Limits at Infinity and Horizontal… — AP® Calculus AB Notes

Every limit so far asked what happens as $x$ approaches a number. Now let $x$ run off forever, to the right or to the left. Where does $f(x)$ head? The answer describes the far ends of a graph, and it settles which of two growing quantities wins in the long run.

Limits at infinity and horizontal asymptotes

$\lim_{x \to \infty} f(x) = L$ means that as $x$ grows without bound, $f(x)$ gets as close to $L$ as you like. $\lim_{x \to -\infty} f(x) = L$ means the same as $x$ decreases without bound. When either limit is a finite number $L$ , the line $y = L$ is a horizontal asymptote: the graph levels off toward it.

The basic fact behind everything here: as $x \to \infty$ , $\frac{1}{x} \to 0$ , and so do $\frac{5}{x^2}$ , $\frac{1}{\sqrt{x}}$ and $e^{-x} = \frac{1}{e^x}$ . A fixed number divided by something growing without bound shrinks to 0.

Left: the graph of f(x) = (2x² + x)/(x² + 1) for x from −10 to 10 with the dashed horizontal asymptote y = 2. The graph crosses the asymptote at x = 2 and then approaches it from above. Right: ln x, x³ and eˣ for x from 0 to 6. At first x³ is above eˣ, but eˣ passes it near x = 4.54 and stays ahead; ln x grows slowest. — **Figure 1.** Left: a graph leveling off at $y = 2$ (after crossing it). Right: which grows faster. LevlPrep original diagram.

A horizontal asymptote is about the far ends only. In the middle the graph may cross it: $\frac{2x^2 + x}{x^2 + 1}$ equals 2 at $x = 2$ and still approaches 2 as $x \to \infty$ . Vertical asymptotes are different: next to one the function is unbounded, and for a rational function the line itself is outside the domain, so the graph runs along it instead of crossing it.

	Vertical asymptote $x = c$	Horizontal asymptote $y = L$
Limit	$\lim_{x \to c^\pm} f(x)$ $= \pm\infty$	$\lim_{x \to \pm\infty} f(x) = L$
$x$ is	approaching a number	growing without bound
$f(x)$ is	growing without bound	approaching a number
Can the graph cross it?	not for a rational function (undefined on the line)	yes, away from the far ends

Rational functions: divide by the dominant power

For large $x$ , only the highest-power terms matter. To make that precise, divide the top and the bottom by the highest power of $x$ in the denominator.

Worked example 1. Find $\lim_{x \to \infty} \frac{3x^2 - x}{2x^2 + 5}$ .

Divide every term by $x^2$ : $\frac{3 - \frac{1}{x}}{2 + \frac{5}{x^2}}$ .

As $x \to \infty$ , $\frac{1}{x} \to 0$ and $\frac{5}{x^2} \to 0$ , so the limit is $\frac{3}{2}$ . The graph has the horizontal asymptote $y = \frac{3}{2}$ .

The same steps give three cases:

Degrees	Limit at $\pm\infty$	Example
top = bottom	ratio of leading coefficients	$\frac{6x^2 - 1}{1 - 3x^2} \to -2$
top < bottom	$0$	$\frac{4x + 1}{x^2 - 3} \to 0$
top > bottom	unbounded; no horizontal asymptote	$\frac{x^3 + 2}{5x^2 + x}$

Square roots: watch the sign

$\sqrt{x^2}$ is $|x|$ , not $x$ . For positive $x$ that makes no difference. For negative $x$ it does: $\sqrt{x^2} = -x$ .

Worked example 2. Find both limits at infinity of $h(x) = \frac{4x}{\sqrt{x^2 + 1}}$ .

For large $|x|$ , $\sqrt{x^2 + 1} \approx \sqrt{x^2} = |x|$ .

As $x \to \infty$ : $|x| = x$ , so $h(x) \approx \frac{4x}{x} = 4$ .

As $x \to -\infty$ : $|x| = -x$ , so $h(x) \approx \frac{4x}{-x} = -4$ .

Two horizontal asymptotes: $y = 4$ and $y = -4$ .

A related trick handles $\infty - \infty$ forms such as $\sqrt{x^2 + 6x} - x$ : multiply by the conjugate to get $\frac{6x}{\sqrt{x^2 + 6x} + x}$ , then divide by $x$ . The limit is $\frac{6}{1 + 1} = 3$ .

Exponentials, logarithms and the race to infinity

Some functions grow without bound, but at very different speeds. In the long run:

slowest $\ln x$ , then $\sqrt{x}$ , then $x$ , then $x^3$ , then $e^x$ fastest

"Slower" means the quotient of the slower function over the faster one approaches 0: $\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}$ $= 0$ and $\lim_{x \to \infty} \frac{x^3}{e^x}$ $= 0$ . A logarithm loses to every positive power of $x$ , a lower power loses to a higher one, and an exponential like $e^x$ beats every power. "Eventually" matters: $x^3$ is bigger than $e^x$ for $x$ between about 1.9 and 4.5, and after that it falls hopelessly behind.

Worked example 3. Find $\lim_{x \to \infty} \frac{3e^{2x} + x}{e^{2x} - 5}$ .

The dominant term in both the top and the bottom is $e^{2x}$ . Divide everything by it:

$\frac{3 + \frac{x}{e^{2x}}}{1 - \frac{5}{e^{2x}}}$

The exponential beats $x$ , so $\frac{x}{e^{2x}} \to 0$ ; and $\frac{5}{e^{2x}} \to 0$ . The limit is 3.

On the left the story changes: $\lim_{x \to -\infty} e^x = 0$ , so $y = 0$ is a horizontal asymptote of $e^x$ on that side only. Bounded oscillations also fade when divided by something growing: $-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$ for $x > 0$ , so by the squeeze theorem $\lim_{x \to \infty} \frac{\sin x}{x}$ $= 0$ .