Answer
$-2$
Work Step by Step
We find the limit:
$\lim_{x\to -1}\dfrac{x^2-4x-5}{x^3+1}\\=\lim_{x\to -1} \dfrac{(x+1)(x-5)}{(x+1)(x^2-x+1)}$
Apply $\lim\limits_{x \to a} \dfrac{A(x)}{B(x)}=\dfrac{\lim\limits_{x \to a} A(x)}{\lim\limits_{x \to a} B(x)}$
$\dfrac{\lim_{x\to -1} (x-5)}{\lim_{x\to -1} x^2-x+1 }\\=\dfrac{(-1)-5}{[(-1)^2-(-1)+1]}\\= -2$
