Answer
The limit $\underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$ is 4.
Work Step by Step
Consider the provided function,
$ f\left( x \right)={{x}^{2}}+9\,\ \text{ for }\ x\ge 0\text{ }$
Follow the steps to find and inverse of the function to get the inverse of $ f\left( x \right)={{x}^{2}}+9\,\ \text{ for }\ x\ge 0\text{ }$
Step 1: Replace $ f\left( x \right)$ with $ y $.
$ y={{x}^{2}}+9\,x\ge 0$
Step 2: Interchange $ y $ and $ x $.
$ x={{y}^{2}}+9\,y\ge 0$
Step 3: Solve for $ y $.
$\begin{align}
& x={{y}^{2}}+9\,y\ge 0 \\
& x-9={{y}^{2}} \\
& \sqrt{x-9}=y\
\end{align}$
Since $ y\ge 0$, so negative root is not considered.
Step 4: Replace $ y $ with ${{f}^{-1}}\left( x \right)$.
${{f}^{-1}}\left( x \right)=\sqrt{x-9}$
Thus, ${{f}^{-1}}\left( x \right)=\sqrt{x-9}$.
Solve the limit $\underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$
$\underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\underset{x\to 25}{\mathop{\lim }}\,\sqrt{x-9}$
Use the limit property $\underset{x\to a}{\mathop{\lim }}\,\sqrt{f\left( x \right)}=\sqrt{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}$
$\begin{align}
& \underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\underset{x\to 25}{\mathop{\lim }}\,\sqrt{x-9} \\
& =\sqrt{\underset{x\to 25}{\mathop{\lim }}\,\left( x-9 \right)}
\end{align}$
Use the limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$
$\begin{align}
& \underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\sqrt{\underset{x\to 25}{\mathop{\lim }}\,\left( x-9 \right)} \\
& =\sqrt{\underset{x\to 25}{\mathop{\lim }}\,x-\underset{x\to 25}{\mathop{\lim }}\,9}
\end{align}$
Use the limit property $\underset{x\to a}{\mathop{\lim }}\,c=c\ \text{where c is a constant}\text{.}$
$\begin{align}
& \underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\sqrt{\underset{x\to 25}{\mathop{\lim }}\,x-\underset{x\to 25}{\mathop{\lim }}\,9} \\
& =\sqrt{25-9} \\
& =\sqrt{16} \\
& =4
\end{align}$
Thus, the limit $\underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$ is 4.
