Answer
a) The limit $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $0$.
b) The limit $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to 0.
c) The limit $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ is 0.
Work Step by Step
(a)
Consider the provided limit,
$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$
This means that it is required to calculate the value of $ f\left( x \right)$ when $ x $ is close to 1 but less than 1.
Since, $ x $ is less than 1, using the first line of the piecewise defined function's equation
$ f\left( x \right)=1-x\ \ \text{if }\ x<1$
$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\left( 1-x \right)$
Now use limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$
$\begin{align}
& \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\left( 1-x \right) \\
& =\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,1-\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,x
\end{align}$
Use the property $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c=\text{constant}$
$\begin{align}
& \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,1-\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,x \\
& =1-1 \\
& =0
\end{align}$
Thus, the limit $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $0$.
b)
Consider the provided limit,
$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
This means that it is required to calculate the value of $ f\left( x \right)$ when $ x $ is close to 1 but greater than 1.
Since, $ x $ is greater than 1, using the third line of the piecewise defined function's equation
$ f\left( x \right)={{x}^{2}}-1\ \ \text{if }\ x>1$
$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)$
Now use limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$
$\begin{align}
& \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right) \\
& =\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{2}}-\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,1
\end{align}$
Use property $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c=\text{constant}$
$\begin{align}
& \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{2}}-\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,1 \\
& =1-1 \\
& =0
\end{align}$
Thus, the limit $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to 0.
(c)
Consider the provided limit,
$\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$
From part (a) $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0$
From part (b) $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=0$
As $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Thus, the limit $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ is 0.
