Answer
For $ f\left( x \right)=\frac{1}{{{x}^{2}}}$, $\underset{x\to -1}{\mathop{\lim }}\,f\left( x \right)=1$.
Work Step by Step
Consider the function,
$ y=\frac{1}{{{x}^{2}}}$
Limits has the following property.
$\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L $ if and only if $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=L $ and $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L $.
For $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$,
Approach the point $ x=-1$ on the graph from the left side; the value of the $ y-$coordinate approaches to $1$. So, $\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1$.
For $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$,
Now approach the point from the right side. As the point is reached, the value of the $ y-$coordinate approaches to $1$. So, $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=1$.
Since, $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1$, the value for the given limit exists and is given by,
$\underset{x\to -1}{\mathop{\lim }}\,\frac{1}{{{x}^{2}}}=1$
