Infinite Limits and Vertical… — AP® Calculus AB Notes

When substitution gives a nonzero number over 0, you already know the function is unbounded near that point. This topic gives that behavior its notation, shows how to find which way the function heads on each side, and connects it to vertical asymptotes.

What an infinite limit says

Look at $f(x) = \frac{1}{(x - 2)^2}$ near 2. At $x = 2.1$ it is 100; at $x = 2.01$ it is 10,000; at $x = 2.001$ it is 1,000,000. The same happens from the left. The values do not settle on any number. They grow beyond every bound. We write

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

Read it as " $f(x)$ increases without bound as $x$ approaches 2." This is an infinite limit. Be careful about what it means: $\infty$ is not a number, so the limit does not exist. The notation tells you how it fails to exist. In the same way, $\lim_{x \to c} f(x) = -\infty$ means the values decrease without bound.

Two graphs with a dashed vertical asymptote at x = 2. Left: y = 1/(x − 2)², which rises without bound on both sides, so the limit as x approaches 2 is +∞ from each side. Right: y = 1/(x − 2), which falls without bound from the left (−∞) and rises without bound from the right (+∞), so the one-sided infinite limits have opposite signs. — **Figure 1.** Left: both sides go up. Right: the sides go opposite ways. LevlPrep original diagram.

Finding the sign on each side

When substitution gives a nonzero number over 0, the size of the quotient blows up. Its sign on each side decides between $\infty$ and $-\infty$ . Work it out from the signs of the pieces.

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

Near 3 the numerator is close to 4, which is positive.

Left of 3 (say $x = 2.99$ ): $x - 3$ is a tiny negative number. Positive over tiny negative is hugely negative, so $\lim_{x \to 3^-} \frac{x + 1}{x - 3}$ $= -\infty$ .

Right of 3 (say $x = 3.01$ ): $x - 3$ is a tiny positive number, so $\lim_{x \to 3^+} \frac{x + 1}{x - 3}$ $= \infty$ .

The sides disagree, so the two-sided limit does not exist and is not $\infty$ or $-\infty$ either. Report the one-sided limits.

A squared factor such as $(x - 4)^2$ is positive on both sides, so it never changes sign. That is why $\frac{1}{(x - 2)^2}$ goes up on both sides.

Worked example 2. Find $\lim_{x \to 4} \frac{x - 6}{(x - 4)^2}$ .

The numerator approaches $-2$ . The denominator is a tiny positive number on both sides. Negative over tiny positive is hugely negative on both sides, so the limit is $-\infty$ .

Vertical asymptotes

The line $x = c$ is a vertical asymptote of the graph of $f$ when at least one of $\lim_{x \to c^-} f(x)$ and $\lim_{x \to c^+} f(x)$ is $\infty$ or $-\infty$ . One side is enough. The graph of $\ln x$ , for instance, has the asymptote $x = 0$ with only a right side: $\lim_{x \to 0^+} \ln x = -\infty$ .

For a rational function, factor and cancel first. A factor that cancels gives a hole; a factor that stays in the denominator gives an asymptote.

Worked example 3. Find the vertical asymptotes of $f(x)$ $= \frac{x^2 - 9}{x^2 - 2x - 3}$ .

Factor: $f(x)$ $= \frac{(x - 3)(x + 3)}{(x - 3)(x + 1)}$ .

At $x = 3$ the factor cancels: $f(x)$ $= \frac{x + 3}{x + 1} \to \frac{6}{4}$ $= \frac{3}{2}$ . A hole, not an asymptote.

At $x = -1$ the numerator approaches 2 and the denominator approaches 0, so $f$ is unbounded. The only vertical asymptote is $x = -1$ .

Other functions have asymptotes too. $\tan x = \frac{\sin x}{\cos x}$ has one wherever $\cos x = 0$ , such as $x = \frac{\pi}{2}$ , with $\lim_{x \to \frac{\pi}{2}^-} \tan x$ $= \infty$ and $\lim_{x \to \frac{\pi}{2}^+} \tan x$ $= -\infty$ .

Writing it correctly

Write $\lim_{x \to c^+} f(x) = \infty$ , never $f(c) = \infty$ . A function's value is a number or undefined.
Write $\lim_{x \to c} f(x) = \infty$ only when both sides go to $\infty$ . If the sides differ, give them separately.
To justify an asymptote, show an infinite one-sided limit. "The denominator is 0" is not enough, because a hole also has a zero denominator.