Infinite Limits: Definition, Asymptotes and Examples

What Are Infinite Limits?

Infinite limits describe a function whose values grow without bound, positively or negatively, as $x$ approaches a finite number $a$. If $f(x)$ can be made larger than any number you choose by taking $x$ close enough to $a$, we write $\lim_{x \to a} f(x) = +\infty$. If $f(x)$ can be made more negative than any number you choose, we write $\lim_{x \to a} f(x) = -\infty$.

The idea comes in one-sided and two-sided versions, just like the ordinary limit of a function:

$$\lim_{x \to a^{-}} f(x) = \pm\infty \qquad \lim_{x \to a^{+}} f(x) = \pm\infty \qquad \lim_{x \to a} f(x) = \pm\infty$$

• $x \to a^{-}$ uses only values below $a$, and $x \to a^{+}$ uses only values above $a$.

• $+\infty$ means "increases without bound" and $-\infty$ means "decreases without bound". Neither is a real number.

• The two-sided statement $\lim_{x \to a} f(x) = +\infty$ requires both one-sided limits to be $+\infty$ (and the same for $-\infty$).

The precise definition replaces the usual $\varepsilon$ with a height $M$. The statement $\lim_{x \to a} f(x) = +\infty$ means: for every $M > 0$ there is a $\delta > 0$ such that

$$0 < \lvert x - a \rvert < \delta \quad \Longrightarrow \quad f(x) > M$$

An infinite limit is a description, not a value. In standard calculus, a limit "exists" only when it is a real number $L$. Writing $\lim_{x \to 0} \frac{1}{x^2} = +\infty$ records exactly how the limit fails to exist: the values climb past every bound. OpenStax makes the same point in Calculus Volume 1, Section 2.2, and courses that use the extended real line (where $+\infty$ and $-\infty$ are added as points) are the only exception.

How Do You Evaluate Infinite Limits?

Most infinite limits come from a quotient whose denominator heads to 0 while the numerator heads to a nonzero number. The job is to decide the size (it blows up) and the sign (plus or minus) on each side.

A Table Of Values Near Zero

A table shows the pattern before any rule does. For $f(x) = \frac{1}{x}$ near 0:

Each step closer to 0 multiplies the size by 10, and the sign matches the sign of $x$. So $\lim_{x \to 0^{+}} \frac{1}{x} = +\infty$ and $\lim_{x \to 0^{-}} \frac{1}{x} = -\infty$. Because the two sides disagree, the two-sided limit has no single description, so it simply does not exist.

The Sign Method

Tables can mislead when the blow-up is slow, so the reliable method is a sign analysis:

1. Substitute $x = a$ first. A result of the form $\frac{c}{0}$ with $c \neq 0$ signals an infinite limit. A result of $\frac{0}{0}$ is different: simplify first, because the limit might be finite.

2. Find the sign of the numerator near $a$. It matches the sign of $c$.

3. Find the sign of the denominator on each side. Test one value just left and one just right of $a$, or reason from the factors.

4. Combine. Same signs give $+\infty$, opposite signs give $-\infty$.

For $\frac{x + 1}{x - 2}$ at $x = 2$: the numerator heads to 3 (positive), the denominator is positive for $x > 2$ and negative for $x < 2$. So the right-hand limit is $+\infty$ and the left-hand limit is $-\infty$.

Writing The Answer: Undefined, Infinity Or Does Not Exist

These three words do different jobs, and mixing them costs marks. "Undefined" describes the function value: $f(2)$ is undefined for $\frac{x + 1}{x - 2}$ because of division by zero. "$+\infty$" or "$-\infty$" describes a limit whose values grow without bound in one direction. "Does not exist" is the umbrella verdict whenever no real-number limit exists.

So the most informative answer is the most specific true one:

• If both sides go to $+\infty$, write $\lim_{x \to a} f(x) = +\infty$.

