Answer
$\lim\limits_{x \to 0}\frac{\frac{1}{x-1} +\frac{1}{x+1}}{x} = -2$
Work Step by Step
Simplify:
$\frac{\frac{1}{x-1} +\frac{1}{x+1}}{x} = \frac{(x+1+x-1)}{x(x^{2}-1)} = \frac{2}{(x^{2}-1)}$
Now,
$\lim\limits_{x \to 0}\frac{\frac{1}{x-1} +\frac{1}{x+1}}{x} = \lim\limits_{x \to 0} \frac{2}{(x^{2}-1)} = \frac{2}{-1} = -2$
