Infinite Limits and Asymptotes

Two different questions

An infinite limit near a finite input describes outputs growing without bound. A limit as $x\to\infty$ describes what happens when the input itself grows. They answer different questions and can occur in the same function.

Writing $f(x)\to+\infty$ means that every finite output threshold is eventually exceeded; it does not mean the function reaches a number called infinity. A vertical asymptote $x=a$ occurs when at least one one-sided limit is infinite. A horizontal asymptote $y=L$ occurs when a finite limit at positive or negative infinity equals $L$ .

Visual guide

VISUAL GUIDEApproach a line without touching the singularity

Left branch
Right branch
Horizontal asymptote
Vertical asymptote

For f(x) = 1 + 1/x, x = 0 is a vertical asymptote and y = 1 is a horizontal asymptote. The two branches are drawn separately because the function is undefined at zero.

Worked example: signs near a pole

For $f(x)=1/(x-2)$ , a denominator approaching zero from below produces arbitrarily large negative values. Therefore $\lim_{x\to2^-}f(x)=-\infty$ , while $\lim_{x\to2^+}f(x)=+\infty$ . The line $x=2$ is a vertical asymptote. The two-sided finite limit does not exist.

For $1/(x-2)^2$ , the denominator is positive on both sides, so both limits are $+\infty$ . Always inspect signs; a small denominator by itself does not determine the direction of divergence.

Worked example: compare leading powers

Consider $g(x)=(3x^2+x)/(2x^2-5)$ . Divide numerator and denominator by $x^2$ :

g(x)=\frac{3+1/x}{2-5/x^2}\longrightarrow\frac32.

This holds in both infinite directions. For rational functions after cancellation, a smaller numerator degree gives limit zero, equal degrees give the ratio of leading coefficients, and a larger numerator degree requires polynomial division or another growth analysis.

A denominator zero can be a removable hole instead of an asymptote. For example, $(x^2-4)/(x-2)$ equals $x+2$ away from $2$ and has finite limit $4$ there.

Slant asymptotes and crossings

Dividing $x^2$ by $x-1$ gives $x+1+1/(x-1)$ . The difference from $y=x+1$ tends to zero at either infinity, so this is a slant asymptote. An asymptote describes approach, not a forbidden line. The function $\sin x/x$ crosses its horizontal asymptote $y=0$ repeatedly while tending to it as $x\to\infty$ .

Practice

Find the vertical and horizontal asymptotes of $2/(x+3)$ .
Evaluate $\lim_{x\to\infty}(x^2+1)/(x^3-2)$ .
Does $(x^2-1)/(x-1)$ have a vertical asymptote at $1$ ?

Show worked solutions

The pole is $x=-3$ , and the end limit is zero, giving $y=0$ .
Dividing by $x^3$ makes the numerator tend to zero and denominator to one, so the limit is zero.
No. Cancellation gives $x+1$ nearby, with finite limit $2$ . The original expression has a hole.