Example 1: Basic Factoring

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

Step 1: Direct substitution gives $\frac{0}{0}$ ✗

Step 2: Factor the numerator: $\lim_{x \to 2} \frac{(x-2)(x+2)}{x - 2}$

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

Step 4: Direct substitution: $= 2 + 2 = 4$

Example 2: Trinomial Factoring

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

Factor numerator: $x^2 - 5x + 6 = (x-2)(x-3)$

$\lim_{x \to 3} \frac{(x-2)(x-3)}{x - 3} = \lim_{x \to 3} (x-2) = 1$

Common Factoring Patterns

Difference of squares: $a^2 - b^2 = (a-b)(a+b)$

Perfect square trinomial: $a^2 \pm 2ab + b^2 = (a \pm b)^2$

Trinomial: $x^2 + bx + c$ → Find two numbers that multiply to $c$ and add to $b$

Why Does This Work?

When we cancel $(x-a)$, we're essentially "filling the hole" in the function. The limit cares about values near $a$, not at $a$, so canceling is valid!

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

Direct substitution: $= -2 + 3 = 1$

Answer: 1