## Precalculus (6th Edition) Blitzer

a) The value of $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ where $f\left( x \right)=\left\{ \begin{matrix} {{x}^{2}}+5\ \text{ if }x<2 \\ {{x}^{3}}+1\text{ if }x\ge 2\text{ } \\ \end{matrix} \right.$ is $9$. b) The value of $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ where $f\left( x \right)=\left\{ \begin{matrix} {{x}^{2}}+5\ \text{ if }x<2 \\ {{x}^{3}}+1\text{ if }x\ge 2\text{ } \\ \end{matrix} \right.$ is $9$. c) The value of $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ where $f\left( x \right)=\left\{ \begin{matrix} {{x}^{2}}+5\ \text{ if }x<2 \\ {{x}^{3}}+1\text{ if }x\ge 2 \\ \end{matrix} \right.$ is $9$.
(a) Consider the provided function, $f\left( x \right)=\left\{ \begin{matrix} {{x}^{2}}+5\ \text{ if }x<2 \\ {{x}^{3}}+1\text{ if }x\ge 2\text{ } \\ \end{matrix} \right.$ It is required to find the value of $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$. 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. Because $x$ is less than 2, use the first line of the piecewise defined function's equation, $f\left( x \right)={{x}^{2}}+5$. Then, $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\left( {{x}^{2}}+5 \right)$ Use limit property, $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,\left( x \right)$ \begin{align} & \underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\left( {{x}^{2}}+5 \right) \\ & =\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,{{x}^{2}}+\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,5 \end{align} Now, use limit property, $\underset{x\to a}{\mathop{\lim }}\,c=c$ 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 }}\,5 \\ & ={{2}^{2}}+5 \\ & =4+5 \\ & =9 \end{align} Thus, $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=9$. (b) Consider the provided function, $f\left( x \right)=\left\{ \begin{matrix} {{x}^{2}}+5\ \text{ if }x<2 \\ {{x}^{3}}+1\text{ if }x\ge 2\text{ } \\ \end{matrix} \right.$ It is required to find the value of $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$. 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. Because $x$ is greater than 2, using the second line of the piecewise defined function's equation, $f\left( x \right)={{x}^{3}}+1$. Then, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{x}^{3}}+1 \right)$ 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}}+1 \right) \\ & =\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,{{x}^{3}}+\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,1 \end{align} Now, use limit property, $\underset{x\to a}{\mathop{\lim }}\,c=c$ 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 }}\,1 \\ & ={{2}^{3}}+1 \\ & =8+1 \\ & =9 \end{align} Thus, $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=9$. (c) Consider the provided function, $f\left( x \right)=\left\{ \begin{matrix} {{x}^{2}}+5\ \text{ if }x<2 \\ {{x}^{3}}+1\text{ if }x\ge 2\text{ } \\ \end{matrix} \right.$ From part (a), $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=9$ From part (b), $\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=9$ As the left-hand limit and right-hand limit are equal, so, $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ exist and is equal to $9$. Hence, the value of $\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ where $f\left( x \right)=\left\{ \begin{matrix} {{x}^{2}}+5\ \text{ if }x<2 \\ {{x}^{3}}+1\text{ if }x\ge 2 \\ \end{matrix} \right.$ is $9$.