1.5 · Infinite limits, limits at infinity, and asymptotes

Two different roles for infinity

In $\displaystyle \lim_{x\to a}f(x)=+\infty$, the input approaches a finite number while the output grows beyond every positive bound. This is an infinite limit; infinity is not a real number that you substitute or reach. A finite real limit does not exist in this situation.

In $\displaystyle \lim_{x\to+\infty}f(x)=L$, the inputs grow without bound and the outputs approach a finite height L. This is a limit at infinity and describes end behavior. Always identify whether infinity is attached to the input or to the output.

$$\lim_{x\to a^+}f(x)=+\infty\quad\mathrm{versus}\quad\lim_{x\to+\infty}f(x)=L$$

Use signs near a vertical asymptote

The line x = a is a vertical asymptote when at least one one-sided limit is +∞ or −∞. For $\displaystyle \frac{1}{x-2}$, the denominator is negative just left of 2 and positive just right of 2. Thus the left-hand limit is −∞ and the right-hand limit is +∞. The two-sided limit does not exist.

For $\displaystyle \frac{1}{(x-2)^2}$, both sides approach +∞ because the denominator is positive and approaches zero. It is valid to write a two-sided infinite limit of +∞, while recognizing that no finite real limit exists. A squared factor can make the signs agree.

$$\lim_{x\to2^-}\frac{1}{x-2}=-\infty,\quad\lim_{x\to2^+}\frac{1}{x-2}=+\infty$$

A zero denominator is a candidate, not a conclusion

First factor the rational expression. A fully canceled denominator factor may produce a removable hole rather than a vertical asymptote: $\displaystyle \frac{x^2-4}{x-2}$ agrees with x + 2 away from 2 and has a finite limit there.

A remaining denominator factor may produce unbounded behavior. For $\displaystyle \frac{x+1}{(x-2)(x+3)}$ near 2, the numerator and x + 3 are positive. The factor x − 2 controls the sign, giving −∞ from the left and +∞ from the right. Check signs on each side instead of simply writing “infinity.”

Find rational end behavior with dominant powers

For $\displaystyle \lim_{x\to+\infty}\frac{3x^2+1}{2x^2-x}$, divide numerator and denominator by x². The expression becomes $\displaystyle \frac{3+1/x^2}{2-1/x}$. As x grows, the terms involving reciprocal powers approach 0, leaving $\displaystyle \frac{3}{2}$. The same limit holds at negative infinity.

After any cancellation, if the numerator degree is smaller than the denominator degree, the limit is 0. If the degrees match, the limit is the ratio of leading coefficients. If the numerator degree is larger, there is no finite horizontal-asymptote limit; use the leading term and the direction of approach to determine signs.

$$\lim_{x\to\pm\infty}\frac{3x^2+1}{2x^2-x}=\frac{3}{2}$$

Horizontal asymptotes describe the ends

If $\displaystyle \lim_{x\to+\infty}f(x)=L$ or $\displaystyle \lim_{x\to-\infty}f(x)=L$, then y = L is a horizontal asymptote for that end. The limits at the two ends can differ. A horizontal asymptote is not a barrier: a curve may cross it at finite inputs.

For f(x) = $\displaystyle \frac{x}{x^2+1}$, the end limits are 0, giving y = 0 as a horizontal asymptote. Yet f(0) = 0, so the curve crosses that asymptote. Vertical and horizontal asymptotes describe different kinds of limiting behavior.

Square roots require an absolute value

The identity $\displaystyle \sqrt{x^2}=|x|$ holds for every real x. For positive x, $\displaystyle \frac{|x|}{x}=1$; for negative x, $\displaystyle \frac{|x|}{x}=-1$. Ignoring this distinction gives incorrect limits at negative infinity.

For $\displaystyle \frac{\sqrt{x^2+9}}{x}$, factor x² from under the radical to get $\displaystyle \frac{|x|}{x}\sqrt{1+\frac{9}{x^2}}$. The radical approaches 1 at either end, but the first factor is 1 as x approaches positive infinity and −1 as x approaches negative infinity. The two horizontal asymptotes are y = 1 and y = −1.

$$\frac{\sqrt{x^2+9}}x=\frac{|x|}{x}\sqrt{1+\frac{9}{x^2}}$$