## Precalculus (6th Edition) Blitzer

a) The limit $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to 3. b) The limit $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to 3. c) The limit $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ is 3.
(a) Consider the provided limit, $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ This means that it is required to calculate the value of $f\left( x \right)$ when $x$ is close to 1 but less than 1. Since, $x$ is less than 1, using the first line of the piecewise defined function's equation $f\left( x \right)=4-x\ \ \text{if }\ x<1$ $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\left( 4-x \right)$ 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 {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\left( 4-x \right) \\ & =\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,4-\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,x \end{align} Use the property $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c=\text{constant}$ \begin{align} & \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,4-\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,x \\ & =4-1 \\ & =3 \end{align} Thus, the limit $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to 3. (b) Consider the provided limit, $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ This means that it is required to calculate the value of $f\left( x \right)$ when $x$ is close to 1 but greater than 1. Since, $x$ is greater than 1, using the third line of the piecewise defined function's equation $f\left( x \right)={{x}^{2}}+2\ \ \text{if }\ x>1$ $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}+2 \right)$ 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 {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}+2 \right) \\ & =\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{2}}+\underset{x\to {{1}^{+}}}{\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 {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{2}}+\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,2 \\ & ={{1}^{2}}+2 \\ & =1+2 \\ & =3 \end{align} Thus, the limit $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to 3. (c) Consider the provided limit, $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ From part (a) $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=3$ From part (b) $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=3$ As $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ Thus, the limit $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ is 3.