Infinite Limits and Vertical Asymptotes
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 2.2: “The Limit of a Function” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Key idea
The notation $\lim_{x \to a} f(x) = \infty$ does not mean the limit exists in the usual sense -- it means the limit fails to exist in a very specific way: $f(x)$ grows without bound as $x \to a$. Writing $\infty$ is a description of the failure, not a finite value.
Vertical asymptotes are the geometric picture of this behavior: the graph of $f$ shoots upward or downward without bound near $x = a$, and no finite height is approached. Understanding infinite limits lets you identify vertical asymptotes precisely and determine their one-sided behavior (does the graph go to $+\infty$ or $-\infty$ from each side?).
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
Infinite limit notation.
- $\lim_{x \to a} f(x) = +\infty$: $f(x)$ grows without bound (positive) as $x \to a$.
- $\lim_{x \to a} f(x) = -\infty$: $f(x)$ decreases without bound (negative) as $x \to a$.
Vertical asymptote. The line $x = a$ is a vertical asymptote of $f$ if \[ \lim_{x \to a^+} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^-} f(x) = \pm\infty. \]
Sign analysis for $\frac{c}{x - a}$ type limits. To determine whether the one-sided limit is $+\infty$ or $-\infty$, track the sign of the numerator and the sign of $(x - a)$ as $x$ approaches $a$ from each side.
Key Concepts
1. What an Infinite Limit Means
Consider $f(x) = \dfrac{1}{x}$. As $x \to 0^+$ (approaching 0 from the right), the denominator is a tiny positive number, so $1/x$ is a huge positive number. As $x \to 0^-$ (from the left), the denominator is a tiny negative number, so $1/x$ is a large negative number.
| $x$ | $1/x$ |
|---|---|
| 0.1 | 10 |
| 0.01 | 100 |
| 0.001 | 1000 |
| $-0.001$ | $-1000$ |
| $-0.01$ | $-100$ |
| $-0.1$ | $-10$ |
The function does not approach any finite number. We write: \[ \lim_{x \to 0^+} \frac{1}{x} = +\infty, \qquad \lim_{x \to 0^-} \frac{1}{x} = -\infty. \] Since the two one-sided limits are different (not even the same infinite value), $\lim_{x \to 0} \frac{1}{x}$ does not exist.
The line $x = 0$ is a vertical asymptote.
2. Sign Analysis for Infinite Limits
To determine the sign of an infinite limit near $x = a$:
- Write $f(x)$ as $\dfrac{N(x)}{D(x)}$ where $D(a) = 0$ and $N(a) \neq 0$.
- Determine the sign of $N(a)$ (it stays the same for $x$ near $a$).
- Determine the sign of $D(x)$ for $x$ just above $a$ (right side) and just below $a$ (left side).
- The sign of $f(x)$ is the sign of $\dfrac{N(a)}{D(x)\text{-sign}}$.
Example 1. Determine $\displaystyle\lim_{x \to 3^+} \frac{2}{x - 3}$ and $\displaystyle\lim_{x \to 3^-} \frac{2}{x-3}$.
Numerator: $2 > 0$ (positive, constant).
For $x > 3$: $x - 3 > 0$ (positive). So $\dfrac{2}{x-3} > 0$, and as $x - 3 \to 0^+$, the fraction $\to +\infty$.
For $x < 3$: $x - 3 < 0$ (negative). So $\dfrac{2}{x-3} < 0$, and the fraction $\to -\infty$.
\[ \lim_{x \to 3^+} \frac{2}{x-3} = +\infty, \qquad \lim_{x \to 3^-} \frac{2}{x-3} = -\infty. \]
Example 2. Determine $\displaystyle\lim_{x \to 0^+} \frac{-3}{x^2}$ and $\displaystyle\lim_{x \to 0^-} \frac{-3}{x^2}$.
Numerator: $-3 < 0$ (negative). Denominator: $x^2 > 0$ for all $x \neq 0$. So $\dfrac{-3}{x^2} < 0$ always.
As $x \to 0$ from either side, $x^2 \to 0^+$, so $\dfrac{-3}{x^2} \to -\infty$.
\[ \lim_{x \to 0^+} \frac{-3}{x^2} = -\infty, \qquad \lim_{x \to 0^-} \frac{-3}{x^2} = -\infty, \qquad \lim_{x \to 0} \frac{-3}{x^2} = -\infty. \]
When both one-sided limits agree (even if the common value is $\pm\infty$), we write the two-sided limit.
