How to Find a Limit by Factoring

Problem

\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Answer

6

A limit asks what value an expression approaches, not what it equals at the point. Factoring exposes the hidden value when direct substitution gives 0/0.

Step-by-step solution

Try direct substitution.
$\frac{3^2 - 9}{3 - 3} = \frac{0}{0}$
Why: The 0/0 form means the expression needs work before you can substitute.
Factor the top as a difference of squares.
$x^2 - 9 = (x - 3)(x + 3)$
Why: The common factor with the bottom becomes visible.
Cancel the common factor.
$\frac{(x - 3)(x + 3)}{x - 3} = x + 3, \quad x \neq 3$
Why: Cancelling is valid for every x except the value that made the factor zero.
Substitute x = 3 into the simplified expression.
$3 + 3 = 6$
Why: The simplified expression is continuous at x = 3.

Common mistakes

Stopping at 0/0 and calling the limit undefined. 0/0 is a signal, not an answer.
Cancelling x - 3 from only part of the top.
Forgetting the restriction for the cancelled factor.

Practice problems

Use the same method. Work each problem on paper, then open the answer to check.

$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

Show answer

$4$
$\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$

Show answer

$2$
$\lim_{x \to 4} \frac{x^2 - 16}{x - 4}$

Show answer

$8$
$\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$

Show answer

$10$
$\lim_{x \to 0} \frac{x^2 + 2x}{x}$

Show answer

$2$