Answer
The value of $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ is $-1$.
Work Step by Step
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, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-1$.
