## Precalculus (6th Edition) Blitzer

The limit $\underset{x\to 8}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$ is 2.
Consider the provided function, $f\left( x \right)={{x}^{2}}+4\,\ \text{ for }\ x\ge 0\text{ }$ Follow the steps to find and inverse of the function to get the inverse of $f\left( x \right)={{x}^{2}}+4\,\ \text{ for }\ x\ge 0\text{ }$ Step 1: Replace $f\left( x \right)$ with $y$. $y={{x}^{2}}+4\,x\ge 0$ Step 2: Interchange $y$ and $x$. $x={{y}^{2}}+4\,y\ge 0$ Step 3: Solve for $y$. \begin{align} & x={{y}^{2}}+4\,y\ge 0 \\ & x-4={{y}^{2}} \\ & \sqrt{x-4}=y\ \end{align} Since $y\ge 0$, so negative root is not considered. Step 4: Replace $y$ with ${{f}^{-1}}\left( x \right)$. ${{f}^{-1}}\left( x \right)=\sqrt{x-4}$ Thus, ${{f}^{-1}}\left( x \right)=\sqrt{x-4}$ Solve the limit $\underset{x\to 8}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$. $\underset{x\to 8}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\underset{x\to 8}{\mathop{\lim }}\,\sqrt{x-4}$ 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 8}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\underset{x\to 8}{\mathop{\lim }}\,\sqrt{x-4} \\ & =\sqrt{\underset{x\to 8}{\mathop{\lim }}\,\left( x-4 \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 8}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\sqrt{\underset{x\to 8}{\mathop{\lim }}\,\left( x-4 \right)} \\ & =\sqrt{\underset{x\to 8}{\mathop{\lim }}\,x-\underset{x\to 8}{\mathop{\lim }}\,4} \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 8}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\sqrt{\underset{x\to 8}{\mathop{\lim }}\,x-\underset{x\to 8}{\mathop{\lim }}\,4} \\ & =\sqrt{8-4} \\ & =\sqrt{4} \\ & =2 \end{align} Thus, the limit $\underset{x\to 8}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$ is 2.