Infinite Limits and Vertical Asymptotes

An infinite limit describes a function whose values grow without bound as x approaches a number. We write $\lim_{x \to a} f(x) = \infty$ (or $-\infty$ ) when f(x) becomes larger (or more negative) than any number you choose. If either one-sided limit at a is infinite, the line x = a is a vertical asymptote.

Before this lesson: One-Sided Limits, Evaluating Limits Algebraically and Graphing Rational Functions

What an infinite limit means

Divide 1 by smaller and smaller positive numbers: $\frac{1}{0.1} = 10$ , $\frac{1}{0.01} = 100$ , $\frac{1}{0.001} = 1000$ . The results grow past any bound. That is the heart of every infinite limit: a nonzero number divided by something that shrinks toward 0.

The symbol $\infty$ is not a number. Writing $\lim_{x \to a} f(x) = \infty$ is a way of saying the limit does not exist, and also describing how it fails: the values blow up in the positive direction.

On a graph, an infinite limit shows up as a vertical asymptote, a vertical line the curve hugs as it shoots up or down.

Graph of f(x) = (x + 1)/(x − 3) with a dashed vertical asymptote at x = 3 and a dashed horizontal line at y = 1. Just left of x = 3 the curve drops down without bound, so the left-hand limit is negative infinity. Just right of x = 3 it rises without bound, so the right-hand limit is positive infinity. — Near x = 3 the curve plunges on the left and soars on the right, so x = 3 is a vertical asymptote.

How to find infinite limits

When direct substitution gives a nonzero number over 0, the one-sided limits are infinite. Their signs decide whether each is $\infty$ or $-\infty$ .

Substitute x = a. Confirm the numerator approaches a nonzero number and the denominator approaches 0. (If both approach 0, simplify first.)
Find the sign of the numerator near a.
Find the sign of the denominator just left of a and just right of a. Write $0^+$ for a tiny positive number and $0^-$ for a tiny negative number.
Divide the signs: same signs give $\infty$ , opposite signs give $-\infty$ .
If both one-sided limits match, the two-sided limit is that infinity; if not, write each side separately.

Numerator near a	Denominator	Limit
Positive	$0^+$	$\infty$
Positive	$0^-$	$-\infty$
Negative	$0^+$	$-\infty$
Negative	$0^-$	$\infty$

Infinite limits examples

Example 1: An even power in the denominator

Find $\lim_{x \to 0} \frac{1}{x^2}$ .

\frac{1}{0^2} = \frac{1}{0}

Substitute: a nonzero number over 0, so the limit is infinite.

x^2 \gt 0

For any x near 0 except 0 itself, $x^2$ is a tiny positive number, $0^+$ , on both sides.

\frac{1}{0^+}

Positive over $0^+$ is positive and huge.

\lim_{x \to 0} \frac{1}{x^2} = \infty

Both sides go to $\infty$ , so the two-sided limit is $\infty$ .

Answer: $\infty$ . The y-axis, x = 0, is a vertical asymptote.

Example 2: Different signs on each side

Find $\lim_{x \to 3^-} \frac{x + 1}{x - 3}$ and $\lim_{x \to 3^+} \frac{x + 1}{x - 3}$ .

x + 1 \to 4

The numerator approaches 4, a positive number.

x - 3 \to 0^-

For x just left of 3, such as 2.99, x − 3 is a tiny negative number.

\lim_{x \to 3^-} \frac{x + 1}{x - 3} = -\infty

Positive over $0^-$ gives $-\infty$ .

x - 3 \to 0^+

For x just right of 3, such as 3.01, x − 3 is a tiny positive number.

\lim_{x \to 3^+} \frac{x + 1}{x - 3} = \infty

Positive over $0^+$ gives $\infty$ .

Answer: $-\infty$ from the left and $\infty$ from the right, as in the graph above. The two-sided limit does not exist, and x = 3 is a vertical asymptote. Check: x = 3.01 gives $\frac{4.01}{0.01} = 401$ .

Example 3: A negative numerator

Find $\lim_{x \to -2} \frac{x - 1}{(x + 2)^2}$ .

x - 1 \to -3

The numerator approaches −3, a negative number.

(x + 2)^2 \to 0^+

A square is positive, so the denominator is $0^+$ on both sides.

\lim_{x \to -2} \frac{x - 1}{(x + 2)^2} = -\infty

Negative over $0^+$ gives $-\infty$ on both sides.

