Infinite Limits and Vertical Asymptotes

Sometimes, as $x$ gets close to a number, $f(x)$ doesn’t settle down at all. It shoots up (or down) past every number you can name. We describe this with an infinite limit, and the graph has a vertical asymptote. This page shows how to find infinite limits by looking at signs, without needing a graph.

Key ideas

What an infinite limit means

$$\lim_{x \to a} f(x) = \infty$$

means $f(x)$ becomes larger than any number you choose, as long as $x$ is close enough to $a$ (but not equal to it). Similarly, $-\infty$ means $f(x)$ becomes more negative than any number.

$\infty$ is not a number, so an infinite limit is really a limit that does not exist. Writing $= \infty$ tells you how it fails: by growing without bound in the positive direction.

Vertical asymptotes

The line $x = a$ is a vertical asymptote of $f$ if at least one of the one-sided limits at $a$ is $\infty$ or $-\infty$:

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

For $\tfrac{1}{x - 2}$ the two sides go opposite ways; for $\tfrac{1}{(x - 2)^2}$ both sides go to $\infty$.

The sign analysis for nonzero over zero

When substituting gives a nonzero number over $0$, the function is blowing up. To decide between $\infty$ and $-\infty$, check the sign on each side:

1. Find the limit of the top (a nonzero number, say positive or negative).
2. Decide whether the bottom is a small positive number ($0^+$) or a small negative number ($0^-$) on that side. Plug in a value just beside $a$ if you’re unsure.
3. Combine the signs: positive over $0^+$ is $\infty$; positive over $0^-$ is $-\infty$; and so on.

A squared factor like $(x - 2)^2$ is positive on both sides, so both one-sided limits have the same sign.

Rational functions

For a rational function, factor first:

• A zero of the denominator whose factor cancels completely gives a hole, not an asymptote.
• A zero of the denominator whose factor doesn’t cancel completely gives a vertical asymptote. For example, $\dfrac{x - 2}{(x - 2)^2} = \dfrac{1}{x - 2}$ still has an asymptote at $x = 2$.

Other functions

• $\displaystyle\lim_{x \to 0^+} \ln x = -\infty$, so $x = 0$ is a vertical asymptote of $y = \ln x$.
• $\tan x = \dfrac{\sin x}{\cos x}$ has vertical asymptotes wherever $\cos x = 0$: $x = \dfrac{\pi}{2} + k\pi$ (radians).

Worked examples

Example 1: A basic infinite limit

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

Solution. The top is $1$ (positive). For $x$ slightly bigger than $3$ (like $3.01$), $x - 3$ is a small positive number. For $x$ slightly smaller (like $2.99$), it’s a small negative number:

$$\lim_{x \to 3^+} \frac{1}{x - 3} = \frac{1}{0^+} = \infty, \qquad \lim_{x \to 3^-} \frac{1}{x - 3} = \frac{1}{0^-} = -\infty$$

The two-sided limit does not exist, and $x = 3$ is a vertical asymptote. (Writing $\tfrac{1}{0^+}$ is shorthand for your reasoning, not real arithmetic.)

Example 2: A squared factor

Find $\displaystyle\lim_{x \to 2} \frac{x + 1}{x^2 - 4x + 4}$.

Solution. The bottom factors as $(x - 2)^2$. Substituting gives $\tfrac{3}{0}$, so the function blows up.

The top approaches $3$ (positive). The bottom $(x - 2)^2$ is positive on both sides of $2$. So both one-sided limits are $\tfrac{3}{0^+} = \infty$:

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

Example 3: Asymptote or hole?

Find the vertical asymptotes of $g(x) = \dfrac{x^2 - 1}{x^2 + x - 2}$, and describe the behaviour of $g$ near each one.

Solution. Factor:

$$g(x) = \frac{(x - 1)(x + 1)}{(x + 2)(x - 1)} = \frac{x + 1}{x + 2} \quad\text{for } x \ne 1$$

At $x = 1$ the factor cancels, so there’s a hole, not an asymptote. At $x = -2$ it doesn’t cancel, so $x = -2$ is a vertical asymptote.

Near $x = -2$, the top $x + 1$ approaches $-1$ (negative).

• Just right of $-2$ (e.g. $-1.99$), $x + 2$ is $0^+$: $\dfrac{-1}{0^+} = -\infty$.
• Just left of $-2$ (e.g. $-2.01$), $x + 2$ is $0^-$: $\dfrac{-1}{0^-} = \infty$.

$$\lim_{x \to -2^+} g(x) = -\infty, \qquad \lim_{x \to -2^-} g(x) = \infty$$

Example 4: Tangent

Find $\displaystyle\lim_{x \to \frac{\pi}{2}^-} \tan x$ and $\displaystyle\lim_{x \to \frac{\pi}{2}^+} \tan x$. (Radians.)

Solution. Write $\tan x = \dfrac{\sin x}{\cos x}$. As $x \to \dfrac{\pi}{2}$, $\sin x \to 1$ and $\cos x \to 0$.

