## Calculus (3rd Edition)

Transforming algebraically and canceling, we have $\frac{x^{3}-8}{x-2}=\frac{(x-2)(x^{2}+2x+2^{2})}{x-2}=x^{2}+2x+4$ Evaluating using continuity, we get $\lim\limits_{x \to 2}\frac{x^{3}-8}{x-2}=\lim\limits_{x \to 2}(x^{2}+2x+4)=2^{2}+2(2)+4=12$