinity describe what happens to $f (x)$ as the input ( $x$ ) approaches $\pm \infty$ . These often reveal horizontal asymptotes, or the long-range/end behavior of the function.

If the limit approaches $L$ as $x$ becomes arbitrarily large or small, then $y = L$ is a horizontal asymptote.

A horizontal asymptote of $f (x)$ is described by:

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

$or$

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

If a problem requires you to find $x \to \pm \infty lim (some rational function)$ , the shortcut is to compare the highest degrees in the numerator and denominator and identify which of these three cases it fits:

Case 1: Degree of top $<$ bottom: $H . A . = 0$

e.g. $x \to \infty lim \frac{x ^{2} - 1}{x ^{4} + 1} = 0$

Case 2: Degree of top $=$ bottom: $H . A . = ratio of leading coefficients$

e.g. $x \to \infty lim \frac{2 x ^{2} + 1}{3 x ^{2} - 2} = \frac{2}{3}$

Case 3: Degree of top $>$ bottom: $H . A . = + \infty$ or $- \infty$

e.g. $x \to \infty lim \frac{x ^{3} - 1}{2 x + 1} = \infty$

This technique of identifying the highest degrees is called dominant term analysis because when $x$ is very large or small, the term with the highest power “dominates” and lower-power terms become negligible in comparison. Rational expressions with limits at infinity can be simplified to their highest degrees; for example,

$x \to \infty lim \frac{x ^{3} - x}{2 x + 1}$

$\approx x \to \infty lim \frac{x ^{3}}{2 x} = x \to \infty lim \frac{x ^{2}}{2} = \infty (DNE)$

Sidenote

About case 3:

Dominant term analysis is most relevant for case 3. The limit can be $+ \infty$ or $- \infty$ depending on what the expression simplifies to.

Examples

1. Evaluate

$x \to - \infty lim \frac{- x ^{3} + x - 1}{3 x ^{2} - 2 x + x ^{4} + 1}$

Solution

Although not written in standard form, the polynomial in the denominator is a 4th degree one, while the one in the numerator is 3rd degree.

Using dominant term analysis, the limit simplifies to

$\approx x \to - \infty lim \frac{- x ^{3}}{x ^{4}}$

$= x \to - \infty lim - \frac{1}{x}$

$= 0$

This is a case 1 situation. Graph the rational function to see that as $x$ gets smaller and smaller, the function starts getting closer and closer to the horizontal asymptote of $y = 0$ .

2. Evaluate

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

Solution

Combine the expression into a single rational function:

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

$= x \to \infty lim \frac{- 5 x ^{2} + 1}{x ^{2}}$

$\approx x \to \infty lim \frac{- 5 x ^{2}}{x ^{2}}$

$= - 5$

This fits case 2, where the limit is just the ratio of the coefficients $\frac{- 5}{1}$ .

Another way to reason through this problem’s original form ( $- 5 + \frac{1}{x ^{2}}$ ) is to consider what happens when $x$ becomes arbitrarily large - the bigger the number on the bottom, the smaller the fraction becomes overall. If $x$ is very large, the function becomes so close to $0$ that the $\frac{1}{x ^{2}}$ effectively doesn’t contribute anything.

3. Evaluate

$x \to - \infty lim \frac{- 2 x ^{4} + x - 1}{- 5 x ^{3} - x}$

Solution

The top is a 4th degree polynomial while the bottom is a 3rd degree one, which fits case 3. With dominant term analysis, the limit simplifies to

$\approx x \to - \infty lim \frac{- 2 x ^{4}}{- 5 x ^{3}}$

$= x \to - \infty lim \frac{2 x}{5}$

$= - \infty (DNE)$

Square roots

Another common type of question involves square roots. For limits at infinity, the general method is to factor out the highest power of $x$ in the square root.

Let

$f (x) = \frac{x}{3 + x ^{2} - 3}$

Evaluate:

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

Solutions

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

(spoiler)

Answer: $1$

Direct substitution results in the indeterminate form $\frac{\infty}{\infty}$ .

In the square root, $x^{2}$ is the highest power. Factor that out:

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

Separate the radical:

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

The next step is important:

$x^{2} = ∣ x ∣$ and not just $x$ . Because no matter what value of $x$ (positive or negative) is put into $x^{2}$ , the output will be positive, which is what the absolute value function does. Then the limit becomes

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

What do we do with limits involving absolute values? Use piecewise functions.

Since $x \to \infty$ and the piece defined for all $x > 0$ is $+ x$ , the limit becomes

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

$x$ must be fully factored in the denominator before canceling:

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

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

As $x$ becomes very large, both $\frac{3}{x ^{2}}$ and $\frac{3}{x}$ effectively become $0$ (negligible). Then the limit is

$\approx x \to \infty lim \frac{1}{0 + 1 - 0} = 1$

b) $x \to - \infty lim f (x)$

(spoiler)

Answer: $- 1$

The process to resolve this is the same as it was in part (a), up to this step:

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

This time, $x \to - \infty$ and the piece defined for all $x < 0$ is $- x$ , so the limit becomes

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

Follow the same process as above (factoring out $x$ and simplifying some of the terms) and the limit this time is $- 1$ .

Graph $f (x)$ in Desmos to confirm visually!

