## Precalculus (6th Edition) Blitzer

a) The value of $\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $1$. b) The value of $\,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $0$. c) The value of $\,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)$ does not exist.
(a) Consider the function f\left( x \right)=\left\{ \begin{align} & 9-2x\text{ if }x<4 \\ & \sqrt{x-4}\text{ if }x\ge 4 \end{align} \right. , As the value of $x$ nears $4$ from the left, the function takes the value $9-2x$. Find the value of $\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ , \begin{align} & \,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,\left( 9-2x \right) \\ & =9-2\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,x \\ & =9-8 \\ & =1 \end{align} Thus, the value of $\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $1$. (b) Consider the function f\left( x \right)=\left\{ \begin{align} & 9-2x\text{ if }x<4 \\ & \sqrt{x-4}\text{ if }x\ge 4 \end{align} \right. , As the value of $x$ nears $4$ from the right, the function takes the value $\sqrt{x-4}$. Find the value of $\,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ , \begin{align} & \,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\,\,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,\sqrt{x-4} \\ & =\sqrt{\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,x-4} \\ & =\sqrt{4-4} \\ & =0 \end{align} Thus, the value of $\,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $0$. (c) Consider the function f\left( x \right)=\left\{ \begin{align} & 9-2x\text{ if }x<4 \\ & \sqrt{x-4}\text{ if }x\ge 4 \end{align} \right. , From the part (a), the value of $\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $1$ and the value of $\,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $0$. Since $\,\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \,\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ , Thus, the value of $\,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)$ does not exist.