• If the sides go to opposite infinities, give each one-sided limit and say the two-sided limit does not exist.

• Never write "the limit is undefined". A limit either exists, is $\pm\infty$, or does not exist.

Infinite Limits Of Reciprocal Powers: Even Versus Odd

The cleanest family of infinite limits is $f(x) = \frac{1}{(x - a)^n}$ for a positive integer $n$, and the parity of $n$ decides everything. An even power makes the denominator positive on both sides of $a$, while an odd power keeps the sign of $x - a$.

$$n \text{ even:} \quad \lim_{x \to a} \frac{1}{(x - a)^n} = +\infty$$

$$n \text{ odd:} \quad \lim_{x \to a^{-}} \frac{1}{(x - a)^n} = -\infty, \qquad \lim_{x \to a^{+}} \frac{1}{(x - a)^n} = +\infty$$

That is why $\frac{1}{x^2}$ shoots up on both sides of 0 while $\frac{1}{x}$ splits. For $\frac{1}{x^2}$ both one-sided limits are $+\infty$, so the two-sided statement $\lim_{x \to 0} \frac{1}{x^2} = +\infty$ is correct, even though no real-number limit exists. A multiplying constant only scales the picture: $\frac{-3}{x^2}$ goes to $-\infty$ on both sides.

How Are Infinite Limits Related To Vertical Asymptotes?

A vertical asymptote is the graphical name for an infinite limit. The line $x = a$ is a vertical asymptote of $f$ if at least one of these holds:

$$\lim_{x \to a^{-}} f(x) = \pm\infty \qquad \text{or} \qquad \lim_{x \to a^{+}} f(x) = \pm\infty$$

One side is enough. The graph of $\ln x$ has a vertical asymptote at $x = 0$ even though only the right-hand limit makes sense there, because $\lim_{x \to 0^{+}} \ln x = -\infty$ (the natural logarithm is only defined for $x > 0$). In the language of discontinuities, every vertical asymptote marks an infinite discontinuity.

For rational functions, the asymptotes sit where the denominator is zero after cancelling common factors:

1. Factor the numerator and the denominator.

2. Cancel any common factor $(x - a)$. Its zero becomes a hole, a removable discontinuity, not an asymptote.

3. Read the asymptotes from the zeros of the denominator that remain.

Trigonometric functions carry asymptotes too. The tangent function blows up at odd multiples of $\frac{\pi}{2}$, with $\lim_{x \to (\pi/2)^{-}} \tan x = +\infty$ and $\lim_{x \to (\pi/2)^{+}} \tan x = -\infty$, and the same sign reasoning handles $\sec x$, $\csc x$ and $\cot x$ (all ratios built from the trigonometric functions sine and cosine).

Infinite Limits Vs Limits At Infinity

Infinite limits and limits at infinity sound alike but ask opposite questions. An infinite limit sends $x$ to a finite number and watches the output explode, while a limit at infinity sends $x$ off to infinity and watches where the output settles.

The two can combine. $\lim_{x \to \infty} x^2 = +\infty$ is an infinite limit at infinity, which says the output grows without bound as the input does. The full method for the right-hand column lives in limits at infinity.

What Are The Rules For Infinite Limits?

Infinite limits follow their own arithmetic, and a few combinations have no fixed answer. Suppose $\lim_{x \to a} f(x) = +\infty$ and $\lim_{x \to a} g(x) = L$, where $L$ is a real number. Then:

• Sum: $\lim_{x \to a} \left[f(x) + g(x)\right] = +\infty$. A finite amount cannot stop unbounded growth.

• Product with $L > 0$: $\lim_{x \to a} f(x),g(x) = +\infty$.

• Product with $L < 0$: $\lim_{x \to a} f(x),g(x) = -\infty$.

• Quotient: $\lim_{x \to a} \frac{g(x)}{f(x)} = 0$. A fixed number divided by a growing one shrinks to 0.

