Infinite Limits, calculus lesson

Slide a candle toward the focal point of a magnifying glass and its sharp image forms farther and farther away, with no upper bound. Near some points a function does the same thing. It does not settle on any number, but it fails in a very specific way, by growing past every bound. Infinite limits give that behavior a precise name.

Key terms

Unbounded one-sided behavior

Definition Let $f$ be defined on some interval $(a, c)$ with $c > a$ . We write

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

if $f(x)$ can be made larger than any given number by taking $x$ sufficiently close to $a$ with $x > a$ . The statements $\lim_{x \to a^-} f(x) = \infty$ , with $f$ defined on some interval $(c, a)$ where $c < a$ , and the versions with $-\infty$ , where $f(x)$ can be made smaller than any given number, are defined in the same way.

In plain words From that side, the outputs shoot up (or down) past every height you could name.

Example Here are some values of $f(x) = 1/x$ near $0$ .

$x$	$-0.01$	$-0.001$	$0.001$	$0.01$
$\frac{1}{x}$	$-100$	$-1000$	$1000$	$100$

For $N > 0$ and $x$ between $0$ and $1/N$ the value $1/x$ exceeds $N$ , and for $x$ between $-1/N$ and $0$ it is below $-N$ , so

\lim_{x \to 0^+} \frac{1}{x} = \infty \quad \text{and} \quad \lim_{x \to 0^-} \frac{1}{x} = -\infty.

Positive and negative infinity

Definition Let $f$ be defined on an open interval containing $a$ , except possibly at $a$ . Then $\lim_{x \to a} f(x) = \infty$ means that both one-sided limits at $a$ are $\infty$ , and $\lim_{x \to a} f(x) = -\infty$ means that both are $-\infty$ . If one side gives $\infty$ and the other $-\infty$ , there is no single two-sided description.

In plain words Both sides have to blow up in the same direction before we write one infinite limit for the point.

Example For $1/x^2$ , both sides of $0$ give large positive values, since $x^2 > 0$ for $x \neq 0$ . So

\lim_{x \to 0} \frac{1}{x^2} = \infty.

For $1/x$ the two sides differ, so we report the right-hand limit $\infty$ and the left-hand limit $-\infty$ separately.

Infinite limits at finite points and at infinity

Definition An infinite limit can happen as $x$ approaches a real number $a$ , as above, or as $x \to \pm\infty$ . For $f$ defined on some interval $(c, \infty)$ , the statement $\lim_{x \to \infty} f(x) = \infty$ means that $f(x)$ can be made larger than any given number by taking $x$ sufficiently large. The versions with $-\infty$ in either place are defined in the same way. Being unbounded is not enough. Every output beyond some point has to be large, not just some of them.

In plain words “Goes to infinity” means stays high eventually, not merely reaches high now and then.

Example As $x \to \infty$ , $x^3 \to \infty$ , and as $x \to -\infty$ , $x^3 \to -\infty$ . In contrast, $x \sin x$ is unbounded as $x \to \infty$ , yet it returns to $0$ at every $x = k\pi$ , so $\lim_{x \to \infty} x \sin x$ is neither $\infty$ nor $-\infty$ , and it does not exist.

Sign analysis

Definition Suppose $\lim_{x \to a^+} g(x) = c$ with $c \neq 0$ and $\lim_{x \to a^+} h(x) = 0$ , and that $h(x) > 0$ for all $x$ in some interval $(a, a + \delta)$ with $\delta > 0$ . Then

\lim_{x \to a^+} \frac{g(x)}{h(x)} = \begin{cases} \infty, & c > 0, \\ -\infty, & c < 0. \end{cases}

If instead $h(x) < 0$ on that interval, the two answers swap. The same rule holds from the left with an interval $(a - \delta, a)$ .

In plain words A nonzero top over a bottom that shrinks to zero blows up. Its sign is the sign of the top times the sign of the bottom on that side.

Example Take $f(x) = (x + 1)/(x - 2)$ at $a = 2$ . The numerator tends to $3 > 0$ . For $x > 2$ the denominator is positive, so $\lim_{x \to 2^+} f(x) = \infty$ . For $x < 2$ it is negative, so $\lim_{x \to 2^-} f(x) = -\infty$ . Compare $(x - 3)/(x - 2)^2$ , where the top tends to $-1$ and the bottom is positive on both sides, so the limit at $2$ is $-\infty$ from both sides.

Infinity versus a real value

Definition The symbols $\infty$ and $-\infty$ are not real numbers. Writing $\lim_{x \to a} f(x) = \infty$ says the limit does not exist as a real number and describes the particular way it fails. The limit laws are stated for real limits, and expressions such as $\infty - \infty$ and $\infty / \infty$ have no fixed value.

In plain words Infinity is a description of runaway growth, not a number you can do arithmetic with.

Example Let $f(x) = 1/x^2 + 1$ and $g(x) = 1/x^2$ . Both tend to $\infty$ at $0$ , and $f(x) - g(x) = 1$ for all $x \neq 0$ . If instead $f(x) = 2/x^2$ , then $f(x) - g(x) = 1/x^2 \to \infty$ . Two cases of the form $\infty - \infty$ give two different answers, so the form alone decides nothing.

Check yourself

Find both one-sided limits at $3$ of the function

f(x) = \frac{2x}{x - 3}.

Show the answer

The numerator tends to $6 > 0$ . For $x < 3$ the denominator is negative, so the left-hand limit is $-\infty$ . For $x > 3$ it is positive, so the right-hand limit is $\infty$ .

Find the limit

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

Show the answer

The numerator tends to $3 > 0$ and the denominator is positive on both sides and tends to $0$ . The limit is $\infty$ .

Find both one-sided limits of $\tan x$ at $\pi/2$ .

Show the answer

Write $\tan x = \sin x / \cos x$ . As $x \to \pi/2$ the numerator tends to $1$ and the denominator to $0$ . Just left of $\pi/2$ the cosine is positive, so the limit is $\infty$ . Just right of it the cosine is negative, so the limit is $-\infty$ .

Infinite Limits

Theory, part 21

Practice, part 22

Key terms

Unbounded one-sided behavior

Positive and negative infinity

Infinite limits at finite points and at infinity

Sign analysis

Infinity versus a real value

Check yourself

How this lesson connects

Builds on

Used later in