3. Vertical Asymptotes of Rational Functions
A rational function $r(x) = p(x)/q(x)$ has a vertical asymptote at $x = a$ if $q(a) = 0$ and $p(a) \neq 0$.
(If both $p(a) = 0$ and $q(a) = 0$, there may be a removable discontinuity or a hole rather than a vertical asymptote -- factor and cancel first to determine which.)
Example 3. Find all vertical asymptotes of $f(x) = \dfrac{x + 1}{x^2 - 4}$.
Factor the denominator: $x^2 - 4 = (x-2)(x+2)$. Zeros at $x = 2$ and $x = -2$.
Check the numerator at each:
- $x = 2$: $p(2) = 3 \neq 0$. Vertical asymptote at $x = 2$.
- $x = -2$: $p(-2) = -1 \neq 0$. Vertical asymptote at $x = -2$.
Both are genuine vertical asymptotes.
Example 4. Find the vertical asymptotes of $g(x) = \dfrac{x^2 - 9}{x - 3}$ and determine any removable discontinuities.
Denominator is zero at $x = 3$. Numerator at $x = 3$: $9 - 9 = 0$. Both vanish -- this is a $\frac{0}{0}$ situation.
Factor: $\dfrac{(x-3)(x+3)}{x-3} = x + 3$ for $x \neq 3$. The $(x-3)$ factor cancels. There is no vertical asymptote at $x = 3$; there is a removable discontinuity (hole) at $(3, 6)$.
4. The Behavior of Common Infinite Limits
These are the building blocks; composite limits can often be analyzed using sign analysis on these.
| Function | $\lim_{x \to 0^+}$ | $\lim_{x \to 0^-}$ |
|---|---|---|
| $\dfrac{1}{x}$ | $+\infty$ | $-\infty$ |
| $\dfrac{1}{x^2}$ | $+\infty$ | $+\infty$ |
| $\dfrac{-1}{x}$ | $-\infty$ | $+\infty$ |
| $\ln x$ | $-\infty$ | (undefined for $x < 0$) |
| $\dfrac{1}{\sqrt{x}}$ | $+\infty$ | (undefined for $x < 0$) |
Memorizing the $1/x$ and $1/x^2$ cases and using sign analysis gives you all the rest.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Concluding $\lim = \infty$ means the limit “exists” | “The limit exists and equals $\infty$” | $\infty$ is not a real number; we say the limit fails to exist, but describe how |
| Not checking both sides | Finding right limit is $+\infty$ and calling that the limit | Check both sides; they may disagree in sign |
| Treating $\frac{0}{0}$ and $\frac{c}{0}$ the same | Concluding $\frac{0}{0}$ gives an infinite limit | $\frac{0}{0}$ is indeterminate; factor first. $\frac{c}{0}$ with $c \neq 0$ gives $\pm\infty$ |
| Sign error in the limit | Getting the wrong sign on $+\infty$ vs $-\infty$ | Carefully track the sign of numerator and denominator separately |
Leveled Practice
Level 1 -- Sign Analysis
Problem 1. Evaluate $\displaystyle\lim_{x \to 5^+} \frac{1}{x - 5}$ and $\displaystyle\lim_{x \to 5^-} \frac{1}{x - 5}$.
Show answer
For $x > 5$: $x - 5 > 0$; so $\frac{1}{x-5} \to +\infty$.
For $x < 5$: $x - 5 < 0$; so $\frac{1}{x-5} \to -\infty$.
Problem 2. Evaluate $\displaystyle\lim_{x \to 0} \frac{5}{x^2}$.
Show answer
$x^2 > 0$ for all $x \neq 0$; numerator is positive. Both one-sided limits are $+\infty$, so $\lim_{x \to 0} \frac{5}{x^2} = +\infty$.
Problem 3. Determine all vertical asymptotes of $f(x) = \dfrac{x - 2}{(x+1)(x-3)}$.
Show answer
Denominator is zero at $x = -1$ and $x = 3$.
At $x = -1$: numerator is $-3 \neq 0$. Vertical asymptote.
