## Precalculus (6th Edition) Blitzer

a) The limit $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $0$. b) The limit $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to 0. c) The limit $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ is 0.
(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 $f\left( x \right)=1-x\ \ \text{if }\ x<1$ $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\left( 1-x \right)$ 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( 1-x \right) \\ & =\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,1-\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 }}\,1-\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,x \\ & =1-1 \\ & =0 \end{align} Thus, the limit $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $0$. 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 $f\left( x \right)={{x}^{2}}-1\ \ \text{if }\ x>1$ $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right)$ 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}}-1 \right) \\ & =\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{2}}-\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,1 \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 }}\,1 \\ & =1-1 \\ & =0 \end{align} Thus, the limit $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to 0. (c) Consider the provided limit, $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ From part (a) $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0$ From part (b) $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=0$ 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 0.