Math Core

So far, limits have described what happens as $x$ approaches a finite number $c$ , and every limit has been a real number or has failed to exist. Now you will use limit notation to describe two kinds of extreme behavior: outputs that grow without bound near a point, and the long-run behavior of a function as $x$ itself grows without bound. These two ideas give precise definitions of vertical and horizontal asymptotes.

Infinite limits and vertical asymptotes

When the outputs of $f$ increase without bound as $x \to c$ , we write $\displaystyle \lim_{x \to c} f(x) = \infty$ ; when they decrease without bound, we write $-\infty$ . These are called infinite limits. Remember that the limit still does not exist as a real number; the symbol $\infty$ describes why.

Definition

Vertical asymptote

The line $x = c$ is a vertical asymptote of the graph of $f$ if at least one of the one-sided limits $\displaystyle \lim_{x \to c^-} f(x)$ or $\displaystyle \lim_{x \to c^+} f(x)$ is $\infty$ or $-\infty$ .

For a quotient, a vertical asymptote usually appears where the denominator approaches $0$ and the numerator approaches a nonzero number. To decide between $\infty$ and $-\infty$ , analyze signs: the size of the quotient becomes huge, and its sign is the sign of the numerator divided by the sign of the (tiny) denominator.

Worked example: Sign analysis at a vertical asymptote

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

Solution. As $x \to 3$ , the numerator approaches $5$ (positive) and the denominator approaches $0$ .

For $x$ slightly less than $3$ , $x - 3$ is a tiny negative number, so the quotient is $\dfrac{\text{positive}}{\text{tiny negative}}$ : large and negative. $\displaystyle \lim_{x \to 3^-} \frac{x + 2}{x - 3} = -\infty$ .
For $x$ slightly greater than $3$ , $x - 3$ is a tiny positive number, so $\displaystyle \lim_{x \to 3^+} \frac{x + 2}{x - 3} = \infty$ .

So $x = 3$ is a vertical asymptote, and the two-sided limit does not exist (not even as $\infty$ , since the sides disagree).

A squared factor behaves differently: $\dfrac{x - 1}{(x - 4)^2}$ has a positive denominator on both sides of $4$ and a numerator near $3$ , so $\displaystyle \lim_{x \to 4} \frac{x - 1}{(x - 4)^2} = \infty$ from both sides.

Limits at infinity and horizontal asymptotes

The notation $\displaystyle \lim_{x \to \infty} f(x) = L$ means that $f(x)$ can be made as close to $L$ as you like by taking $x$ large enough. Similarly for $x \to -\infty$ .

Definition

Horizontal asymptote

The line $y = L$ is a horizontal asymptote of the graph of $f$ if $\displaystyle \lim_{x \to \infty} f(x) = L$ or $\displaystyle \lim_{x \to -\infty} f(x) = L$ .

A function can have at most two horizontal asymptotes, one on each end. Unlike vertical asymptotes, a graph may cross a horizontal asymptote; the asymptote only describes end behavior.

y = (2x + 1)/(x − 1) has a vertical asymptote x = 1 and a horizontal asymptote y = 2.Open in grapher →

The basic building block is

\lim_{x \to \pm\infty} \frac{1}{x^n} = 0 \quad \text{for any } n > 0,

because dividing $1$ by a huge number gives a number close to $0$ . The limit laws hold for limits at infinity as well.

Rational functions: divide by the highest power

To find the limit of a rational function as $x \to \pm\infty$ , divide the numerator and denominator by the highest power of $x$ in the denominator, then use $\dfrac{1}{x^n} \to 0$ .

Worked example: Equal degrees

Evaluate $\displaystyle \lim_{x \to \infty} \frac{3x^2 - 5x + 1}{6x^2 + x - 4}$ .

Solution. Divide every term by $x^2$ :

\lim_{x \to \infty} \frac{3 - \dfrac{5}{x} + \dfrac{1}{x^2}}{6 + \dfrac{1}{x} - \dfrac{4}{x^2}} = \frac{3 - 0 + 0}{6 + 0 - 0} = \frac{1}{2}.

The horizontal asymptote is $y = \dfrac{1}{2}$ .

Repeating this process for every case gives a shortcut worth memorizing.

End behavior of rational functions

For $f(x) = \dfrac{p(x)}{q(x)}$ with leading terms $ax^m$ (numerator) and $bx^n$ (denominator):

Degrees	$\displaystyle \lim_{x \to \pm\infty} f(x)$	Horizontal asymptote
$m$ less than $n$ (bottom-heavy)	$0$	$y = 0$
$m = n$ (equal)	$\dfrac{a}{b}$	$y = \dfrac{a}{b}$
$m$ greater than $n$ (top-heavy)	$\infty$ or $-\infty$	none

In the top-heavy case the sign comes from the leading terms. For instance, $\dfrac{x^3 + 1}{4x^2 + x}$ behaves like $\dfrac{x^3}{4x^2} = \dfrac{x}{4}$ for large $x$ , so it approaches $\infty$ as $x \to \infty$ and $-\infty$ as $x \to -\infty$ .

Square roots and x → −∞

When a square root is involved, remember that $\sqrt{x^2} = \lvert x \rvert$ , which equals $-x$ when $x$ is negative.

Worked example: A radical at negative infinity

Evaluate $\displaystyle \lim_{x \to -\infty} \frac{\sqrt{9x^2 + 4}}{2x - 1}$ .

Solution. For $x < 0$ , $\sqrt{9x^2 + 4} = \sqrt{x^2}\sqrt{9 + \dfrac{4}{x^2}} = -x\sqrt{9 + \dfrac{4}{x^2}}$ . Divide numerator and denominator by $x$ :

\frac{-x\sqrt{9 + \frac{4}{x^2}}}{2x - 1} = \frac{-\sqrt{9 + \frac{4}{x^2}}}{2 - \frac{1}{x}} \;\longrightarrow\; \frac{-\sqrt{9}}{2} = -\frac{3}{2}.

As $x \to \infty$ the same function approaches $+\dfrac{3}{2}$ , so the graph has two different horizontal asymptotes, $y = \dfrac{3}{2}$ and $y = -\dfrac{3}{2}$ .

Exponential and other functions

Some other end behaviors to know:

$\displaystyle \lim_{x \to -\infty} e^x = 0$ and $\displaystyle \lim_{x \to \infty} e^x = \infty$ , so $\displaystyle \lim_{x \to \infty} e^{-x} = 0$ .
$\displaystyle \lim_{x \to \infty} \ln x = \infty$ and $\displaystyle \lim_{x \to 0^+} \ln x = -\infty$ (a vertical asymptote at $x = 0$ ).
$\displaystyle \lim_{x \to \infty} \frac{\sin x}{x} = 0$ by the squeeze theorem: for $x > 0$ , $-\dfrac{1}{x} \le \dfrac{\sin x}{x} \le \dfrac{1}{x}$ , and both bounds approach $0$ . This graph crosses its horizontal asymptote $y = 0$ infinitely many times.

Common mistake

"The degree rule" only applies to limits as $x \to \pm\infty$ . It says nothing about limits at a finite point. Also, with square roots, check the sign carefully as $x \to -\infty$ : forgetting that $\sqrt{x^2} = -x$ for negative $x$ is the most common error, and it flips the sign of the answer.

Tip

For end behavior, only the "biggest" terms matter. As a quick check, keep just the leading term on top and bottom: $\dfrac{3x^2 - 5x + 1}{6x^2 + x - 4} \approx \dfrac{3x^2}{6x^2} = \dfrac{1}{2}$ for large $\lvert x \rvert$ .