## Precalculus (6th Edition) Blitzer

The limit $\underset{x\to 4}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$ is $\frac{5}{2}$.
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 $\frac{5}{2}$.