Factoring Method for Limits

Use factoring to simplify and evaluate limits with indeterminate forms

Solving Limits by Factoring

When direct substitution gives you $\frac{0}{0}$ , factoring is often your best friend!

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

Factor both the numerator and denominator
Cancel the common factor (x - a)
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

| Expression | Factored Form | |------------|---------------| | $x^2 - a^2$ | $(x - a)(x + a)$ | | $x^2 + bx + c$ | Find two numbers that multiply to c, add to b | | $x^3 - a^3$ | $(x - a)(x^2 + ax + a^2)$ | | $x^3 + a^3$ | $(x + a)(x^2 - ax + a^2)$ |

Strategy Summary

Always try direct substitution first
If you get $\frac{0}{0}$ , factor!
Look for common factors to cancel
Try direct substitution again on the simplified form
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)

📚 Practice Problems

1Problem 1easy

❓ Question:

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

💡 Show Solution

Step 1: Try direct substitution

$\frac{5^2 - 25}{5 - 5} = \frac{0}{0}$

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

2Problem 2medium

❓ Question:

Evaluate $\lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$ (This is important for derivatives!)

💡 Show Solution

Step 1: Expand the numerator

$(x + h)^2 - x^2 = x^2 + 2xh + h^2 - x^2 = 2xh + h^2$

Step 2: Rewrite the limit

$\lim_{h \to 0} \frac{2xh + h^2}{h}$

Step 3: Factor out h from the numerator

$\lim_{h \to 0} \frac{h(2x + h)}{h}$

Step 4: Cancel the common factor

$\lim_{h \to 0} (2x + h)$

Step 5: Direct substitution

$= 2x + 0 = 2x$

Answer: 2x

Note: This limit is actually the derivative of $f(x) = x^2$ !

3Problem 3easy

❓ Question:

Evaluate lim(x→3) (x² - 9)/(x - 3)

💡 Show Solution

Step 1: Try direct substitution: (3² - 9)/(3 - 3) = 0/0 (indeterminate)

Step 2: Factor the numerator: x² - 9 = (x - 3)(x + 3)

Step 3: Simplify: (x² - 9)/(x - 3) = [(x - 3)(x + 3)]/(x - 3) = x + 3 (for x ≠ 3)

Step 4: Evaluate the limit: lim(x→3) (x + 3) = 3 + 3 = 6

Answer: 6

4Problem 4medium

❓ Question:

Find lim(x→2) (x² - 5x + 6)/(x² - 4)

💡 Show Solution

Step 1: Check direct substitution: (4 - 10 + 6)/(4 - 4) = 0/0 (indeterminate)

Step 2: Factor numerator: x² - 5x + 6 = (x - 2)(x - 3)

Step 3: Factor denominator: x² - 4 = (x - 2)(x + 2)

Step 4: Simplify: [(x - 2)(x - 3)]/[(x - 2)(x + 2)] = (x - 3)/(x + 2) for x ≠ 2

Step 5: Evaluate: lim(x→2) (x - 3)/(x + 2) = (2 - 3)/(2 + 2) = -1/4

Answer: -1/4

5Problem 5hard

❓ Question:

Evaluate lim(h→0) [(2 + h)² - 4]/h

💡 Show Solution

Step 1: Expand (2 + h)²: (2 + h)² = 4 + 4h + h²

Step 2: Substitute into expression: [(4 + 4h + h²) - 4]/h = [4h + h²]/h

Step 3: Factor numerator: [4h + h²]/h = [h(4 + h)]/h

Step 4: Cancel common factor: h(4 + h)/h = 4 + h (for h ≠ 0)

Step 5: Evaluate limit: lim(h→0) (4 + h) = 4 + 0 = 4

Step 6: Note: This is the form of a derivative! It's the derivative of x² at x = 2

Answer: 4

🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore other calculus topics