Evaluating a limit by factoring

Direct substitution at $x = 2$ gives $\frac{2^4-16}{2-2} = \frac{0}{0}$, an indeterminate form, so we need to factor. Notice $x^4 - 16$ is a difference of squares: $x^4 - 16 = (x^2-4)(x^2+4)$. Factoring again, $x^2 - 4 = (x-2)(x+2)$, so $x^4 - 16 = (x-2)(x+2)(x^2+4)$.

Substituting this factorization into the original expression gives $\frac{(x-2)(x+2)(x^2+4)}{x-2}.$ For $x \neq 2$, the factor $(x-2)$ cancels, leaving $f(x) = (x+2)(x^2+4)$, which is now continuous at $x=2$. Direct substitution gives $(2+2)(2^2+4) = 4 \cdot 8 = 32$.