## Calculus (3rd Edition)

By substituting, we have $$\lim _{x \rightarrow 1 } \frac{x^3-2x}{x-1}=\frac{1-2}{1-1}=\frac{1}{0}$$ which means that the limit does not exist. The one-sided limits are infinite and given by $$\lim _{x \rightarrow 1^+ } \frac{x^3-2x}{x-1}= \infty, \quad \lim _{x \rightarrow 1^- } \frac{x^3-2x}{x-1}= \infty.$$