Evaluating a limit by factoring to resolve a 0/0 form

Substituting $x = 3$ directly gives $\frac{3^2-9}{3-3} = \frac{0}{0}$, which is indeterminate, so direct substitution fails and we need another approach. The numerator $x^2 - 9$ is a difference of squares, so it factors as $(x-3)(x+3)$.

Rewriting the expression gives $\frac{(x-3)(x+3)}{x-3},$ and since $x \neq 3$ in the limiting process, we can cancel the common factor $(x-3)$, leaving $x+3$. Now direct substitution works: $\lim_{x\to 3}(x+3) = 3+3 = 6$.

The factor canceled was x minus three. It appeared in the denominator directly, and in the factored numerator as one of the two binomial factors from the difference of squares. Canceling it left the simplified expression x plus three, which allowed direct substitution to succeed.