Ta có
$\underset{x \to 1}{\lim} \dfrac{(x+1)^3 - 8 - (\sqrt{x + 63} - 8)}{x-1}$
$= \underset{x \to 1}{\lim} \dfrac{(x+1 - 2)[(x+1)^2 + 4 + 2(x+1)] - \frac{x+63 - 64}{\sqrt{x+63} + 8}}{x-1}$
$= \underset{x \to 1}{\lim} \left( x^2 + 4x +7 - \dfrac{1}{\sqrt{x+63} + 8}\right)$
$= 1 + 4 + 7 - \dfrac{1}{8 + 8} = \dfrac{191}{16}$.
Vậy
$\underset{x \to 1}{\lim} \dfrac{(x+1)^3 - 8 - \sqrt{x + 63} - 8}{x-1} = \dfrac{191}{16}$.