One-Sided Limits, Infinite Limits & Vertical Asymptotes

The little superscript is everything: $x\to a^-$ means approach from the left (values just below $a$); $x\to a^+$ means from the right (just above). And when a denominator shrinks to zero while the top stays nonzero, $f$ shoots off to $\pm\infty$ — that’s a vertical asymptote.

Nonzero over zero means a vertical asymptote

If direct substitution gives $\frac{k}{0}$ with $k\neq0$, the function has a vertical asymptote at $x=a$. To get the sign of the infinity, test the sign of the denominator just on each side of $a$.

[Diagram — see the figure in the print workbook.]

Saying $\lim_{x\to a}f(x)=\infty$ is a special kind of DNE: the limit does not exist (no finite value), but we report $\infty$ to describe how it fails.

Worked example 1

For $f(x)=\begin{cases}x^2,&x<1\\ 2x+1,&x\ge1\end{cases}$, find $\lim_{x\to1^-}f$, $\lim_{x\to1^+}f$, $\lim_{x\to1}f$.

1. 1.Left uses the $x<1$ piece: $x^2\to 1$.
2. 2.Right uses the $x\ge1$ piece: $2x+1\to 3$.
3. 3.$1\neq3$ ⇒ two-sided limit DNE.

Answer: $\lim_{x\to1^-}=1,\ \lim_{x\to1^+}=3,\ \lim_{x\to1}$ DNE.

Worked example 2

Evaluate $\displaystyle\lim_{x\to2^+}\frac{1}{x-2}$ and $\displaystyle\lim_{x\to2^-}\frac{1}{x-2}$.

1. 1.From the right, $x-2$ is a tiny positive number ⇒ $\frac{1}{0^+}=+\infty$.
2. 2.From the left, $x-2$ is a tiny negative number ⇒ $\frac{1}{0^-}=-\infty$.

Answer: $\lim_{x\to2^+}=+\infty,\quad \lim_{x\to2^-}=-\infty$.

Worked example 3

Find all vertical asymptotes of $f(x)=\dfrac{x+1}{x^2-x-6}$.

1. 1.Factor the bottom: $x^2-x-6=(x-3)(x+2)$.
2. 2.Top is $x+1$; it shares no factor with the bottom.
3. 3.Zeros of the denominator that don’t cancel give asymptotes: $x=3$ and $x=-2$.

Answer: Vertical asymptotes at $x=3$ and $x=-2$.

Worked example 4

Does $g(x)=\dfrac{x-3}{x^2-9}$ have a vertical asymptote at $x=3$?

1. 1.Factor: $\dfrac{x-3}{(x-3)(x+3)}=\dfrac{1}{x+3}$ for $x\neq3$.
2. 2.The $(x-3)$ cancels, so $x=3$ is a hole, not an asymptote.
3. 3.$\lim_{x\to3}g=\frac{1}{6}$. The genuine asymptote is at $x=-3$.

A zero of the denominator gives an asymptote only if it does not cancel. If it cancels, you get a hole.

Answer: No — $x=3$ is a removable hole; $x=-3$ is the asymptote.

Worked example 5

Evaluate $\displaystyle\lim_{x\to0^+}\ln x$.

1. 1.As $x$ approaches $0$ from the right, $\ln x$ decreases without bound.
2. 2.($\ln$ is undefined for $x\le0$, so only the right side exists.)

Answer: $-\infty$

Worked example 6

For $f(x)=\dfrac{x-1}{(x-1)^2}$, evaluate $\lim_{x\to1^-}f$ and $\lim_{x\to1^+}f$.

1. 1.Simplify: $\dfrac{x-1}{(x-1)^2}=\dfrac{1}{x-1}$.
2. 2.Right: $\frac{1}{0^+}=+\infty$. Left: $\frac{1}{0^-}=-\infty$.

Answer: $\lim_{x\to1^+}=+\infty,\ \lim_{x\to1^-}=-\infty$.

Common trap: Guessing the sign of $\infty$

$\frac{k}{0}$ is not automatically $+\infty$. You must check the sign of the denominator on the side you’re approaching from. For $\frac{1}{x-2}$ at $x\to2^-$, the denominator is a tiny negative number, so the result is $-\infty$. Make a quick sign test — don’t guess.

Hack: The sign-test thumb

For $\frac{k}{0}$, ignore magnitudes and just track signs. Pick a number a hair to the left of $a$ and a hair to the right, drop each into the denominator, and read off the sign. Tiny $+$ on the bottom → the answer has the sign of the top; tiny $-$ → flip it. Ten seconds, no calculator.