Calculus I · Limits and Continuity · lesson

Vertical Asymptotes and One-Sided Sign Analysis

Learn Vertical Asymptotes and One-Sided Sign Analysis with plain-language explanations, guided examples, worked homework methods, interactive checks, and exam-s

Course progress27 of 47 core pages27 of 47

Where this chapter fits

Chapter 4: Infinite behavior

Read vertical, horizontal, and slant asymptotes through sign analysis and dominant-term reasoning.

Reading lens: Is the function growing without bound near a finite input, or settling into end behavior as the input grows? Keep that question in view while reading Vertical Asymptotes and One-Sided Sign Analysis; the worked mathematics is evidence for the idea, not a substitute for it.

This page connects Infinite Limits Explained to Limits at Infinity and Horizontal Asymptotes. Read the explanation first, predict each example’s next move, and only then compare the written solution.

Concept

Learning objectives

Find vertical asymptote candidates, cancel removable factors before classifying behavior, and use a sign chart to determine each one-sided infinite limit.

Vertical Asymptotes and Sign Analysis

The line $x=a$ is a vertical asymptote if the function becomes unbounded on at least one side of $a$ .

For a rational function

f(x)=\frac{p(x)}{q(x)},

zeros of $q(x)$ are candidates. But first simplify common factors.

Common mistake

A denominator zero does not automatically create a vertical asymptote. If the zero factor cancels with the numerator, the graph may have a removable hole instead.

Guided walkthrough

Identify the sign before saying infinity

Evaluate

\lim_{x\to2^-}\frac1{x-2}.

Show worked solution

As $x\to2^-$ , $x-2$ is a tiny negative number. Dividing $1$ by a tiny negative number gives a very large negative number. Therefore,

\boxed{\lim_{x\to2^-}\frac1{x-2}=-\infty}.

Interactive checkinfinite-sign-01

Evaluate $\lim_{x\to1^+}\frac1{x-1}$ .

Your work stays on this device. No account or AI grader is used.

Show hint

From the right, the denominator is positive and tiny.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Method

One-Sided Sign Chart

To analyze $p(x)/q(x)$ near $x=a$ :

• Factor numerator and denominator. • Cancel any common factor, while remembering the corresponding hole. • Determine the sign of every remaining factor just left of $a$ . • Determine the sign just right of $a$ . • Combine signs and decide whether the magnitude grows without bound.

Worked example

A sign chart with several factors

Find both one-sided limits of

f(x)=\frac{x+2}{(x-3)(x+1)}

at $x=3$ .

Show worked solution

Near $x=3$ , the factors $x+2$ and $x+1$ are positive. Only $x-3$ changes sign.

From the left of $3$ :

x+2>0,\qquad x-3<0,\qquad x+1>0.

The quotient is negative. Its denominator approaches zero in magnitude, so

\boxed{\lim_{x\to3^-}f(x)=-\infty}.

From the right of $3$ :

x+2>0,\qquad x-3>0,\qquad x+1>0.

The quotient is positive, so

\boxed{\lim_{x\to3^+}f(x)=+\infty}.

Thus $x=3$ is a vertical asymptote.

Worked example

Even multiplicity

Analyze

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

near $x=-2$ .

Show worked solution

Near $-2$ , the numerator $x-4$ is negative. The denominator $(x+2)^2$ is positive on both sides and approaches zero.

Therefore, on both sides the quotient is negative with unbounded magnitude:

\boxed{\lim_{x\to-2^-}g(x)=-\infty},

\boxed{\lim_{x\to-2^+}g(x)=-\infty}.

The two one-sided infinite behaviors agree, so it is also common to write

\lim_{x\to-2}g(x)=-\infty.

This does not mean the limit is a real number; it means both sides decrease without bound.

Worked example

Exam-level: hole versus asymptote

Find and classify every discontinuity of

h(x)=\frac{(x-1)(x+4)}{(x-1)(x-3)}.

Show worked solution

The original denominator is zero at $x=1$ and $x=3$ . Cancel the common factor for $x\ne1$ :

h(x)=\frac{x+4}{x-3}.

At $x=1$ , the simplified formula is finite:

\frac{1+4}{1-3}=-\frac52.

Therefore $x=1$ is a removable hole, not a vertical asymptote.

At $x=3$ , the remaining denominator is zero while the numerator is $7\ne0$ . Thus $x=3$ is a vertical asymptote.

For signs near $3$ , the numerator is positive. The denominator $x-3$ is negative from the left and positive from the right:

\lim_{x\to3^-}h(x)=-\infty, \qquad \lim_{x\to3^+}h(x)=+\infty.