Answer: $-\infty$ . Both branches of the graph head down along x = −2.

Example 4: Asymptote or hole?

Find the vertical asymptotes of $f(x) = \frac{x^2 - 1}{x^2 - x - 2}$ and describe the limits there.

f(x) = \frac{(x - 1)(x + 1)}{(x - 2)(x + 1)}

Factor the top and the bottom (see factoring trinomials).

f(x) = \frac{x - 1}{x - 2}

Cancel x + 1, for $x \ne -1$ .

\lim_{x \to -1} f(x) = \frac{-1 - 1}{-1 - 2} = \frac{2}{3}

At x = −1 the factor canceled, so there is a hole, not an asymptote.

x - 1 \to 1

At x = 2 the numerator approaches 1, which is positive.

\lim_{x \to 2^-} f(x) = -\infty

Left of 2 the denominator is $0^-$ : positive over negative.

\lim_{x \to 2^+} f(x) = \infty

Right of 2 the denominator is $0^+$ : positive over positive.

Answer: the only vertical asymptote is x = 2, with $-\infty$ on the left and $\infty$ on the right. At x = −1 the graph has a hole at $\left(-1, \frac{2}{3}\right)$ .

Example 5: Word problem with pollution cleanup

The cost, in thousands of dollars, to remove p percent of a pollutant from a lake is $C(p) = \frac{50p}{100 - p}$ . Find $\lim_{p \to 100^-} C(p)$ and explain what it means.

50p \to 5000

The numerator approaches 50(100) = 5000, which is positive.

100 - p \to 0^+

For p just under 100, such as 99.9, 100 − p is a tiny positive number.

\lim_{p \to 100^-} C(p) = \infty

Positive over $0^+$ gives $\infty$ .

C(99) = \frac{4950}{1} = 4950

Check: removing 99% costs $4,950,000.

C(99.9) = \frac{4995}{0.1} = 49950

Removing 99.9% costs $49,950,000.

Answer: $\infty$ . The cost grows without bound as p approaches 100, so removing all of the pollutant is impossible at any price.

Common mistakes

Mistake	Example of the error	Fix
Treating ∞ as a number	$\infty - \infty = 0$	∞ describes growth without bound; it does not follow ordinary arithmetic.
Calling every zero denominator an asymptote	Saying x = −1 is an asymptote of Example 4	If the factor cancels, it is a hole. Simplify first.
Getting the sign wrong on one side	$\lim_{x \to 3^-} \frac{x + 1}{x - 3} = \infty$	Left of 3, x − 3 is negative, so the limit is $-\infty$ .
Writing a two-sided limit when the sides differ	$\lim_{x \to 3} \frac{x + 1}{x - 3} = \infty$	The sides are $-\infty$ and $\infty$ , so give each side separately.
Forgetting that squares are positive	Using $0^-$ for $(x + 2)^2$ left of −2	A square is never negative: $0^+$ on both sides.
Stopping at nonzero/0 and writing “undefined”	“The limit is 4/0, undefined”	Do the sign analysis to say whether it is $\infty$ or $-\infty$ .

Related methods

Trig and log functions

Infinite limits also come from other functions. $\tan x = \frac{\sin x}{\cos x}$ blows up where $\cos x = 0$ , so $\lim_{x \to \pi/2^-} \tan x = \infty$ . The logarithm has $\lim_{x \to 0^+} \ln x = -\infty$ (see graphing logarithmic functions).

Infinite limits versus limits at infinity

An infinite limit has x approaching a number and y blowing up, which gives a vertical asymptote. A limit at infinity has x growing without bound, which can give a horizontal asymptote. That is the next lesson.

Practice problems

Find $\lim_{x \to 4} \frac{3}{(x - 4)^2}$ .
Find $\lim_{x \to 1^-} \frac{2}{x - 1}$ and $\lim_{x \to 1^+} \frac{2}{x - 1}$ .
Find $\lim_{x \to -3^+} \frac{x}{x + 3}$ .
Find $\lim_{x \to -3^-} \frac{x}{x + 3}$ .
Find $\lim_{x \to 0} \frac{-5}{x^4}$ .
Find the vertical asymptotes of $f(x) = \frac{x + 2}{x^2 - 9}$ .
Does $g(x) = \frac{x^2 - 4}{x - 2}$ have a vertical asymptote at x = 2? Explain.
Find $\lim_{x \to \pi/2^-} \tan x$ .
Find $\lim_{x \to 0^+} \ln x$ .
A pool loses 20 liters per minute through a leak. With a pump that adds r liters per minute, the time to add 600 liters is $T(r) = \frac{600}{r - 20}$ minutes. Find $\lim_{r \to 20^+} T(r)$ and explain it.

