Evaluating a 0/0 limit using L'Hopital's Rule

Substituting $x=2$ gives $\frac{2^3-8}{2^2-4} = \frac{0}{0}$, which is an indeterminate form, so direct substitution fails and L'Hopital's Rule applies. Differentiate the numerator and denominator separately: the derivative of $x^3-8$ is $3x^2$, and the derivative of $x^2-4$ is $2x$.

This gives the new limit $\lim_{x\to 2} \frac{3x^2}{2x}$. Substituting $x=2$ now gives $\frac{3(4)}{2(2)} = \frac{12}{4} = 3$, which is no longer indeterminate, so the original limit equals $3$.