Calculus I · Limits and Continuity · lesson

Evaluating Limits by Factoring

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}.

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

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Factor $x^2-9$ .

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Worked example

Factor a quadratic

Evaluate

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

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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}.

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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.

After the explanation

Use the section idea

Reading lens

What did direct substitution reveal, and which algebraic move removes the obstacle without changing nearby behavior?

Mental model

Substitution is a diagnostic first move: a real number usually finishes the problem, while an indeterminate form asks for a structural rewrite.

Decision

Match the obstacle to the algebra—factor polynomial zeros, rationalize radicals, combine complex fractions, and split absolute values into one-sided cases.

Common trap

Zero over zero is not an answer, and cancellation is legal only for factors after the expression has been rewritten as a product.

Check yourself

You are ready to move on when you can justify why each rewrite preserves nearby values even if the original expression is undefined at the target.

Source & rights

Original instruction with traceable references.

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