Show answers

(x - 4)^2 \to 0^+

The numerator is 3, and the square is positive on both sides.

\lim_{x \to 4} \frac{3}{(x - 4)^2} = \infty

Positive over $0^+$ .

Answer: $\infty$

x - 1 \to 0^-

Left of 1 the denominator is negative.

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

Positive over $0^-$ .

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

Right of 1 the denominator is $0^+$ .

Answer: $-\infty$ and $\infty$

x \to -3

The numerator approaches −3, negative.

x + 3 \to 0^+

Right of −3, such as −2.99, x + 3 is positive.

\lim_{x \to -3^+} \frac{x}{x + 3} = -\infty

Negative over $0^+$ .

Answer: $-\infty$

x + 3 \to 0^-

Left of −3, such as −3.01, x + 3 is negative; the numerator is still near −3.

\lim_{x \to -3^-} \frac{x}{x + 3} = \infty

Negative over $0^-$ is positive.

Answer: $\infty$

x^4 \to 0^+

An even power is positive on both sides.

\lim_{x \to 0} \frac{-5}{x^4} = -\infty

Negative over $0^+$ .

Answer: $-\infty$

x^2 - 9 = (x - 3)(x + 3)

Factor the denominator. It is 0 at x = 3 and x = −3.

3 + 2 = 5

At x = 3 the numerator is 5, not 0, so x = 3 is an asymptote.

-3 + 2 = -1

At x = −3 the numerator is −1, not 0, so x = −3 is an asymptote.

Answer: x = 3 and x = −3

\frac{x^2 - 4}{x - 2} = \frac{(x + 2)(x - 2)}{x - 2} = x + 2

Factor and cancel, for $x \ne 2$ .

\lim_{x \to 2} g(x) = 2 + 2 = 4

The limit is a number.

Answer: no. The factor x − 2 cancels, so there is a hole at (2, 4), not an asymptote.

\sin x \to 1

Near $\frac{\pi}{2}$ , the sine is close to 1.

\cos x \to 0^+

Just left of $\frac{\pi}{2}$ (in Quadrant I), the cosine is small and positive (see the unit circle).

\lim_{x \to \pi/2^-} \tan x = \infty

Positive over $0^+$ .

Answer: $\infty$

\ln 0.01 \approx -4.6

Try small inputs.

\ln 0.000001 \approx -13.8

The outputs keep decreasing without bound.

Answer: $-\infty$

10.

600 \gt 0

The numerator is a positive constant.

r - 20 \to 0^+

For r just over 20, r − 20 is a tiny positive number.

\lim_{r \to 20^+} T(r) = \infty

Positive over $0^+$ .

Answer: $\infty$ . When the pump barely beats the leak, the time to add 600 liters grows without bound.

Infinite limits FAQ

What is an infinite limit?

It is a limit where the function values grow without bound, positively or negatively, as x approaches a number. It is written with $\infty$ or $-\infty$ .

Does an infinite limit exist?

No. Writing $\infty$ describes how the limit fails to exist; the values do not settle on a real number.

How do you find a vertical asymptote with limits?

The line x = a is a vertical asymptote if at least one one-sided limit at a is $\infty$ or $-\infty$ . For rational functions, look for zeros of the denominator that do not cancel.

How do you know if the limit is positive or negative infinity?

Check the sign of the numerator and the sign of the tiny denominator on that side. Same signs give $\infty$ ; opposite signs give $-\infty$ .

Summary

$\lim_{x \to a} f(x) = \infty$ means f(x) grows without bound as x approaches a; the limit still does not exist as a number.
Nonzero over 0 after substituting signals an infinite limit.
Use sign analysis on each side: same signs give $\infty$ , opposite signs give $-\infty$ .
An infinite one-sided limit at a means x = a is a vertical asymptote.
Simplify first: a canceled factor makes a hole, not an asymptote.

Previous lesson: Evaluating Limits Algebraically

Next lesson: Limits at Infinity

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