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

Direct substitution of $x=4$ gives $\frac{16-16}{4-4} = \frac{0}{0}$, an indeterminate form, so we cannot evaluate the limit this way. Instead, factor the numerator as a difference of squares: $x^2 - 16 = (x-4)(x+4)$.

For $x \neq 4$, the expression simplifies since $\frac{(x-4)(x+4)}{x-4} = x+4$. Since this simplified function agrees with $g(x)$ everywhere near $x=4$ (except at $x=4$ itself), the limits are equal, so by the limit laws (sum rule applied to $x+4$) we get $\lim_{x\to 4}(x+4) = 4+4 = 8$.