Answer
$\lim\limits_{x\to-2}\frac{x+2}{x^3+8}=\frac{1}{12}$
Work Step by Step
$\lim\limits_{x\to-2}\frac{x+2}{x^3+8}$
$=\lim\limits_{x\to-2}\frac{x+2}{(x+2)(x^2-2x+4)}$
(remember $(a^3+b^3)=(a+b)(a^2-ab+b^2)$
$=\lim\limits_{x\to-2}\frac{1}{x^2-2x+4}$
(divide both numerator and denominator by $x+2$)
$=\frac{1}{(-2)^2-2\times(-2)+4}$
$=\frac{1}{12}$
