## Calculus (3rd Edition)

The limit does not exist. It approaches $-\infty$ as $x$ approaches $-1^+$.
By substituting, we have $$\lim _{x \rightarrow -1^+} \frac{1}{x+1}=\frac{1}{-1+1}=\frac{1}{0} .$$ So the limit does not exist. It approaches $\infty$ as $x$ approaches $-1^+$. Graphing confirms that the one-sided limits are infinite.