Answer
$\lim\limits_{x\to-1^-}\dfrac{x+1}{x^4-1}=-\dfrac{1}{4}$.
Work Step by Step
$\lim\limits_{x\to-1^-}\dfrac{x+1}{x^4-1}=\lim\limits_{x\to-1^-}\dfrac{x+1}{(x^2-1)(x^2+1)}$
$=\lim\limits_{x\to-1^-}\dfrac{(x+1)}{(x+1)(x-1)(x^2+1)}=\lim\limits_{x\to-1^-}\dfrac{1}{(x-1)(x^2+1)}$
$=\dfrac{1}{((-1^-)-1)((-1^-)^2+1)}=-\dfrac{1}{4}.$
