Answer
$\lim\limits_{x \to -1} \frac{sin(x+1)}{x^2-1} = -\frac{1}{2}$
Work Step by Step
From the squeeze theorem:
$\lim\limits_{x \to 0} \frac{sin(x)}{x}=1$.
$\lim\limits_{x \to -1} \frac{sin(x+1)}{x^2-1}=\lim\limits_{x \to -1} \frac{sin(x+1)}{(x+1)(x-1)}=\lim\limits_{x \to -1} \frac{sin(x+1)}{(x+1)} *\lim\limits_{x \to -1}\frac{1}{x-1} = 1* \lim\limits_{x \to -1}\frac{1}{x-1} = -\frac{1}{2}$.