At $x = 3$: numerator is $1 \neq 0$. Vertical asymptote.
Vertical asymptotes at $x = -1$ and $x = 3$.
Level 2 -- One-Sided Behavior
Problem 4. For $f(x) = \dfrac{x - 2}{(x+1)(x-3)}$, determine whether each one-sided limit is $+\infty$ or $-\infty$ at each asymptote.
Show answer
At $x = -1$:
Numerator at $x = -1$: $-1 - 2 = -3 < 0$.
For $x \to -1^+$: $(x+1) \to 0^+$, $(x-3) \to -4 < 0$. Product $\to 0^-$. Fraction: $\frac{-3}{0^-} \to +\infty$.
For $x \to -1^-$: $(x+1) \to 0^-$, $(x-3) \to -4 < 0$. Product $\to 0^+$. Fraction: $\frac{-3}{0^+} \to -\infty$.
At $x = 3$:
Numerator at $x = 3$: $1 > 0$.
For $x \to 3^+$: $(x+1) \to 4 > 0$, $(x-3) \to 0^+$. Product $\to 0^+$. Fraction: $\frac{1}{0^+} \to +\infty$.
For $x \to 3^-$: $(x+1) \to 4 > 0$, $(x-3) \to 0^-$. Product $\to 0^-$. Fraction: $\frac{1}{0^-} \to -\infty$.
Level 3 -- Analysis and Reasoning
Problem 5. The function $h(x) = \dfrac{x^2 - 1}{x^2 - 3x + 2}$ -- find all vertical asymptotes and removable discontinuities.
Show answer
Factor: numerator $= (x-1)(x+1)$; denominator $= (x-1)(x-2)$.
Cancel $(x-1)$: $h(x) = \dfrac{x+1}{x-2}$ for $x \neq 1$.
At $x = 1$: both factors cancel. Removable discontinuity (hole) at $x = 1$; the limit is $\frac{1+1}{1-2} = -2$.
At $x = 2$: denominator is zero after cancellation, numerator $= 3 \neq 0$. Vertical asymptote at $x = 2$.
Common Misconceptions
a vertical asymptote is a wall the graph cannot cross.
This is the asymptote-as-wall error. The graph of $f(x) = \frac{1}{(x-2)^2}$ shoots upward near $x = 2$ and does not cross the line $x = 2$ there, which reinforces the wall image. However, a vertical asymptote only describes behavior at the specific value $x = a$; it places no restriction on the graph elsewhere. Some functions with vertical asymptotes do cross the asymptote line at other $x$-values. The asymptote records where the function grows without bound, not a boundary that blocks the graph everywhere.
writing $\lim_{x \to a} f(x) = \infty$ means the limit exists.
This is the limit-as-unreachable-barrier error. The notation $= \infty$ resembles a completed equation, so students sometimes conclude the limit exists and equals infinity. Infinity is not a real number, and the standard definition of a limit requires the function to get arbitrarily close to a specific finite value. Writing $\lim_{x \to a} f(x) = \infty$ is shorthand for describing how the limit fails: the values exceed every bound. For $f(x) = 1/x^2$ near $x = 0$, no finite number is ever approached, so the limit does not exist.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
A vertical asymptote is what happens when the denominator drives to zero while the numerator stays away from zero. The fraction becomes a very large number divided by a very small number -- the result grows without bound.
Track two things: the sign of the numerator (does it stay positive or negative near $a$?) and the sign of the denominator (is it approaching zero from the positive or negative side?). The sign of the limit is the sign of their quotient. The magnitude always goes to infinity -- it is only the sign that you need to determine.
When both numerator and denominator approach zero ($\frac{0}{0}$), the analysis is more subtle: cancel common factors first, then re-examine. The vertical asymptote may not exist at all (it may be a removable hole instead).
Connections
Within Calculus I (MATH161)
- Horizontal asymptotes (limits at infinity): The companion concept. As $x \to \infty$ instead of $x \to a$, we ask about the end behavior. Vertical asymptotes are about vertical blowup; horizontal asymptotes are about leveling off.
- Continuity: A function with a vertical asymptote at $a$ is not continuous at $a$. This is the most severe type of discontinuity.
- Related rates and optimization: Infinite limits appear in context problems when a physical quantity becomes unbounded (e.g., force as objects get infinitely close).
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