Answer
$2$
Work Step by Step
\begin{align*} \lim _{x \rightarrow 1}\frac{x^3-x}{x-1}&=\lim _{x \rightarrow 1}\frac{x(x^2-1)}{x-1}\\&=\lim _{x \rightarrow 1}\frac{x(x+1)(x-1)}{x-1}
\\ &=\lim _{x \rightarrow 1} x(x+1) \\ &=1(1+1)\\ &=2. \end{align*}