Connecting Infinite Limits and Vertical Asymptotes

Evaluate $\dfrac{1}{(x-1)^2}$ at $x = 1.01$ and you get $10{,}000$; at $x = 1.001$ you get $1{,}000{,}000$. As $x$ closes in on 1 the values pass every number you could name, so they do not approach any number, and the limit at 1 does not exist. "Does not exist" hides a lot, though: these values all grow in one direction without bound, which is very different from values that jump or oscillate. Infinite limit notation records exactly that behavior, and on a graph it shows up as a vertical asymptote.

What an infinite limit says

Writing

means that $f(x)$ can be made larger than any number you choose, by taking $x$ close enough to $a$ (with $x \neq a$). Pick $1{,}000$, and there is a window around $a$ where every value of $f$ exceeds $1{,}000$; pick a million, and a narrower window works. $\lim\limits_{x \to a} f(x) = -\infty$ means the same with "smaller than any negative number." The one-sided versions, $x \to a^-$ (inputs to the left of $a$) and $x \to a^+$ (inputs to the right), restrict attention to one side.

The symbol $\infty$ is not a number, and an infinite limit is a way of saying how a limit fails to exist. Writing $\lim\limits_{x \to 1} \frac{1}{(x-1)^2} = \infty$ is correct and more informative than writing "does not exist," but it does not mean the limit exists. For the same reason, you can't do arithmetic with it as if it were a value: two expressions that each go to $\infty$ can have a difference that goes to anything.

A two-sided infinite limit needs both sides to agree. If $f \to \infty$ from both sides of $a$, write $\lim\limits_{x \to a} f(x) = \infty$. If one side goes to $\infty$ and the other to $-\infty$, as $\frac{1}{x - 3}$ does at 3, state the two one-sided limits and say the two-sided limit does not exist; writing it as $\infty$ is wrong.

One unbounded side is enough. The graph then runs up or down along the line $x = a$, getting closer to it without reaching it, and that is what an asymptote is.

Nonzero divided by almost zero

Most vertical asymptotes come from a quotient whose denominator approaches 0 while its numerator does not. Dividing a fixed number by smaller and smaller numbers gives larger and larger results: $\frac{3}{0.01} = 300$ and $\frac{3}{0.0001} = 30{,}000$. So the size of the quotient blows up, and only its sign is left to decide. The sign comes from the signs of the numerator and the denominator on the side you are approaching from.

To analyze a quotient $\frac{N(x)}{D(x)}$ at a point $a$ where $D(a) = 0$:

1. Check the numerator. If $N(a) \neq 0$, the quotient is unbounded near $a$ and $x = a$ is a vertical asymptote. If $N(a) = 0$ too, the form is $\frac{0}{0}$ and you must simplify first; the limit may be a finite number, giving a hole instead of an asymptote (topic 1.13).
2. Find the sign on each side. Factor the denominator and decide the sign of each factor for $x$ slightly less than $a$, then slightly more. Only the factor that is 0 at $a$ changes sign; every other factor keeps the sign it has at $a$.
3. Write the one-sided limits. Positive over a small positive or negative over a small negative gives $\infty$; mixed signs give $-\infty$.

The power of the zero factor decides whether the two sides agree. An odd power such as $(x - a)$ or $(x - a)^3$ changes sign at $a$, so the two one-sided limits have opposite signs. An even power such as $(x - a)^2$ is positive on both sides, so the two sides match. For instance, with $r(x) = \dfrac{x - 6}{(x + 2)^2}$, the numerator is $-8$ near $x = -2$ and the denominator is a small positive number on both sides, so

a genuine two-sided infinite limit.

Asymptotes beyond rational functions

The same reasoning applies to any function that has a quotient or a logarithm inside it.

• Logarithms. As $x \to 0^+$, $\ln x \to -\infty$, because reaching a very negative output only needs a tiny positive input ($\ln 10^{-6} \approx -13.8$). The line $x = 0$ is a vertical asymptote of $y = \ln x$, and only the right-hand limit makes sense, since $\ln x$ is undefined for $x \le 0$.
• Tangent. $\tan x = \frac{\sin x}{\cos x}$. As $x \to \frac{\pi}{2}^-$, $\sin x \to 1$ and $\cos x$ is a small positive number, so $\tan x \to \infty$; from the right, $\cos x$ is a small negative number and $\tan x \to -\infty$. That happens at every odd multiple of $\frac{\pi}{2}$.
• One side only. For $e^{1/x}$, as $x \to 0^+$ the exponent $\frac{1}{x} \to \infty$ and $e^{1/x} \to \infty$, but as $x \to 0^-$ the exponent goes to $-\infty$ and $e^{1/x} \to 0$. The right side alone makes $x = 0$ a vertical asymptote.
• A zero denominator without an asymptote. $\frac{\sin x}{x}$ is undefined at 0, but its limit there is 1 (topic 1.8), so its graph has a hole at $(0, 1)$, not an asymptote.

A function that is continuous at $a$ has the finite limit $f(a)$ there, so it cannot have a vertical asymptote at $a$. Vertical asymptotes can only occur where a function is not continuous, and they are the infinite discontinuities of topic 1.10. To find all of them, list the inputs where the function is undefined or its formula breaks (zero denominators, a logarithm's argument reaching 0, odd multiples of $\frac{\pi}{2}$ for $\tan$) and test each one.

Worked examples

Worked exampleExample 1: one-sided limits from signs

Let $f(x) = \dfrac{2x + 1}{x^2 - 4x}$. Find the vertical asymptotes of the graph of $f$, and find both one-sided limits at each.

Show solution

The denominator factors as $x(x - 4)$, which is 0 at $x = 0$ and $x = 4$. Check the numerator at each: $2(0) + 1 = 1$ and $2(4) + 1 = 9$. Neither is 0, so both are vertical asymptotes, and only the signs remain.

Near $x = 0$, the numerator is near 1 (positive) and $x - 4$ is near $-4$ (negative). The factor $x$ carries the sign change.

• $x \to 0^-$: $x < 0$, so the denominator is $(-)(-) = (+)$, small. Positive over small positive: $\lim\limits_{x \to 0^-} f(x) = \infty$.
• $x \to 0^+$: $x > 0$, so the denominator is $(+)(-) = (-)$, small. Positive over small negative: $\lim\limits_{x \to 0^+} f(x) = -\infty$.

Near $x = 4$, the numerator is near 9 and $x$ is near 4, both positive. The factor $x - 4$ carries the sign change.

• $x \to 4^-$: $x - 4 < 0$, so $\lim\limits_{x \to 4^-} f(x) = -\infty$.
• $x \to 4^+$: $x - 4 > 0$, so $\lim\limits_{x \to 4^+} f(x) = \infty$.

The graph below shows all four behaviors: up along $x = 0$ on the left and down on the right, down along $x = 4$ on the left and up on the right.

Where students lose points

On the exam

To justify that $x = a$ is a vertical asymptote, give one one-sided limit at $a$ that is $\infty$ or $-\infty$; a statement about the denominator being 0 is not enough on its own. On calculator-active questions, a graphing calculator can draw a near-vertical segment across an asymptote or skip a narrow spike, so decide asymptotes from the formula, using the graph only to confirm.