1.14 - Connecting Infinite Limits and Vertical Asymptotes

Introduction

Earlier we identified vertical asymptotes as non-removable discontinuities that happen when a denominator equals zero and the factor doesn’t cancel. Now we’ll figure out the behavior near those asymptotes: does the function head toward positive or negative infinity?

Identifying Vertical Asymptotes

Definition Using Limits

The line

x = a

is a vertical asymptote of

f

f(x)

goes to

+\infty

-\infty

x

approaches

a

from at least one side:

\lim_{x \to a^-} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^+} f(x) = \pm\infty

This works for any function, not just fractions:

$y = \ln x$ has a vertical asymptote at $x = 0$ , since $\lim_{x \to 0^+} \ln x = -\infty$ . There’s no left side here (the log isn’t defined for $x \leq 0$ ), and one side is enough.

$y = \tan x$ has a vertical asymptote at $x = \frac{\pi}{2}$ , since $\lim_{x \to (\pi/2)^-} \tan x = +\infty$ and $\lim_{x \to (\pi/2)^+} \tan x = -\infty$ .

For rational functions, the factoring steps below find these quickly.

Review: How to Find Them

Step 1: Factor both the numerator and denominator completely

Step 2: Cancel any common factors (those create holes, not asymptotes)

Step 3: Set the remaining denominator factors equal to zero and solve

Those

x

-values are your vertical asymptotes.

Example 1: Basic Identification

Find all vertical asymptotes:

f(x) = \frac{x^2 - 4}{x^2 - 5x + 6}

Factor the numerator:

x^2 - 4 = (x - 2)(x + 2)

Factor the denominator:

x^2 - 5x + 6 = (x - 2)(x - 3)

Factored form:

f(x) = \frac{(x - 2)(x + 2)}{(x - 2)(x - 3)}

The factor

(x - 2)

cancels, so there’s a hole at $x = 2$ .

The factor

(x - 3)

doesn’t cancel, so there’s a vertical asymptote at $x = 3$ .

Answer: vertical asymptote at

x = 3

Understanding Behavior Near Vertical Asymptotes

Near a vertical asymptote, the function shoots off toward infinity. The question is which direction.

1.14 - Connecting Infinite Limits and Vertical Asymptotes

For this graph at

x = 3

From the left: $\lim_{x \to 3^-} f(x) = -\infty$

From the right: $\lim_{x \to 3^+} f(x) = +\infty$

Two-sided limit:

\lim_{x \to 3} f(x) = \text{DNE (does not exist)}

Since the one-sided limits differ, the two-sided limit doesn’t exist.

Determining Infinite Limits Without a Graph

To find out whether a function heads toward

+\infty

-\infty

near a vertical asymptote, plug in values very close to the asymptote and look at the signs.

Strategy

Identify the vertical asymptote (where the denominator is zero)

Choose a test value extremely close to it

Evaluate to see whether the result is positive or negative

Determine the direction: positive means $+\infty$ , negative means $-\infty$

Example 2: Right-Hand Limit

Find

\lim_{x \to 5^+} \dfrac{x - 2}{x - 5}

Vertical asymptote:

x = 5

(denominator is zero, nothing cancels)

Approaching from the right means using values slightly greater than 5. Choose

x = 5.001

\frac{5.001 - 2}{5.001 - 5} = \frac{3.001}{0.001}

Numerator: $3.001$ (positive)

Denominator: $0.001$ (positive, very small)

Result: $\frac{\text{positive}}{\text{tiny positive}}$ , a very large positive number

Answer:

\lim_{x \to 5^+} \dfrac{x - 2}{x - 5} = +\infty

Example 3: Left-Hand Limit

Find

\lim_{x \to 5^-} \dfrac{x - 2}{x - 5}

Approaching from the left means using values slightly less than 5. Choose

x = 4.999

\frac{4.999 - 2}{4.999 - 5} = \frac{2.999}{-0.001}

Numerator: $2.999$ (positive)

Denominator: $-0.001$ (negative, very small in absolute value)

Result: $\frac{\text{positive}}{\text{tiny negative}}$ , a very large negative number

Answer:

\lim_{x \to 5^-} \dfrac{x - 2}{x - 5} = -\infty

Example 4: When Both Sides Match

Find

\lim_{x \to 2} \dfrac{x + 1}{(x - 2)^2}

Vertical asymptote:

x = 2

. Notice the denominator is squared.

From the right ( $x \to 2^+$ ): choose

x = 2.001

\frac{2.001 + 1}{(2.001 - 2)^2} = \frac{3.001}{(0.001)^2} = \frac{3.001}{0.000001}

Numerator: $3.001$ (positive)

Denominator: $(0.001)^2 = 0.000001$ (positive, extremely small)

Result: $\frac{\text{positive}}{\text{tiny positive}} = +\infty$

From the left ( $x \to 2^-$ ): choose

x = 1.999

\frac{1.999 + 1}{(1.999 - 2)^2} = \frac{2.999}{(-0.001)^2} = \frac{2.999}{0.000001}

Numerator: $2.999$ (positive)

Denominator: $(-0.001)^2 = 0.000001$ (positive, because it’s squared)

Result: $\frac{\text{positive}}{\text{tiny positive}} = +\infty$

Two-sided limit: since both one-sided limits equal

+\infty

\lim_{x \to 2} \frac{x + 1}{(x - 2)^2} = +\infty

Key insight: when the denominator is squared (or raised to any even power), it’s always positive no matter which side you come from. That often means both sides go to the same infinity.