Two combinations stay open. When $L = 0$, the product $f(x),g(x)$ has the form $0 \cdot \infty$, and when two functions both go to $+\infty$, their difference has the form $\infty - \infty$. Both are indeterminate forms: the answer depends on how fast each part moves, so you have to rewrite the expression before deciding. Example 5 below works one.

Examples Of Infinite Limits

Example 1

Find $\lim_{x \to 3^{+}} \frac{5}{x - 3}$ and $\lim_{x \to 3^{-}} \frac{5}{x - 3}$.

Substituting gives $\frac{5}{0}$, so both limits are infinite. The numerator is positive.

For $x > 3$, $x - 3 > 0$:

$$\lim_{x \to 3^{+}} \frac{5}{x - 3} = +\infty$$

For $x < 3$, $x - 3 < 0$:

$$\lim_{x \to 3^{-}} \frac{5}{x - 3} = -\infty$$

Final answer: $+\infty$ from the right, $-\infty$ from the left. The two-sided limit does not exist, and $x = 3$ is a vertical asymptote.

Example 2

Find $\lim_{x \to 1} \frac{x - 4}{(x - 1)^2}$.

The tempting move is to spot the zero denominator and the even power, remember that $\frac{1}{x^2}$ goes to $+\infty$, and stop there:

$$\frac{x - 4}{(x - 1)^2} \to \frac{\text{something}}{0^{+}} \quad \text{so the answer is } +\infty\text{?}$$

Test a value before trusting that. At $x = 1.1$, the numerator is $-2.9$ and the denominator is $0.01$, so the quotient is $-290$. The values are going down, not up. The even power fixed the sign of the denominator, but the numerator was never checked.

The full sign analysis:

$$\lim_{x \to 1} (x - 4) = -3 \quad \text{(negative)}$$

$$(x - 1)^2 > 0 \quad \text{on both sides of } 1$$

$$\frac{\text{negative}}{\text{small positive}} \to -\infty$$

Final answer: $\lim_{x \to 1} \frac{x - 4}{(x - 1)^2} = -\infty$, from both sides.

Example 3

For $f(x) = \frac{x^2 - 4}{x^2 - x - 2}$, find the vertical asymptotes and the one-sided limits there.

Factor both parts:

$$f(x) = \frac{(x - 2)(x + 2)}{(x - 2)(x + 1)} = \frac{x + 2}{x + 1}, \quad x \neq 2$$

The factor $(x - 2)$ cancels, so $x = 2$ is a hole, with $\lim_{x \to 2} f(x) = \frac{4}{3}$. The remaining zero $x = -1$ gives the asymptote. Near $-1$ the numerator heads to $1$ (positive):

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

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

Final answer: one vertical asymptote, $x = -1$ ($+\infty$ from the right, $-\infty$ from the left), and a hole at $\left(2, \frac{4}{3}\right)$.

Example 4

Find $\lim_{x \to \pi^{-}} \cot x$ and $\lim_{x \to \pi^{+}} \cot x$.

Write $\cot x = \frac{\cos x}{\sin x}$. Near $\pi$, $\cos x \to -1$ and $\sin x \to 0$.

For $x$ just below $\pi$ (second quadrant), $\sin x > 0$:

$$\lim_{x \to \pi^{-}} \frac{\cos x}{\sin x} = \frac{-1}{0^{+}} \to -\infty$$

For $x$ just above $\pi$ (third quadrant), $\sin x < 0$:

$$\lim_{x \to \pi^{+}} \frac{\cos x}{\sin x} = \frac{-1}{0^{-}} \to +\infty$$

Final answer: $-\infty$ from the left and $+\infty$ from the right, so $x = \pi$ is a vertical asymptote of $\cot x$.

Example 5

Find $\lim_{x \to 0^{+}} \left(\frac{1}{x} + \ln x\right)$.