Not all problems involving limits at infinity are rational functions. For example, to evaluate

$x \to \infty lim (x^{2} + 4 x - x)$

First use conjugates to rewrite the expression in rational form.

Solution

(spoiler)

Answer: 2

Direct substitution results in the indeterminate form
$\infty - \infty$ (which is $\neq = 0$ ). Then

$x \to \infty lim (x^{2} + 4 x - x) \cdot Multiplied by conjugate \frac{x ^{2} + 4 x + x}{x ^{2} + 4 x + x}$

$= x \to \infty lim \frac{( x ^{2} + 4 x ) - x ^{2}}{x ^{2} + 4 x + x}$

$= x \to \infty lim \frac{4 x}{x ^{2} + 4 x + x}$

Then follow a similar process to the previous example, factoring out the highest power of $x$ in the square root.

$x \to \infty lim \frac{4 x}{x ^{2} ( 1 + \frac{4}{x} ) + x}$

$= x \to \infty lim \frac{4 x}{( x \cdot 1 + \frac{4}{x} ) + x}$

$= x \to \infty lim \frac{4 x}{x \cdot ( 1 + \frac{4}{x} + 1 )}$

$\approx x \to \infty lim \frac{4}{1 + 0 + 1} = 2$

As a challenge problem, find

$x \to - \infty lim (x^{2} + 4 x - x)$

instead. The answer is, interestingly, $+ \infty$ .

Hint:

(spoiler)

While $\frac{4}{x}$ might as well be treated as $0$ for very negative values of $x$ , think of whether it’s actually a very small positive or negative number and what that contributes to the limit expression!

Key points

Infinite limits occur when $f (x) \to \pm \infty$ as $x \to finite #$
- Expressed as $x \to a lim f (x) = \pm \infty$
- Typically indicates that $x = a$ is a vertical asymptote.
Limits at infinity are about what $f (x)$ does as $x \to \pm \infty$
- Expressed as $x \to \pm \infty lim f (x) = L$ .
- If $L$ is finite, then $y = L$ is a horizontal asymptote.
- If $L = \pm \infty$ , there is no horizontal asymptote ( $f (x)$ rises or falls forever).
Use the 3 cases (dominant term analysis) as a shortcut to determine horizontal asymptotes of rational functions.
For limits at infinity with square roots, turn into a rational expression with conjugates and then factor out the highest power of $x$ in the square root.

What you’ll learn:

How to evaluate limits approaching vertical asymptotes (infinite limits)
Evaluate limits at infinity (horizontal asymptotes)
Dominant term analysis for rational functions
Handling limits at infinity involving square roots

Infinite limits & vertical asymptotes

An infinite limit occurs when a function approaches $\pm \infty$ as the input ( $x$ ) approaches a finite value. When the limit

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

$or$

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

Sidenote

Constant over zero

If direct substitution in a limit problem results in a non-zero constant over a denominator of zero, this is not indeterminate - instead, it signals a vertical asymptote at $x = a$ .

Examples

1. Evaluate

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

Solution

Direct substitution results in $\frac{1}{0}$ , which is not an indeterminate form and suggests a vertical asymptote at $x = 2$ .

$f (x)$ would become $1$ divided by a negative number very close to 0, and overall $f (x)$ becomes a very negative number. This is confirmed by a table of values:

$x$	$f (x)$
$1.9$	$- 10$
$1.99$	$- 100$
$1.999$	$- 1000$
$1.9999$	$- 10000$

From the left:

As $x \to 2^{-}$ , $f (x)$ quickly becomes smaller and smaller. So we can estimate that

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

From the right:

On the other hand, inputting a value of $x$ that is slightly larger than $2$ results in $1$ over a very small positive number, and the overall value of $f (x)$ is overwhelmingly large. Then

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

So the overall limit does not exist.

2. Find the vertical asymptotes of $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

Solution

(spoiler)

Answer: $x = 2 only$

This function is undefined when the denominator $= 0$ .

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

However, directly substituting $x = 1$ into the function results in $\frac{0}{0}$ . This indeterminate form means $x = 1$ is not a vertical asymptote. In fact, by taking the limit

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

$= x \to 1 lim \frac{( x - 1 )}{Factor & cancel ( x - 1 ) ( x - 2 )}$

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

$= - 1$

we see that the function approaches $- 1$ as $x$ approaches $1$ . $f (1)$ is still undefined but there is a hole at $x = 1$ rather than an asymptote.

Limits at infinity & horizontal asymptotes

If the limit approaches $L$ as $x$ becomes arbitrarily large or small, then $y = L$ is a horizontal asymptote.

A horizontal asymptote of $f (x)$ is described by:

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

$or$

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

Case 1: Degree of top $<$ bottom: $H . A . = 0$

e.g. $x \to \infty lim \frac{x ^{2} - 1}{x ^{4} + 1} = 0$

Case 2: Degree of top $=$ bottom: $H . A . = ratio of leading coefficients$

e.g. $x \to \infty lim \frac{2 x ^{2} + 1}{3 x ^{2} - 2} = \frac{2}{3}$

Case 3: Degree of top $>$ bottom: $H . A . = + \infty$ or $- \infty$

e.g. $x \to \infty lim \frac{x ^{3} - 1}{2 x + 1} = \infty$