\[ \begin{split} &\lim_{x \to -1}\frac{\sin (x+1)}{1-x^2}=\\[16pt] &=\lim_{x \to -1}\frac{\sin (x+1)}{(1-x)(1+x)}=\\[16pt] &=\lim_{x \to -1}\frac{\sin (x+1)}{1+x}\cdot \lim_{x \to -1}\frac{1}{1-x}=\\[16pt] &=1\cdot \frac{1}{2}=\frac{1}{2} \end{split} \]