As $x \to 0^{+}$, $\frac{1}{x} \to +\infty$ and $\ln x \to -\infty$. That is the indeterminate form $\infty - \infty$, so combine into one fraction:

$$\frac{1}{x} + \ln x = \frac{1 + x\ln x}{x}$$

The product $x \ln x$ tends to 0 as $x \to 0^{+}$ (the factor $x$ shrinks faster than $\ln x$ grows), so

$$\lim_{x \to 0^{+}} (1 + x\ln x) = 1$$

The numerator heads to 1 and the denominator to $0^{+}$:

$$\lim_{x \to 0^{+}} \frac{1 + x\ln x}{x} = +\infty$$

Final answer: $+\infty$. The $\frac{1}{x}$ term wins the race.

Example 6

The Lorentz factor is $\gamma(v) = \frac{1}{\sqrt{1 - v^2/c^2}}$. Find $\lim_{v \to c^{-}} \gamma(v)$ and compute $\gamma$ at $v = 0.99c$, $0.999c$ and $0.9999c$.

Write $\beta = \frac{v}{c}$, so $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$. As $\beta \to 1^{-}$, $1 - \beta^2 \to 0^{+}$, so its square root heads to $0^{+}$:

$$\lim_{\beta \to 1^{-}} \frac{1}{\sqrt{1 - \beta^2}} = +\infty$$

The values show the climb:

$$\gamma(0.99c) = \frac{1}{\sqrt{1 - 0.9801}} \approx 7.0888$$

$$\gamma(0.999c) \approx 22.3663$$

$$\gamma(0.9999c) \approx 70.7124$$

Final answer: $\lim_{v \to c^{-}} \gamma(v) = +\infty$. Only the left-hand limit is meaningful, because speeds above $c$ make the square root undefined.

Why Do Infinite Limits Matter? "No Object With Mass Reaches Light Speed"

Infinite limits matter because they mark the places where a model, a graph or an integral changes character. An infinite limit is a warning sign that a formula has hit a wall. Reading that warning correctly is useful well beyond the limit chapter.

• Physical limits. The Lorentz factor in Example 6 is why particle accelerators can push particles ever closer to light speed but never to it: the energy needed has an infinite limit as $v \to c^{-}$. OpenStax uses the same relativity formula as a worked example in its limits chapter.

• Graph sketching. Every vertical asymptote of a graph comes from an infinite limit, so finding them is a standard step in curve sketching. Knowing the sign on each side tells you which way each branch bends.

• Areas next to an asymptote. An infinite limit does not always mean infinite area. Even though $\lim_{x \to 0^{+}} \frac{1}{\sqrt{x}} = +\infty$, the area $\int_{0}^{1} \frac{1}{\sqrt{x}},dx$ equals 2, while $\int_{0}^{1} \frac{1}{x},dx$ diverges. Those questions belong to improper integrals, and every one of them starts by spotting an infinite limit.

• Where models stop. An inverse-square law such as $\frac{k}{r^2}$ predicts an unbounded value as $r \to 0^{+}$. The infinite limit is the formula's way of saying it no longer describes the physics at that scale.

Infinite Limits: Common Confusions Cleared Up

Almost every lost mark on infinite limits comes from either treating $\infty$ like a number or skipping the sign check. These three errors cover most of them.

Mistake 1: Doing Arithmetic With Infinity

Where it slips in:

Sums and differences of two pieces that both blow up, as in Example 5.

Don't do this:

Do not write $\infty - \infty = 0$ or $\frac{\infty}{\infty} = 1$.

The correct way:

Treat $\infty - \infty$ and $\frac{\infty}{\infty}$ as indeterminate forms. Combine into a single fraction, factor, or compare growth rates before deciding.

The memorizer is most exposed here, having learned "anything minus itself is zero" as a reflex. It helps to remember that $\infty$ is shorthand for a process, two processes can run at different speeds, and only the algebra reveals which one wins.

