Answer
The limit $\underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$ is $\frac{-17}{3}$.
Work Step by Step
Consider the provided function, $ f\left( x \right)=\frac{2x+3}{x+4}$.
Follow the steps to find and inverse of the function to get the inverse of $ f\left( x \right)=\frac{2x+3}{x+4}$
Step 1 : Replace $ f\left( x \right)$ with $ y $
$ y=\frac{2x+3}{x+4}$
Step 2 : Interchange $ y $ and $ x $
$ x=\frac{2y+3}{y+4}$
Step 3 : Solve for $ y $
$ x=\frac{2y+3}{y+4}$
Multiply by $ y+4$ on both sides:
$\begin{align}
& x\left( y+4 \right)=\frac{2y+3}{y+4}\left( y+4 \right) \\
& xy+4x=2y+3
\end{align}$
Subtract $2y $ from both sides:
$\begin{align}
& xy+4x-2y=2y+3-2y \\
& xy-2y+4x=3
\end{align}$
Subtract $4x $ from both sides:
$\begin{align}
& xy-2y+4x-4x=3-4x \\
& xy-2y=3-4x \\
& y\left( x-2 \right)=3-4x
\end{align}$
Divide by $ x-2$ on both sides;
$\begin{align}
& \frac{y\left( x-2 \right)}{\left( x-2 \right)}=\frac{3-4x}{\left( x-2 \right)} \\
& y=\frac{3-4x}{\left( x-2 \right)} \\
& =\frac{-4x+3}{\left( x-2 \right)}
\end{align}$
Replace $ y $ with ${{f}^{-1}}\left( x \right)$,
Thus, ${{f}^{-1}}\left( x \right)=\frac{-4x+3}{x-2}$.
Now, solve the limit $\underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$.
$\underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\underset{x\to 5}{\mathop{\lim }}\,\frac{-4x+3}{x-2}$
Use the quotient property of limits:
$\begin{align}
& \underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\underset{x\to 5}{\mathop{\lim }}\,\frac{-4x+3}{x-2} \\
& =\frac{\underset{x\to 5}{\mathop{\lim }}\,\left( -4x+3 \right)}{\underset{x\to 5}{\mathop{\lim }}\,\left( x-2 \right)}
\end{align}$
Use the limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ in the numerator and $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ in the denominator.
$\begin{align}
& \underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\frac{\underset{x\to 5}{\mathop{\lim }}\,\left( -4x+3 \right)}{\underset{x\to 5}{\mathop{\lim }}\,\left( x-2 \right)} \\
& =\frac{\underset{x\to 5}{\mathop{\lim }}\,\left( -4x \right)+\underset{x\to 5}{\mathop{\lim }}\,3}{\underset{x\to 5}{\mathop{\lim }}\,x-\underset{x\to 5}{\mathop{\lim }}\,2}
\end{align}$
Use the limit property $\underset{x\to a}{\mathop{\lim }}\,c=c\ \text{where c is a constant}\text{.}$
$\begin{align}
& \underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\frac{\underset{x\to 5}{\mathop{\lim }}\,\left( -4x \right)+\underset{x\to 5}{\mathop{\lim }}\,3}{\underset{x\to 5}{\mathop{\lim }}\,x-\underset{x\to 5}{\mathop{\lim }}\,2} \\
& =\frac{\left( -4\times 5 \right)+3}{5-2} \\
& =\frac{-20+3}{3} \\
& =\frac{-17}{3}
\end{align}$
Thus $\underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=-\frac{17}{3}$
