Chapter 11 - Mid-Chapter Check Point - Page 1163: 19

a) The value of $\,\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $32$. b) The value of $\,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $32$. c) The value of $\,\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ is $32$.

(a) Consider the function $f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{4}}-16}{x-2}\text{ if }x\ne 2 \\ & 32\text{ if }x=2 \end{align} \right.$ As the value of $x$ nears $2$ from the left, the function takes the value $\frac{{{x}^{4}}-16}{x-2}$. Find the value of $\,\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ , $\begin{align} & \,\,\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\frac{{{x}^{4}}-16}{x-2} \\ & =\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\frac{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+4 \right)}{x-2} \\ & =\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\frac{\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)}{x-2} \\ & =\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\left( x+2 \right)\left( {{x}^{2}}+4 \right) \end{align}$ Now, take the limit inside and calculate the value of $\,\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ , $\begin{align} & \,\,\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\left( \underset{x\to {{2}^{-}}}{\mathop{\lim }}\,x+2 \right)\left( \underset{x\to {{2}^{-}}}{\mathop{\lim }}\,{{x}^{2}}+4 \right) \\ & =\left( 2+2 \right)\left( {{2}^{2}}+4 \right) \\ & =\left( 4 \right)\left( 8 \right) \\ & =32 \end{align}$ Thus, the value of $\,\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $32$. (b) Consider the function $f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{4}}-16}{x-2}\text{ if }x\ne 2 \\ & 32\text{ if }x=2 \end{align} \right.$ As the value of $x$ nears $2$ from the right, the function takes the value $\frac{{{x}^{4}}-16}{x-2}$. Find the value of $\,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ , $\begin{align} & \,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{4}}-16}{x-2} \\ & =\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\frac{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+4 \right)}{x-2} \\ & =\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\frac{\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)}{x-2} \\ & =\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\left( x+2 \right)\left( {{x}^{2}}+4 \right) \end{align}$ Now, take the limit inside and calculate the value of $\,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ , $\begin{align} & \,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\left( \,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,x+2 \right)\left( \,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,{{x}^{2}}+4 \right) \\ & =\left( 2+2 \right)\left( {{2}^{2}}+4 \right) \\ & =\left( 4 \right)\left( 8 \right) \\ & =32 \end{align}$ Thus, the value of $\,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $32$. (c) Consider the function $f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{4}}-16}{x-2}\text{ if }x\ne 2 \\ & 32\text{ if }x=2 \end{align} \right.$ From part (a), the value of $\,\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is $32$ and the value of $\,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is $32$. Since $\,\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ , Thus, the value of $\,\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ exists and is equal to $32$.