Calculus I · Limits and Continuity · lesson

Limits With Absolute Values and Piecewise Functions

Learn Limits With Absolute Values and Piecewise Functions with plain-language explanations, guided examples, worked homework methods, interactive checks, and ex

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Where this chapter fits

Chapter 2: Finite limits and algebra

Turn indeterminate forms into solvable expressions using substitution, limit laws, factoring, conjugates, and piecewise reasoning.

Reading lens: What did direct substitution reveal, and which algebraic move removes the obstacle without changing nearby behavior? Keep that question in view while reading Limits With Absolute Values and Piecewise Functions; the worked mathematics is evidence for the idea, not a substitute for it.

This page connects Limits With Complex Fractions to Finite Limit Decision Tree. Read the explanation first, predict each example’s next move, and only then compare the written solution.

Concept

Learning objectives

Rewrite absolute values on each side of a target and evaluate one-sided limits before deciding whether a two-sided limit exists.

Absolute Values and Piecewise Rules

The absolute value is piecewise:

|u|= \left\{\matrix{ u,\quad u\ge0,\quad -u,\quad u<0. }\right.

The sign of the inside expression determines which rule applies.

Guided walkthrough

$|x|/x$

Evaluate

\lim_{x\to0}\frac{|x|}{x}.

Show worked solution

For $x>0$ , $|x|=x$ , so

\frac{|x|}{x}=\frac{x}{x}=1.

Therefore,

\lim_{x\to0^+}\frac{|x|}{x}=1.

For $x<0$ , $|x|=-x$ , so

\frac{|x|}{x}=\frac{-x}{x}=-1.

Therefore,

\lim_{x\to0^-}\frac{|x|}{x}=-1.

The sides disagree, so

\boxed{\lim_{x\to0}\frac{|x|}{x}=\mathrm{DNE}}.

Worked example

A shifted absolute value

Evaluate

\lim_{x\to3}\frac{|x-3|}{x-3}.

Show worked solution

When $x>3$ , the inside $x-3$ is positive, so

|x-3|=x-3

and the quotient equals $1$ .

When $x<3$ , the inside $x-3$ is negative, so

|x-3|=-(x-3)

and the quotient equals $-1$ .

Thus,

\lim_{x\to3^-}\frac{|x-3|}{x-3}=-1,

\lim_{x\to3^+}\frac{|x-3|}{x-3}=1.

Therefore the two-sided limit does not exist.

Worked example

An absolute value whose limit does exist

Evaluate

\lim_{x\to2}\frac{|x-2|}{\sqrt{|x-2|}}.

Show worked solution

For $x\ne2$ , $|x-2|>0$ , and

\frac{|x-2|}{\sqrt{|x-2|}}=\sqrt{|x-2|}.

As $x\to2$ , the distance $|x-2|\to0$ . Therefore,

\sqrt{|x-2|}\to0,

\boxed{0}.

Absolute value does not automatically cause a limit to fail. It causes one-sided analysis when the formula changes sign or form across the target.

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.

Chapter 2: Finite limits and algebra

Turn indeterminate forms into solvable expressions using substitution, limit laws, factoring, conjugates, and piecewise reasoning.

This page connects Limits With Complex Fractions to Finite Limit Decision Tree. Read the explanation first, predict each example’s next move, and only then compare the written solution.

Absolute Values and Piecewise Rules

The absolute value is piecewise:

|u|= \left\{\matrix{ u,\quad u\ge0,\quad -u,\quad u<0. }\right.

The sign of the inside expression determines which rule applies.

Guided walkthrough

$|x|/x$

Evaluate

\lim_{x\to0}\frac{|x|}{x}.

Show worked solution

For $x>0$ , $|x|=x$ , so

\frac{|x|}{x}=\frac{x}{x}=1.

Therefore,

\lim_{x\to0^+}\frac{|x|}{x}=1.

For $x<0$ , $|x|=-x$ , so

\frac{|x|}{x}=\frac{-x}{x}=-1.

Therefore,

\lim_{x\to0^-}\frac{|x|}{x}=-1.

The sides disagree, so

\boxed{\lim_{x\to0}\frac{|x|}{x}=\mathrm{DNE}}.

Worked example

A shifted absolute value

Evaluate

\lim_{x\to3}\frac{|x-3|}{x-3}.

Show worked solution

When $x>3$ , the inside $x-3$ is positive, so

|x-3|=x-3

and the quotient equals $1$ .

When $x<3$ , the inside $x-3$ is negative, so

|x-3|=-(x-3)

and the quotient equals $-1$ .

Thus,

\lim_{x\to3^-}\frac{|x-3|}{x-3}=-1,

\lim_{x\to3^+}\frac{|x-3|}{x-3}=1.

Therefore the two-sided limit does not exist.

Worked example

An absolute value whose limit does exist

Evaluate

\lim_{x\to2}\frac{|x-2|}{\sqrt{|x-2|}}.

Show worked solution

For $x\ne2$ , $|x-2|>0$ , and

\frac{|x-2|}{\sqrt{|x-2|}}=\sqrt{|x-2|}.

As $x\to2$ , the distance $|x-2|\to0$ . Therefore,

\sqrt{|x-2|}\to0,

\boxed{0}.

Absolute value does not automatically cause a limit to fail. It causes one-sided analysis when the formula changes sign or form across the target.

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.

Limits With Absolute Values and Piecewise Functions

Chapter 2: Finite limits and algebra

Learning objectives

Absolute Values and Piecewise Rules

\(|x|/x\)

A shifted absolute value

An absolute value whose limit does exist

Original instruction with traceable references.

Limits With Absolute Values and Piecewise Functions

Chapter 2: Finite limits and algebra

Learning objectives

Absolute Values and Piecewise Rules

\(|x|/x\)

A shifted absolute value

An absolute value whose limit does exist

Original instruction with traceable references.

Chapter 2: Finite limits and algebra

Absolute Values and Piecewise Rules

Related resources

Original instruction with traceable references.

Chapter 2: Finite limits and algebra

Absolute Values and Piecewise Rules

Related resources

Original instruction with traceable references.