Just left of $\dfrac{\pi}{2}$ (in the first quadrant), $\cos x$ is small and positive. Just right (in the second quadrant), $\cos x$ is small and negative:

$$\lim_{x \to \frac{\pi}{2}^-} \tan x = \frac{1}{0^+} = \infty, \qquad \lim_{x \to \frac{\pi}{2}^+} \tan x = \frac{1}{0^-} = -\infty$$

So $x = \dfrac{\pi}{2}$ is a vertical asymptote of $y = \tan x$.

Common mistakes

Saying a limit that “equals infinity” exists. If a question asks whether $\displaystyle\lim_{x \to a} f(x)$ exists and the answer is $\infty$, the limit does not exist. Saying it’s $\infty$ is a description, not a value.

Assuming nonzero/0 is always positive infinity. The sign depends on the side. In Example 1, the left side goes to $-\infty$. Always check each side.

Calling a hole an asymptote. In Example 3, $x = 1$ makes the denominator $0$, but the factor cancels. That’s a hole. Factor before deciding.

Getting the sign of 0⁺ or 0⁻ wrong. If you’re unsure, substitute a number just beside $a$ (like $a + 0.01$) and look at the sign of each factor.

Treating ∞ − ∞ as 0. In Practice 8, both $\tfrac{1}{x}$ and $\tfrac{1}{x^2}$ go to $\infty$ as $x \to 0^+$, but their difference doesn’t go to $0$. Combine into one fraction first.

Practice

1. (Warm-up) Find $\displaystyle\lim_{x \to 5^-} \frac{2}{x - 5}$.

Solution

The top is $2$ (positive). For $x$ just less than $5$, $x - 5$ is $0^-$:

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

2. (Warm-up) Find $\displaystyle\lim_{x \to -1} \frac{3}{(x + 1)^2}$.

Solution

The top is $3$ and $(x + 1)^2$ is a small positive number on both sides, so

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

3. (Warm-up) Find the vertical asymptote of $f(x) = \dfrac{x + 4}{x - 7}$.

Solution

The denominator is $0$ at $x = 7$, and the top is $11 \ne 0$ there. The vertical asymptote is $x = 7$.

4. (Core) Find $\displaystyle\lim_{x \to 2^+} \frac{x + 3}{x^2 - 4}$ and $\displaystyle\lim_{x \to 2^-} \frac{x + 3}{x^2 - 4}$.

Solution

Factor the bottom: $(x - 2)(x + 2)$. The top approaches $5$, and $x + 2$ approaches $4$ (both positive).

Right of $2$: $x - 2$ is $0^+$, so the bottom is $0^+$ and the limit is $\infty$.

Left of $2$: $x - 2$ is $0^-$, so the bottom is $0^-$ and the limit is $-\infty$.

5. (Core) Find the vertical asymptotes and holes of $f(x) = \dfrac{x^2 - 9}{x^2 - 2x - 3}$.

Solution

$$f(x) = \frac{(x - 3)(x + 3)}{(x - 3)(x + 1)} = \frac{x + 3}{x + 1} \quad\text{for } x \ne 3$$

At $x = 3$ the factor cancels: a hole at height $\dfrac{3 + 3}{3 + 1} = \dfrac{3}{2}$, so at $\left(3, \tfrac{3}{2}\right)$.

At $x = -1$ the top is $2 \ne 0$: a vertical asymptote $x = -1$.

6. (Core) Find $\displaystyle\lim_{x \to 0} \frac{x - 1}{x^2}$.

Solution

The top approaches $-1$ (negative) and $x^2$ is $0^+$ on both sides:

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

7. (Core) Find $\displaystyle\lim_{x \to \frac{\pi}{2}^-} \sec x$. (Radians.)

Solution

$\sec x = \dfrac{1}{\cos x}$. Just left of $\dfrac{\pi}{2}$, $\cos x$ is small and positive, so

$$\lim_{x \to \frac{\pi}{2}^-} \sec x = \frac{1}{0^+} = \infty$$

8. (Challenge) Find $\displaystyle\lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{x^2}\right)$.

Solution

Both terms go to $\infty$, which tells you nothing (it’s an $\infty - \infty$ form). Combine over $x^2$:

$$\frac{1}{x} - \frac{1}{x^2} = \frac{x - 1}{x^2}$$

The top approaches $-1$ and the bottom is $0^+$, so

$$\lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{x^2}\right) = -\infty$$

9. (Challenge) Write a rational function with vertical asymptotes $x = 2$ and $x = -3$, a hole at $x = 1$, and $\displaystyle\lim_{x \to 2^+} f(x) = \infty$.

Solution

Answers vary. One example:

$$f(x) = \frac{x - 1}{(x - 1)(x - 2)(x + 3)}$$

The factor $x - 1$ cancels, giving a hole at $x = 1$. The factors $x - 2$ and $x + 3$ don’t cancel, giving the two asymptotes.

Check the sign near $2^+$: for $x \ne 1$, $f(x) = \dfrac{1}{(x - 2)(x + 3)}$. Just right of $2$, $x - 2$ is $0^+$ and $x + 3 \approx 5 \gt 0$, so $f(x) \to \infty$.