Answer
a) The value of $\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $1$.
b) The value of $\,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $0$.
c) The value of $\,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)$ does not exist.
Work Step by Step
(a)
Consider the function $f\left( x \right)=\left\{ \begin{align}
& 9-2x\text{ if }x<4 \\
& \sqrt{x-4}\text{ if }x\ge 4
\end{align} \right.$ ,
As the value of $x$ nears $4$ from the left, the function takes the value $9-2x$.
Find the value of $\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ ,
$\begin{align}
& \,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,\left( 9-2x \right) \\
& =9-2\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,x \\
& =9-8 \\
& =1
\end{align}$
Thus, the value of $\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $1$.
(b)
Consider the function $f\left( x \right)=\left\{ \begin{align}
& 9-2x\text{ if }x<4 \\
& \sqrt{x-4}\text{ if }x\ge 4
\end{align} \right.$ ,
As the value of $x$ nears $4$ from the right, the function takes the value $\sqrt{x-4}$.
Find the value of $\,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ ,
$\begin{align}
& \,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\,\,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,\sqrt{x-4} \\
& =\sqrt{\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,x-4} \\
& =\sqrt{4-4} \\
& =0
\end{align}$
Thus, the value of $\,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $0$.
(c)
Consider the function $f\left( x \right)=\left\{ \begin{align}
& 9-2x\text{ if }x<4 \\
& \sqrt{x-4}\text{ if }x\ge 4
\end{align} \right.$ ,
From the part (a), the value of $\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $1$ and the value of $\,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $0$.
Since $\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ ,
Thus, the value of $\,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)$ does not exist.