Mistake 2: Skipping The Sign Check

Where it slips in:

Any $\frac{c}{0}$ result, especially with an even power in the denominator, as in Example 2.

Don't do this:

Do not answer $+\infty$ just because the denominator heads to 0.

The correct way:

Find the sign of the numerator and the sign of the denominator on each side, then combine them.

The rusher sees the zero denominator, writes $+\infty$, and moves on. The first instinct on $\frac{x - 4}{(x - 1)^2}$ is to think about the squared denominator and forget the numerator entirely, and testing a single value such as $x = 1.1$ catches it in seconds.

Mistake 3: Calling Every Zero Of The Denominator An Asymptote

Where it slips in:

Rational functions with a common factor in the numerator and denominator, as in Example 3.

Don't do this:

Do not list $x = 2$ as an asymptote of $\frac{x^2 - 4}{x^2 - x - 2}$ just because the denominator is 0 there.

The correct way:

Factor and cancel first. A cancelled factor gives a hole with a finite limit. Only the surviving zeros give infinite limits.

The second-guesser often gets the factoring right and then doubts it, adding $x = 2$ back to the asymptote list "to be safe". Checking the limit at the cancelled point, $\frac{4}{3}$ here, settles the doubt, because a finite limit cannot sit on an asymptote.

Mistake 2 has an expensive real-world cousin: forgetting what happens when a divisor reaches zero. On 21 September 1997, the US Navy cruiser USS Yorktown lost propulsion after a crew member entered a zero into a database field and the ship's software divided by it. Atlantic Fleet officials said the ship was dead in the water for about 2 hours and 45 minutes.

Practice Problems On Infinite Limits

1. Find $\lim_{x \to -4^{+}} \frac{3}{x + 4}$ and $\lim_{x \to -4^{-}} \frac{3}{x + 4}$.

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

3. For $h(x) = \frac{x^2 - 9}{x^2 - 2x - 3}$, find every vertical asymptote and every hole, with the one-sided limits at each asymptote.

4. Find $\lim_{x \to (\pi/2)^{-}} \sec x$ and $\lim_{x \to (\pi/2)^{+}} \sec x$.

5. Find $\lim_{x \to 0^{+}} e^{1/x}$ and $\lim_{x \to 0^{-}} e^{1/x}$. Is $x = 0$ a vertical asymptote?

Answer to Question 1: $+\infty$ from the right and $-\infty$ from the left.

Answer to Question 2: $+\infty$, since the numerator heads to 3 and the denominator is positive on both sides.

Answer to Question 3: $h(x) = \frac{(x - 3)(x + 3)}{(x - 3)(x + 1)}$, so there is a hole at $\left(3, \frac{3}{2}\right)$ and an asymptote at $x = -1$, with $+\infty$ from the right and $-\infty$ from the left.

Answer to Question 4: $+\infty$ from the left and $-\infty$ from the right.

Answer to Question 5: $+\infty$ from the right and 0 from the left. Yes, $x = 0$ is a vertical asymptote, because one infinite one-sided limit is enough.

What To Remember About Infinite Limits

• Infinite limits describe outputs that grow without bound, positively or negatively, as $x$ approaches a finite number $a$.

• Writing $\pm\infty$ describes how the limit fails to exist and is not a real-number value.

• A $\frac{c}{0}$ form with $c \neq 0$ signals an infinite limit, and the signs of the numerator and denominator on each side decide $+\infty$ or $-\infty$.

• Every infinite one-sided limit gives a vertical asymptote, while a cancelled factor gives a hole instead.

• Infinite limits send $x$ to a number, while limits at infinity send $x$ to infinity.

Where To Go From Here

Work through practice problems 1 to 5; if Question 3 trips you up, return to the section on vertical asymptotes and redo Example 3.

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Read More

• Limit laws

• Standard limits

• Continuity of a function

• Types of discontinuity

• Graphing functions