Answer
$\lim_{x\to-1}\dfrac{x^{2}-x-2}{x^{3}-x}=-\dfrac{3}{2}$
The graph is shown below:
Work Step by Step
$\lim_{x\to-1}\dfrac{x^{2}-x-2}{x^{3}-x}$
Try to evaluate the limit applying direct substitution:
$\lim_{x\to-1}\dfrac{x^{2}-x-2}{x^{3}-x}=\dfrac{(-1)^{2}-(-1)-2}{(-1)^{3}-(-1)}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Factor the numerator and the denominator of the function and simplify:
$\lim_{x\to-1}\dfrac{x^{2}-x-2}{x^{3}-x}=\lim_{x\to-1}\dfrac{(x-2)(x+1)}{x(x^{2}-1)}=...$
$...=\lim_{x\to-1}\dfrac{(x-2)(x+1)}{x(x-1)(x+1)}=\lim_{x\to-1}\dfrac{x-2}{x(x-1)}=...$
Try to evaluate the limit using direct substitution again:
$\lim_{x\to-1}\dfrac{x-2}{x(x-1)}=\dfrac{-1-2}{(-1)(-1-1)}=\dfrac{-3}{(-1)(-2)}=-\dfrac{3}{2}$
