Chapter 11 - Section 11.2 - Finding Limits Using Properties of Limits - Exercise Set - Page 1153: 46

a) The limit $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $10$. b) The limit $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $10$. c) The limit $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ is 10.

(a) Consider the provided limit, $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ This means that it is required to calculate the value of $ f\left( x \right)$ when $ x $ is close to 2 but less than 2. Since, $ x $ is less than 2, using the first line of the piecewise defined function's equation $ f\left( x \right)={{x}^{2}}+6\ \ \text{if }\ x<2$ $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\left( {{x}^{2}}+6 \right)$ Now use limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$, $\begin{align} & \underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\left( {{x}^{2}}+6 \right) \\ & =\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,{{x}^{2}}+\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,6 \end{align}$ Use the property $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c=\text{constant}$ $\begin{align} & \underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,{{x}^{2}}+\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,6 \\ & ={{2}^{2}}+6 \\ & =4+6 \\ & =10 \end{align}$ Thus, the limit $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $10$. (b) Consider the provided limit, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ This means that it is required to calculate the value of $ f\left( x \right)$ when $ x $ is close to 2 but greater than 2. Since, $ x $ is greater than 2, using the second line of the piecewise defined function's equation $ f\left( x \right)={{x}^{3}}+2\ \ \text{if }\ x\ge 2$ $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{x}^{3}}+2 \right)$ Now use limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ $\begin{align} & \underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{x}^{3}}+2 \right) \\ & =\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,{{x}^{3}}+\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,2 \end{align}$ Use the property $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c=\text{constant}$ $\begin{align} & \underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,{{x}^{3}}+\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,2 \\ & ={{2}^{3}}+2 \\ & =8+2 \\ & =10 \end{align}$ Thus, the limit $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $10$. (c) Consider the provided limit, $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ Use the second equation of the piecewise defined function $ f\left( x \right)={{x}^{3}}+2\ \text{if }x\ge 2$ $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{3}}+2 \right)$ Now use the limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ $\begin{align} & \underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{3}}+2 \right) \\ & =\underset{x\to 2}{\mathop{\lim }}\,{{x}^{3}}+\underset{x\to 2}{\mathop{\lim }}\,2 \end{align}$ Use property $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c=\text{constant}$ $\begin{align} & \underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 2}{\mathop{\lim }}\,{{x}^{3}}+\underset{x\to 2}{\mathop{\lim }}\,2 \\ & ={{2}^{3}}+2 \\ & =8+2 \\ & =10 \end{align}$ Thus, the limit $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ is 10.