## Precalculus (6th Edition) Blitzer

The value of $\underset{x\to -1}{\mathop{\lim }}\,f\left( x \right)$ does not exist.
The value of $\underset{x\to -1}{\mathop{\lim }}\,f\left( x \right)$ exists only if the values of $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ and $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ exist and are equal. It is seen from the graph that, the value of $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $2$ and the value of $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $1$. Since $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, Thus, the value of $\underset{x\to -1}{\mathop{\lim }}\,f\left( x \right)$ does not exist.