Source & rights

Original instruction with traceable references.

The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.

The verified handoff declares original composition and requires owner provenance review. BetterGrades-original material remains separate from public-domain references; no source textbook PDF is published here.

Calculus I · Limits and Continuity · lesson

Vertical Asymptotes and One-Sided Sign Analysis

Learn Vertical Asymptotes and One-Sided Sign Analysis with plain-language explanations, guided examples, worked homework methods, interactive checks, and exam-s

Course progress27 of 47 core pages27 of 47

Where this chapter fits

Chapter 4: Infinite behavior

Read vertical, horizontal, and slant asymptotes through sign analysis and dominant-term reasoning.

Concept

Learning objectives

Find vertical asymptote candidates, cancel removable factors before classifying behavior, and use a sign chart to determine each one-sided infinite limit.

Vertical Asymptotes and Sign Analysis

The line $x=a$ is a vertical asymptote if the function becomes unbounded on at least one side of $a$ .

For a rational function

f(x)=\frac{p(x)}{q(x)},

zeros of $q(x)$ are candidates. But first simplify common factors.

Common mistake

A denominator zero does not automatically create a vertical asymptote. If the zero factor cancels with the numerator, the graph may have a removable hole instead.

Guided walkthrough

Identify the sign before saying infinity

Evaluate

\lim_{x\to2^-}\frac1{x-2}.

Show worked solution

As $x\to2^-$ , $x-2$ is a tiny negative number. Dividing $1$ by a tiny negative number gives a very large negative number. Therefore,

\boxed{\lim_{x\to2^-}\frac1{x-2}=-\infty}.

Interactive checkinfinite-sign-01

Evaluate $\lim_{x\to1^+}\frac1{x-1}$ .

Your work stays on this device. No account or AI grader is used.

Show hint

From the right, the denominator is positive and tiny.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Method

One-Sided Sign Chart

To analyze $p(x)/q(x)$ near $x=a$ :

Worked example

A sign chart with several factors

Find both one-sided limits of

f(x)=\frac{x+2}{(x-3)(x+1)}

at $x=3$ .

Show worked solution

Near $x=3$ , the factors $x+2$ and $x+1$ are positive. Only $x-3$ changes sign.

From the left of $3$ :

x+2>0,\qquad x-3<0,\qquad x+1>0.

The quotient is negative. Its denominator approaches zero in magnitude, so

\boxed{\lim_{x\to3^-}f(x)=-\infty}.

From the right of $3$ :

x+2>0,\qquad x-3>0,\qquad x+1>0.

The quotient is positive, so

\boxed{\lim_{x\to3^+}f(x)=+\infty}.

Thus $x=3$ is a vertical asymptote.

Worked example

Even multiplicity

Analyze

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

near $x=-2$ .

Show worked solution

Near $-2$ , the numerator $x-4$ is negative. The denominator $(x+2)^2$ is positive on both sides and approaches zero.

Therefore, on both sides the quotient is negative with unbounded magnitude:

\boxed{\lim_{x\to-2^-}g(x)=-\infty},

\boxed{\lim_{x\to-2^+}g(x)=-\infty}.

The two one-sided infinite behaviors agree, so it is also common to write

\lim_{x\to-2}g(x)=-\infty.

This does not mean the limit is a real number; it means both sides decrease without bound.

Worked example

Exam-level: hole versus asymptote

Find and classify every discontinuity of

h(x)=\frac{(x-1)(x+4)}{(x-1)(x-3)}.

Show worked solution

The original denominator is zero at $x=1$ and $x=3$ . Cancel the common factor for $x\ne1$ :

h(x)=\frac{x+4}{x-3}.

At $x=1$ , the simplified formula is finite:

\frac{1+4}{1-3}=-\frac52.

Therefore $x=1$ is a removable hole, not a vertical asymptote.

At $x=3$ , the remaining denominator is zero while the numerator is $7\ne0$ . Thus $x=3$ is a vertical asymptote.

For signs near $3$ , the numerator is positive. The denominator $x-3$ is negative from the left and positive from the right:

\lim_{x\to3^-}h(x)=-\infty, \qquad \lim_{x\to3^+}h(x)=+\infty.

Source & rights

Original instruction with traceable references.

The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.

Chapter 4: Infinite behavior

Vertical Asymptotes and Sign Analysis

Related resources

Original instruction with traceable references.

Chapter 4: Infinite behavior

Vertical Asymptotes and Sign Analysis

Related resources

Original instruction with traceable references.