Answer
The function $f\left( x \right)=\left\{ \begin{align}
& \frac{{{\left( x+3 \right)}^{2}}-9}{x}\text{ if }x\ne 0 \\
& 6\text{ if }x=0
\end{align} \right.$
is continuous at $0$.
Work Step by Step
Find the value of $f\left( x \right)$ at $a=0$.
From the definition of the function, for $a=0$, the function takes the value $6$.
Thus, $f\left( 0 \right)=6$
The function is defined at the point $a=0$.
Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ ,
First find the left hand limit of $\,f\left( x \right)$ ,
As x nears $0$ from the left, the function $\,f\left( x \right)$ takes the value $\frac{{{\left( x+3 \right)}^{2}}-9}{x}$.
Thus, the value of $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is,
$\begin{align}
& \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{{{\left( x+3 \right)}^{2}}-9}{x} \\
& =\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{{{x}^{2}}+6x+9-9}{x} \\
& =\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{{{x}^{2}}+6x}{x} \\
& =\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( x+6 \right)
\end{align}$
Taking the limit inside,
$\begin{align}
& \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,x+6 \\
& =0+6 \\
& =6
\end{align}$
Thus, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=6$
Now find the right hand limit of $\,f\left( x \right)$.
As x nears $0$ from the right, the function $\,f\left( x \right)$ takes the value $\frac{{{\left( x+3 \right)}^{2}}-9}{x}$.
Thus, the value of $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is,
$\begin{align}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{\left( x+3 \right)}^{2}}-9}{x} \\
& =\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{2}}+6x+9-9}{x} \\
& =\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{2}}+6x}{x} \\
& =\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( x+6 \right)
\end{align}$
Taking the limit inside,
$\begin{align}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x+6 \\
& =0+6 \\
& =6
\end{align}$
Thus, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=6$
The left hand limit and right hand limit are equal, that is $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=6=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Thus, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=6$
From the above steps, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=6=f\left( 0 \right)$
Since the function satisfies all the properties of being continuous,
Hence, the function $f\left( x \right)=\left\{ \begin{align}
& \frac{{{\left( x+3 \right)}^{2}}-9}{x}\text{ if }x\ne 0 \\
& 6\text{ if }x=0
\end{align} \right.$
is continuous at $0$.
