## Precalculus (6th Edition) Blitzer

The value of $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ is $-1$.
The value of $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ exists only if the values of $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ and $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ exist and are equal. It is seen from the table that, as the value of x nears $0$ from the left, the value of $f\left( x \right)$ nears $-1$ Thus $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-1$. It is seen from the table that, as the value of x nears $0$ from the right, the value of $f\left( x \right)$ nears $-1$. Thus $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=-1$. Since $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$. Thus, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-1$.