Resposta:

Resolvamos o limite dado por meio da Regra de L'Hopital:

[tex]\lim_{x \to 4} \frac{x^2-3x-4}{x^2-2x-8}\\\\ = \lim_{x \to 4} \frac{\frac{d}{dx}\left(x^2-3x-4\right)}{\frac{d}{dx}\left(x^2-2x-8\right)}\\\\= \lim_{x \to 4} \frac{2x -3}{2x -2}\\\\= \frac{2 \cdot 4 - 3}{2 \cdot 4 - 2}\\\\= \boxed{\frac{5}{6}.}[/tex]