Factoring Method for Limits

The Problem

You try to evaluate $\lim_{x \to a} \frac{f(x)}{g(x)}$ and get:

$\frac{f(a)}{g(a)} = \frac{0}{0}$

This means both the numerator and denominator have (x - a) as a factor.

The Solution

1. Factor both the numerator and denominator
2. Cancel the common factor (x - a)
3. Re-evaluate using direct substitution

The key insight: $(x - a)$ is causing the problem, so eliminate it!

Example 1: Basic Factoring

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

Step 1: Try direct substitution $\frac{2^2 - 4}{2 - 2} = \frac{0}{0}$ ← Indeterminate!

Step 2: Factor the numerator $x^2 - 4 = (x - 2)(x + 2)$

Step 3: Rewrite and cancel $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x - 2} = \lim_{x \to 2} (x + 2)$

Step 4: Now use direct substitution $= 2 + 2 = 4$

Answer: 4

Why Can We Cancel?

We're evaluating the limit as x approaches 2, not at x = 2.

Since $x \neq 2$ during the approach, we can safely divide by $(x - 2)$.

Example 2: Factor Both Parts

Find $\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 5x + 6}$

Step 1: Check for $\frac{0}{0}$ $\frac{9 - 9}{9 - 15 + 6} = \frac{0}{0}$ ✓

Step 2: Factor everything

• Numerator: $x^2 - 9 = (x - 3)(x + 3)$
• Denominator: $x^2 - 5x + 6 = (x - 3)(x - 2)$

Step 3: Cancel common factor $\lim_{x \to 3} \frac{(x-3)(x+3)}{(x-3)(x-2)} = \lim_{x \to 3} \frac{x + 3}{x - 2}$

Step 4: Direct substitution $= \frac{3 + 3}{3 - 2} = \frac{6}{1} = 6$

Answer: 6

Common Factoring Patterns

Strategy Summary

1. Always try direct substitution first
2. If you get $\frac{0}{0}$, factor!
3. Look for common factors to cancel
4. Try direct substitution again on the simplified form
5. Success! ✓

What If Factoring Doesn't Work?

If you still get $\frac{0}{0}$ after factoring, try:

• Rationalizing (for radicals)
• Multiplying by conjugates
• L'Hôpital's Rule (advanced)

This is indeterminate, so we need to factor.

Step 2: Factor the numerator

$x^2 - 25 = (x - 5)(x + 5)$

This is a difference of squares pattern.

Step 3: Rewrite and cancel

$\lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \lim_{x \to 5} \frac{(x-5)(x+5)}{x - 5}$

Cancel the common factor $(x - 5)$:

$= \lim_{x \to 5} (x + 5)$

Step 4: Direct substitution

$= 5 + 5 = 10$

Answer: 10