Limits at Infinity, Infinite Limits and Asymptotes

Quick answer

A limit at infinity describes the end behavior of a function: if f(x) approaches L as x grows without bound, the line y = L is a horizontal asymptote. For rational functions, divide by the highest power of x in the denominator; only the leading terms survive, so the degrees decide the answer. An infinite limit — values growing without bound as x approaches a — marks a vertical asymptote x = a. Exponentials outgrow every power, and every power outgrows the logarithm.

What you'll learn

Evaluate limits as x → ∞ and x → −∞, including rational and radical functions
Find horizontal asymptotes from limits at infinity
Find vertical asymptotes from infinite limits, telling them apart from holes
Compare the growth of logarithms, powers and exponentials

Limits at infinity

$\displaystyle\lim_{x \to \infty} f(x) = L$ means $f(x)$ can be made as close to $L$ as we like by taking $x$ large enough. The line $y = L$ is then a horizontal asymptote. The same idea with $x \to -\infty$ describes the left end of the graph.

The basic fact behind almost everything here:

\lim_{x \to \pm\infty} \frac{1}{x^k} = 0 \quad \text{for every } k > 0

y = (2x + 1)/(x − 1)
y = 2
x = 1

A horizontal and a vertical asymptote

Rational functions: divide by the highest power

For $\displaystyle\lim_{x \to \infty} \frac{3x^2 - 5}{2x^2 + x}$ , divide the top and bottom by $x^2$ , the highest power in the denominator:

\frac{3x^2 - 5}{2x^2 + x} = \frac{3 - \tfrac{5}{x^2}}{2 + \tfrac{1}{x}} \;\longrightarrow\; \frac{3 - 0}{2 + 0} = \frac{3}{2}

Carried out in general, this gives a rule based on the degrees:

Degrees	Limit at $\pm\infty$	Horizontal asymptote
top < bottom	$0$	$y = 0$
top = bottom	ratio of leading coefficients	$y = \tfrac{a}{b}$
top > bottom	$\pm\infty$	none

Why only the leading terms matter

Far out, the leading term dwarfs the rest. At $x = 1000$ , $3x^2 - 5$ is $2\,999\,995$ : the $-5$ changes it by less than two parts in a million.

Dividing by the highest power turns that intuition into algebra. Every other term ends up with $x$ in a denominator and goes to $0$ , leaving only the leading coefficients. At infinity, a polynomial behaves like its leading term, so a rational function behaves like the ratio of its leading terms.

Infinite limits and vertical asymptotes

$\displaystyle\lim_{x \to a} f(x) = \infty$ means the values grow without bound as $x \to a$ — the limit does not exist, and the symbol says how it fails. If either one-sided limit at $a$ is $\pm\infty$ , the line $x = a$ is a vertical asymptote.

For a rational function in lowest terms, the vertical asymptotes sit exactly at the zeros of the denominator. A zero that cancels with the numerator is a hole instead: the limit exists there.

Growth rates

Logarithms, powers and exponentials all go to $\infty$ , at wildly different speeds:

\ln x \;\ll\; x^p \;\ll\; e^x \qquad (p > 0, \; x \to \infty)

Here $f \ll g$ means $\tfrac{f}{g} \to 0$ . So $\tfrac{x^{100}}{e^x} \to 0$ and $\tfrac{\ln x}{\sqrt{x}} \to 0$ , even though the tops look huge for moderate $x$ . L’Hôpital’s rule, later in the course, proves these.

Worked examples

Common mistakes

Practice problems

Find $\lim_{x \to \infty} \tfrac{5x^3 - x}{2x^3 + 7}$ .

Answer

$\tfrac{5}{2}$

Full solution

Equal degrees, so the limit is the ratio of leading coefficients. Dividing by $x^3$ confirms it: $\tfrac{5 - x^{-2}}{2 + 7x^{-3}} \to \tfrac{5}{2}$ .
Find $\lim_{x \to -\infty} \tfrac{x^2 + 1}{x - 4}$ .

