## Precalculus (6th Edition) Blitzer

a) The limit $\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $8$. b) The limit $\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $8$. c) The limit $\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)$ is 8.
(a) Consider the provided limit, $\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ This means that it is required to calculate the value of $f\left( x \right)$ when $x$ is close to 4 but not equal to 4. Since, $x$ is not equal to 4, using the first line of the piecewise defined function's equation $f\left( x \right)=\frac{{{x}^{2}}-16}{x-4}\ \text{if }\ x\ne 4$ Now use the property ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ solve the limit to get the result \begin{align} & \underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,\frac{{{x}^{2}}-16}{x-4} \\ & =\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,\frac{\left( x-4 \right)\left( x+4 \right)}{x-4} \end{align} Cancel out $x-4$ from both the numerator and denominator. \begin{align} & \underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,\frac{\left( x-4 \right)\left( x+4 \right)}{\left( x-4 \right)} \\ & =\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,\left( x+4 \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 {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,\left( x+4 \right) \\ & =\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,x+\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,4 \end{align} Again, use $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c=\text{constant}$ \begin{align} & \underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,x+\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,4 \\ & =4+4 \\ & =8 \end{align} Thus, the limit $\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $8$. (b) Consider the provided limit, $\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ This means that it is required to calculate the value of $f\left( x \right)$ when $x$ is close to 4 but greater than 4. Because $x$ is greater than 4, using the first line of the piecewise defined function's equation $f\left( x \right)=\frac{{{x}^{2}}-16}{x-4}\ \ \text{if }\ x\ne 4$ Use the property ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to solve the limit to get the result \begin{align} & \underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{2}}-16}{x-4} \\ & =\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,\frac{\left( x-4 \right)\left( x+4 \right)}{x-4} \end{align} Cancel out $x-4$ from both the numerator and denominator. \begin{align} & \underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,\frac{\left( x-4 \right)\left( x+4 \right)}{\left( x-4 \right)} \\ & =\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,\left( x+4 \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 {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,\left( x+4 \right) \\ & =\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,x+\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,4 \end{align} Now use $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c=\text{constant}$ \begin{align} & \underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,x+\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,4 \\ & =4+4 \\ & =8 \end{align} Thus, the limit $\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is equal to $8$. (c) Consider the provided limit, $\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)$ From part (a) $\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=8$ From part (b) $\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)=8$ 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 $\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)$ is 8.