Infinite Limits and Vertical Asymptotes

So far our limits have been numbers. Near a vertical asymptote, which you met in Estimating Limits from Graphs and Tables, the values do not settle down at all: they grow without bound. One-sided limits let us describe exactly how.

Key idea

$\displaystyle\lim_{x \to a^+} f(x) = \infty$ means $f(x)$ becomes larger than any number you name once $x$ is close enough to $a$ on the right. Then the line $x = a$ is a vertical asymptote. The symbol $\infty$ describes how the limit fails to be a number; it is not a number itself.

The usual source is division by something tiny. If the top of a fraction stays near a nonzero number $c$ while the bottom shrinks to 0, the fraction's size blows up. Its sign comes from the signs of $c$ and of the tiny bottom.

$y = \frac{1}{x-2}$: from the right the values shoot up to $\infty$, from the left down to $-\infty$.

Why it works

Take $\frac{1}{x - 2}$ with $x = 2.01, 2.001, 2.0001$: the outputs are $100, 1000, 10000$. Each step closer multiplies the output by 10, and there is no ceiling. On the left, $x - 2$ is a tiny negative number, so the outputs are $-100, -1000, \ldots$ The sign analysis is just the rule for dividing signed numbers.

A three-question method

Does the bottom go to 0 while the top does not? Then the limit is $\infty$ or $-\infty$ (on each side).
What is the sign of the top near $a$?
What is the sign of the bottom on the side you approach from? Combine the two signs.

Worked example: One side at a time

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

The top tends to $4$, which is positive. The bottom tends to 0.
For $x > 3$ the bottom is a tiny positive number: positive over tiny positive is huge and positive.
For $x < 3$ the bottom is a tiny negative number: positive over tiny negative is huge and negative.

Answer: $\infty$ from the right and $-\infty$ from the left; $x = 3$ is a vertical asymptote.

Worked example: A square in the bottom

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

$x^2$ is a tiny positive number on both sides of 0.
$-2$ divided by a tiny positive number is huge and negative, from both sides.
The sides agree, so we can write the two-sided statement.

Answer: $\displaystyle\lim_{x \to 0} \frac{-2}{x^2} = -\infty$

Worked example: Hole or asymptote?

Where does $f(x) = \dfrac{x - 1}{x^2 - 1}$ have a vertical asymptote?

The bottom is 0 at $x = 1$ and $x = -1$.
Factor: $f(x) = \frac{x - 1}{(x - 1)(x + 1)} = \frac{1}{x + 1}$ for $x \ne 1$. At $x = 1$ the limit is $\frac{1}{2}$: a hole, not an asymptote.
At $x = -1$ the top of $\frac{1}{x+1}$ stays 1 while the bottom goes to 0.

Answer: Only at $x = -1$.

Common mistake

After writing $\lim = \infty$, people sometimes treat $\infty$ like a number: "$\infty - \infty = 0$" or "$\frac{\infty}{\infty} = 1$." Both are false in general. For example, near 0, $\frac{1}{x^2} - \frac{1}{x^4}$ looks like "$\infty - \infty$," but it equals $\frac{x^2 - 1}{x^4}$, which tends to $-\infty$. Rule of thumb: $\infty$ is a direction, so combine the expressions algebraically first, then take the limit.

Infinite Limits and Vertical Asymptotes

A three-question method

Practice

Builds on

Leads to

Sources and further reading