Answer
$$\frac{-5}{3}$$
Work Step by Step
Since we have
$$
\lim _{x \rightarrow -5} \frac{ x^{2}-25 }{5-4x-x^2}= \frac{0}{0}.
$$
Then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow -5} \frac{ x^{2}-25 }{5-4x-x^2}= \lim _{x \rightarrow -5} \frac{ 2x }{ -4-2x}=\frac{-10}{6}=-\frac{5}{3}.
$$
