Calculus I · Limits and Continuity · lesson

Evaluating Limits by Factoring

Learn Evaluating Limits by Factoring with plain-language explanations, guided examples, worked homework methods, interactive checks, and exam-style practice.

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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 Evaluating Limits by Factoring; the worked mathematics is evidence for the idea, not a substitute for it.

This page connects What 0/0 Means in a Limit to Evaluating Radical Limits With Conjugates. Read the explanation first, predict each example’s next move, and only then compare the written solution.

Concept

Learning objectives

Evaluate finite limits by factoring a numerator or denominator, cancelling a common nonzero factor, and then substituting.

Factoring and Cancellation

Guided walkthrough

Why cancellation is legal in a limit

Evaluate

\lim_{x\to2}\frac{x^2-4}{x-2}.

Show worked solution

Direct substitution gives $0/0$ , so factor:

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

For every nearby input $x\ne2$ ,

\frac{(x-2)(x+2)}{x-2}=x+2.

Now substitute:

2+2=4.

Therefore,

\boxed{\lim_{x\to2}\frac{x^2-4}{x-2}=4}.

We did not claim that the original expression is defined at $x=2$ . We only used the simplified expression for nearby values $x\ne2$ , exactly the values a limit studies.

Interactive checkfactor-flow-01

Evaluate $\lim_{x\to3}\frac{x^2-9}{x-3}$ .

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

Show hint

Factor $x^2-9$ .

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Worked example

Factor a quadratic

Evaluate

\lim_{x\to3}\frac{x^2-5x+6}{x-3}.

Show worked solution

Substitution gives $0/0$ . Factor the numerator:

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

Cancel the common factor for $x\ne3$ :

\frac{(x-2)(x-3)}{x-3}=x-2.

Then

\lim_{x\to3}(x-2)=3-2=\boxed{1}.

Worked example

Difference of cubes

Evaluate

\lim_{x\to2}\frac{x^3-8}{x-2}.

Show worked solution

Use

a^3-b^3=(a-b)(a^2+ab+b^2).

Thus,

x^3-8=(x-2)(x^2+2x+4).

For $x\ne2$ ,

\frac{x^3-8}{x-2}=x^2+2x+4.

Now substitute:

2^2+2(2)+4=4+4+4=\boxed{12}.

Worked example

Exam-level: factor more than once

Evaluate

\lim_{x\to1}\frac{x^4-1}{x^2-1}.

Show worked solution

Substitution gives $0/0$ . Factor both differences of squares:

x^4-1=(x^2-1)(x^2+1),

\frac{x^4-1}{x^2-1}=x^2+1,\qquad x\ne\pm1.

Therefore,

\lim_{x\to1}(x^2+1)=1+1=\boxed{2}.

An alternative full factorization is

x^4-1=(x-1)(x+1)(x^2+1),

x^2-1=(x-1)(x+1).

Both routes lead to the same simplification.

Exam note

Cancel factors, not terms. From

\frac{x^2+2x}{x}

you may factor the numerator and cancel:

\frac{x(x+2)}{x}=x+2.

You may not "cancel the $x$ " from only one term and write $x^2+2$ . Addition prevents cancellation until a common factor has been extracted.

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.

Factoring and Cancellation

Guided walkthrough

Why cancellation is legal in a limit

Evaluate

\lim_{x\to2}\frac{x^2-4}{x-2}.

Show worked solution

Direct substitution gives $0/0$ , so factor:

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

For every nearby input $x\ne2$ ,

\frac{(x-2)(x+2)}{x-2}=x+2.

Now substitute:

2+2=4.

Therefore,

\boxed{\lim_{x\to2}\frac{x^2-4}{x-2}=4}.

We did not claim that the original expression is defined at $x=2$ . We only used the simplified expression for nearby values $x\ne2$ , exactly the values a limit studies.

Interactive checkfactor-flow-01

Evaluate $\lim_{x\to3}\frac{x^2-9}{x-3}$ .

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

Show hint

Factor $x^2-9$ .

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Worked example

Factor a quadratic

Evaluate

\lim_{x\to3}\frac{x^2-5x+6}{x-3}.

Show worked solution

Substitution gives $0/0$ . Factor the numerator:

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

Cancel the common factor for $x\ne3$ :

\frac{(x-2)(x-3)}{x-3}=x-2.

Then

\lim_{x\to3}(x-2)=3-2=\boxed{1}.

Worked example

Difference of cubes

Evaluate

\lim_{x\to2}\frac{x^3-8}{x-2}.

Show worked solution

Use

a^3-b^3=(a-b)(a^2+ab+b^2).

Thus,

x^3-8=(x-2)(x^2+2x+4).

For $x\ne2$ ,

\frac{x^3-8}{x-2}=x^2+2x+4.

Now substitute:

2^2+2(2)+4=4+4+4=\boxed{12}.

Worked example

Exam-level: factor more than once

Evaluate

\lim_{x\to1}\frac{x^4-1}{x^2-1}.

Show worked solution

Substitution gives $0/0$ . Factor both differences of squares:

x^4-1=(x^2-1)(x^2+1),

\frac{x^4-1}{x^2-1}=x^2+1,\qquad x\ne\pm1.

Therefore,

\lim_{x\to1}(x^2+1)=1+1=\boxed{2}.

An alternative full factorization is

x^4-1=(x-1)(x+1)(x^2+1),

x^2-1=(x-1)(x+1).

Both routes lead to the same simplification.

Exam note

Cancel factors, not terms. From

\frac{x^2+2x}{x}

you may factor the numerator and cancel:

\frac{x(x+2)}{x}=x+2.

You may not "cancel the $x$ " from only one term and write $x^2+2$ . Addition prevents cancellation until a common factor has been extracted.

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 2: Finite limits and algebra

Factoring and Cancellation

Related resources

Original instruction with traceable references.

Chapter 2: Finite limits and algebra

Factoring and Cancellation

Related resources

Original instruction with traceable references.