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

Substituting $x=2$ directly gives $\frac{2^3-8}{2^2-4} = \frac{0}{0}$, an indeterminate form, so we cannot use direct substitution and must simplify algebraically first. Factor the numerator using the difference of cubes formula: $x^3 - 8 = (x-2)(x^2+2x+4)$. Factor the denominator using the difference of squares formula: $x^2 - 4 = (x-2)(x+2)$.

Since $x \neq 2$ as $x \to 2$, the common factor $(x-2)$ can be canceled, giving $\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} = \frac{x^2+2x+4}{x+2}.$ Now apply the quotient rule for limits, since both numerator and denominator have finite limits with a nonzero denominator limit at $x=2$: $\lim_{x\to 2}\frac{x^2+2x+4}{x+2} = \frac{2^2+2(2)+4}{2+2} = \frac{12}{4} = 3.$