Answer

$-\infty$

Full solution

The top has the higher degree, so the fraction is unbounded. For large negative $x$ it behaves like $\tfrac{x^2}{x} = x$ , which goes to $-\infty$ .
Find the horizontal asymptote of $y = \tfrac{6 - 2x^2}{x^2 + 1}$ .

Answer

$y = -2$

Full solution

Equal degrees; the leading coefficients are $-2$ and $1$ , so the limit at both ends is $-2$ .
Find $\lim_{x \to \infty} \tfrac{\sqrt{9x^2 + 2}}{3x - 1}$ .

Answer

$1$

Full solution

For $x > 0$ , $\tfrac{\sqrt{9x^2 + 2}}{x} = \sqrt{9 + \tfrac{2}{x^2}} \to 3$ , and $\tfrac{3x - 1}{x} \to 3$ . The limit is $\tfrac{3}{3} = 1$ .
Find the vertical asymptotes and holes of $y = \tfrac{x^2 - 4}{x^2 - 5x + 6}$ .

Answer

Vertical asymptote $x = 3$ ; hole at $x = 2$ .

Full solution

$\tfrac{(x - 2)(x + 2)}{(x - 2)(x - 3)}$ . The factor $x - 2$ cancels, leaving a hole at $x = 2$ (with limit $\tfrac{4}{-1} = -4$ ). The zero at $x = 3$ remains: a vertical asymptote.
Find $\lim_{x \to 0^+} \ln x$ and $\lim_{x \to \infty} \ln x$ .

Answer

$-\infty$ and $\infty$

Full solution

$\ln x$ drops without bound as $x$ shrinks to $0$ : $\ln(10^{-k}) = -k \ln 10$ . It also grows without bound, though slowly: $\ln(10^k) = k \ln 10$ .
Find $\lim_{x \to \infty} \left(\sqrt{x^2 + x} - x\right)$ .

Answer

$\tfrac{1}{2}$

Full solution

Multiply by the conjugate over itself: $\tfrac{(x^2 + x) - x^2}{\sqrt{x^2 + x} + x} = \tfrac{x}{\sqrt{x^2 + x} + x}$ .

Divide by $x$ : $\tfrac{1}{\sqrt{1 + 1/x} + 1} \to \tfrac{1}{2}$ .
Find $\lim_{x \to \infty} \tfrac{e^x}{x^3}$ .

Answer

$\infty$

Full solution

Exponentials outgrow every power, so $\tfrac{x^3}{e^x} \to 0$ and its reciprocal grows without bound.
Find $\lim_{x \to \infty} \tfrac{\sin x}{x}$ .

Answer

$0$

Full solution

$-\tfrac{1}{x} \le \tfrac{\sin x}{x} \le \tfrac{1}{x}$ for $x > 0$ , and both bounds approach $0$ . By the squeeze theorem the limit is $0$ , so $y = 0$ is a horizontal asymptote that the graph crosses infinitely often.
A student says $\lim_{x \to \infty} (x^2 - x) = \infty - \infty = 0$ . What went wrong?

Hint

Factor out $x^2$ .

Answer

$\infty - \infty$ is indeterminate. The limit is $\infty$ .

Full solution

$x^2 - x = x^2\left(1 - \tfrac{1}{x}\right)$ . The bracket approaches $1$ and $x^2$ grows without bound, so the product goes to $\infty$ .

Frequently asked questions

How do I find the limit of a rational function at infinity?

Divide the top and bottom by the highest power of x in the denominator. Every term with x in a denominator goes to 0, and the leading terms decide the result.

What is a horizontal asymptote?

A line y = L with f(x) → L as x → ∞ or as x → −∞. A graph can cross its horizontal asymptote; it only has to approach it far out.

What is the difference between a vertical asymptote and a hole?

At a vertical asymptote the values grow without bound. At a hole the limit exists and only the point is missing, typically where a factor cancels.

Is infinity a number?

No. Writing lim f(x) = ∞ says the values grow without bound. The limit does not exist; the symbol describes how it fails.

Which grows faster, x^100 or e^x?

e^x. Every exponential with base above 1 eventually outgrows every power of x, however large the exponent.

What to learn next

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