## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$-\dfrac{2}{3}$
Use the limit properties: (a) $\lim\limits_{x \to a} l(x)=l (a)$, where $a$ is a constant. We have: $\lim\limits_{x \to a} \dfrac{k(x)}{l(x)}=\dfrac{\lim\limits_{x \to a} k(x)}{\lim\limits_{x \to a} l(x)}$ $\lim\limits_{x\to -1^-}\dfrac{x^2-1}{x^3+1}=\lim\limits_{x\to -1^-}\dfrac{(x+1)(x-1)}{(x+1)(x^2-x+1)} \\=\dfrac{\lim\limits_{x\to -1^-} (x+1)(x-1)}{\lim\limits_{x\to -1^-}(x+1)(x^2-x+1)} \\=\dfrac{(-1)-1}{(-1)^2-(-1)+1} \\ =-\dfrac{2}{3}$