[Figure: FiveStudy logo]
FiveStudy
Calculus BC · Unit 1 · 1.14
AP Calculus · Limits and Continuity
Infinite Limits and Vertical Asymptotes
A zero denominator is a question, not an answer. Cancel first, then check the sign from each side.
key idea$\lim_{x\to c}f=\pm\infty\iff x=c\text{ is a vertical asymptote}$
Reading the graph
An infinite limit is a statement about how a limit fails: the values grow without bound rather than settling. The limit does not exist — writing $\pm\infty$ records the reason. Every such point is a vertical asymptote.
Three functions with a zero denominator at $x=1$. The first two have vertical asymptotes; the third does not, because its factor cancels and leaves a hole. A zero denominator alone guarantees nothing.sign of the denominator near $x=1$
| $x\to1^{-}$ | at $x=1$ | $x\to1^{+}$ |
|---|
| $(x-1)^{2}$ | + | 0 | + |
| $(x-1)$ | − | 0 | + |
| verdict | even power → same sign both sides | denominator vanishes | odd power → sign flips |
Even power$(x-c)^{2}$ — same sign both sides, so both branches go the same way
Odd power$(x-c)$ — sign flips, so the branches go opposite ways
Cancelsno asymptote at all — a removable hole
| What you find at $x=c$ | One-sided behaviour | Conclusion |
|---|
| $\dfrac{k}{0}$ with an even-power factor | both sides $+\infty$ (or both $-\infty$) | vertical asymptote; you may write $\lim=\pm\infty$ |
| $\dfrac{k}{0}$ with an odd-power factor | $-\infty$ one side, $+\infty$ the other | vertical asymptote; the two-sided limit is DNE |
| $\dfrac00$ that cancels | both sides approach a finite value | no asymptote — a removable hole |
| denominator zero and numerator zero, factor remains | still unbounded | vertical asymptote after simplifying |
The decisive question is never “is the denominator zero?” but does the factor survive cancellation, and is its power even or odd?
Worked examples
EX 1
Even power — both sides agree
$\dfrac{1}{(x-1)^{2}}$Find $\lim\limits_{x\to1}\dfrac{1}{(x-1)^{2}}$ and identify any vertical asymptote.
1
Substitute
$\dfrac{1}{0}$ — not indeterminate, so the magnitude blows up.
2
Check the sign from each side
$(x-1)^{2}$ is a square, so it is positive whether $x$ is just below or just above $1$.
3
Conclude the behaviour
Numerator $1\gt 0$ over a small positive number, from both sides: $\lim_{x\to1}\dfrac{1}{(x-1)^{2}}=+\infty$.
4
Name the asymptote
$x=1$ is a vertical asymptote. Note the limit still does not exist — $+\infty$ describes the failure.
EX 2
Odd power — the sides disagree
$\dfrac{1}{x-1}$Find the one-sided limits of $\dfrac{1}{x-1}$ at $x=1$ and state the two-sided limit.
1
Approach from the right
For $x$ slightly above $1$, $x-1$ is a small positive, so the quotient $\to+\infty$.
2
Approach from the left
For $x$ slightly below $1$, $x-1$ is a small negative, so the quotient $\to-\infty$.
3
Compare
The two sides disagree, so $\lim_{x\to1}\dfrac{1}{x-1}$ does not exist — and here you may not write $\pm\infty$ for the two-sided limit.
4
Name the asymptote
$x=1$ is still a vertical asymptote. An asymptote does not require the two sides to agree.
EX 3
Zero denominator with no asymptote
$\dfrac{x^{2}-1}{x-1}$Does $f(x)=\dfrac{x^{2}-1}{x-1}$ have a vertical asymptote at $x=1$?
1
Substitute
$\dfrac{0}{0}$ — indeterminate, so a factor cancels.
2
Factor and cancel
$\dfrac{(x-1)(x+1)}{x-1}=x+1$ for $x\neq1$.
3
Take the limit
$\lim_{x\to1}(x+1)=2$ — finite.
4
Conclude
No asymptote. There is a removable hole at $(1,2)$. The denominator was zero, but the offending factor cancelled.
EX 4
Find every asymptote
$\dfrac{x+3}{x^{2}-x-6}$Locate all vertical asymptotes of $g(x)=\dfrac{x+3}{x^{2}-x-6}$.
1
Factor both parts
$g(x)=\dfrac{x+3}{(x-3)(x+2)}$. The denominator vanishes at $x=3$ and $x=-2$.
2
Check each for cancellation
The numerator $x+3$ matches neither factor, so nothing cancels.
3
Test $x=3$
$\dfrac{6}{0}$ — unbounded. Vertical asymptote at $x=3$.
4
Test $x=-2$
$\dfrac{1}{0}$ — unbounded. Vertical asymptote at $x=-2$.
5
State the answer
Two vertical asymptotes: $x=3$ and $x=-2$. The root $x=-3$ of the numerator gives an $x$-intercept, not an asymptote.
Watch out
An infinite limit does not exist
$\lim f=+\infty$ is a description of unbounded growth, not a value. If asked whether the limit exists, the answer is no.
Zero denominator is not enough
Check whether the factor cancels first. $\dfrac{x^{2}-1}{x-1}$ has a zero denominator at $x=1$ and no asymptote.
Even vs odd power decides the sides
$(x-c)^{2}$ keeps the same sign on both sides, so both branches go the same way. $(x-c)$ flips, so they go opposite ways.
Do not write $\pm\infty$ for a two-sided limit that disagrees
If one side is $+\infty$ and the other $-\infty$, the correct answer is DNE, not $\infty$.