Answer
$\lim\limits_{x \to -1}\dfrac{x^{2}-1}{x+1}=-2$
Work Step by Step
$\lim\limits_{x \to -1}\dfrac{x^{2}-1}{x+1}$
Factor the numerator and simplify:
$\lim\limits_{x \to -1}\dfrac{x^{2}-1}{x+1}=\lim\limits_{x \to -1}\dfrac{(x-1)(x+1)}{x+1}=\lim\limits_{x \to -1}x-1$
Apply direct substitution to evaluate the limit:
$\lim\limits_{x \to -1}x-1=-1-1=-2$
