Answer
$-\frac{1}{4}.$
Work Step by Step
Since we have
$$
\lim _{x \rightarrow -2} \frac{ x^3+2x^2-x-2 }{x^4+2x^3-4x-8}=\frac{-8+8+2-2}{16-16+8-8}= \frac{0}{0}.
$$
Then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow -2} \frac{ x^3+2x^2-x-2 }{x^4+2x^3-4x-8}=\lim _{x \rightarrow -2} \frac{ 3x^2+4x -1 }{4x^3+6x^2-4 }\\
= \frac{ 12-8-1}{ -32+24-4}=-\frac{3}{12}=-\frac{1}{4}.
$$
