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