About the notation: writing

\lim_{x \to 2} \frac{x + 1}{(x - 2)^2} = +\infty

doesn’t mean the limit exists. Infinity isn’t a real number, so the limit still does not exist. The

+\infty

just describes how it fails: the function grows without bound from both sides. (This is the “unbounded” case from 1.3.)

Working with Factored Forms

When the function is already factored, the analysis gets easier.

Example 5: Complete Analysis

Find all one-sided limits at the vertical asymptotes of

g(x) = \frac{x - 4}{(x - 1)^2}

Identify the vertical asymptote: the denominator is zero when

x = 1

, and nothing cancels, so there’s a vertical asymptote at $x = 1$ .

Find $\lim_{x \to 1^+} g(x)$ : substitute

x = 1.001

\frac{1.001 - 4}{(1.001 - 1)^2} = \frac{-2.999}{(0.001)^2}

Numerator: negative

Denominator: positive (squared)

Result: negative

Answer:

\lim_{x \to 1^+} g(x) = -\infty

Find $\lim_{x \to 1^-} g(x)$ : substitute

x = 0.999

\frac{0.999 - 4}{(0.999 - 1)^2} = \frac{-3.001}{(-0.001)^2}

Numerator: negative

Denominator: positive (squared)

Result: negative

Answer:

\lim_{x \to 1^-} g(x) = -\infty

Two-sided limit: since both sides approach

-\infty

\lim_{x \to 1} g(x) = -\infty

Summary: The Sign Analysis Method

Instead of computing exact decimals, you can just analyze the signs directly.

Quick Method

Factor completely

For each factor, decide whether it’s positive or negative near the asymptote

Combine the signs to get the overall sign

Tiny denominator plus non-zero numerator means infinity

Example 6: Sign Analysis

Find

\lim_{x \to 3^-} \dfrac{(x - 1)(x - 5)}{x - 3}

x \to 3^-

(approaching from the left, so

x

is slightly less than 3, like

x = 2.999

Numerator factors:

$(x - 1)$ : when $x \approx 3$ , this is $3 - 1 = 2$ , so positive

$(x - 5)$ : when $x \approx 3$ , this is $3 - 5 = -2$ , so negative

Product: (positive)(negative) = negative

Denominator:

$(x - 3)$ : when $x$ is slightly less than 3, this is negative (like $2.999 - 3 = -0.001$ ), so negative

Overall:

\frac{\text{negative}}{\text{negative}} = \text{positive}

A tiny denominator means a huge magnitude.

Answer:

\lim_{x \to 3^-} \dfrac{(x - 1)(x - 5)}{x - 3} = +\infty

Practice Problems

Problem 1

Find

\lim_{x \to 4^+} \dfrac{x + 1}{x - 4}

Answer:

Numerator at

x = 4

: positive

Denominator as

x \to 4^+

: positive (approaching from the right)

Result:

+\infty

Problem 2

Find

\lim_{x \to -2^-} \dfrac{3}{(x + 2)^2}

Answer:

Numerator: positive (constant)

Denominator: positive (squared, so always positive)

Result:

+\infty

Problem 3

Find

\lim_{x \to 0} \dfrac{-5}{x^2}

Answer:

From both sides, the denominator is positive (squared) and the numerator is negative.

Both sides give

-\infty

, so

\lim_{x \to 0} \frac{-5}{x^2} = -\infty

Key Takeaways

Vertical asymptotes happen where the denominator is zero and the factors don’t cancel

Using limits: $x = a$ is a vertical asymptote if $f(x) \to \pm\infty$ as $x \to a$ from at least one side. This works for any function, like $\ln x$ at $x = 0$

Near asymptotes, functions approach $+\infty$ or $-\infty$

To figure out which, substitute values very close to the asymptote

Sign analysis: if the result is positive you get $+\infty$ , if negative you get $-\infty$

Even powers (like squared terms) are always positive

If both one-sided limits equal the same infinity, the two-sided limit equals that infinity

Writing $= +\infty$ or $= -\infty$ describes unbounded behavior. The limit still doesn’t exist as a real number

If the one-sided limits differ, the two-sided limit does not exist (DNE)

Positive numerator over tiny positive denominator gives $+\infty$

Positive numerator over tiny negative denominator gives $-\infty$

Negative numerator over tiny positive denominator gives $-\infty$

Negative numerator over tiny negative denominator gives $+\infty$

Work with factored forms when you can, since it makes sign analysis simpler

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