Answer
$12$
Work Step by Step
Given: $\lim\limits_{x\to-2}\dfrac{x^3+8}{x+2}$
This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule.
$\lim_{x\to-2}\dfrac{\dfrac{d(x^3+8)}{dx}}{\dfrac{d(x+2)}{dx}}=\lim\limits_{x\to-2}\dfrac{3x^2}{1}$
or, $=\lim\limits_{x\to-2}(3x^2)$
or, $=3\times(-2)^2$
or, $=3\times4$
or, $=12$
