## Precalculus (6th Edition) Blitzer

a) The limit $\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $6$. b) The limit $\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $6$. c) The limit $\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)$ is 6.
(a) Consider the provided limit, $\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ This means that it is required to calculate the value of $f\left( x \right)$ when $x$ is close to 3 but less than 3. Because $x$ is less than 3, using the first line of the piecewise defined function's equation $f\left( x \right)=\frac{{{x}^{2}}-9}{x-3}\ \text{if }\ x\ne 3$ Use ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to solve the limit to get the result \begin{align} & \underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3} \\ & =\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\frac{\left( x-3 \right)\left( x+3 \right)}{x-3} \end{align} Cancel out $x-3$ from both the numerator and denominator. \begin{align} & \underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\frac{\left( x-3 \right)\left( x+3 \right)}{\left( x-3 \right)} \\ & =\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\left( x+3 \right) \end{align} 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 {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\left( x+3 \right) \\ & =\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,x+\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,3 \end{align} Use $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c=\text{constant}$ \begin{align} & \underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,x+\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,3 \\ & =3+3 \\ & =6 \end{align} Thus, the limit $\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $6$. (b) Consider the provided limit, $\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ This means that it is required to calculate the value of $f\left( x \right)$ when $x$ is close to 3 but not equal to 3. Since, $x$ not equal to 3, using the first line of the piecewise defined function's equation $f\left( x \right)=\frac{{{x}^{2}}-9}{x-3}\ \ \text{if }\ x\ne 3$ Now use ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to solve the limit to get the result \begin{align} & \underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3} \\ & =\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\frac{\left( x-3 \right)\left( x+3 \right)}{x-3} \end{align} Cancel out $x-3$ from both the numerator and denominator. \begin{align} & \underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\frac{\left( x-3 \right)\left( x+3 \right)}{\left( x-3 \right)} \\ & =\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\left( x+3 \right) \end{align} 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 {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\left( x+3 \right) \\ & =\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,x+\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,3 \end{align} Use property $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c=\text{constant}$ \begin{align} & \underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,x+\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,3 \\ & =3+3 \\ & =6 \end{align} Thus, the limit $\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $6$. (c) Consider the provided limit, $\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)$ From part (a) $\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=6$ From part (b) $\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=6$ 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 $\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)$ is 6.