Chapter 12, Problem 10RE

a.

To determine

ToEvaluate:The limit $\underset{x\to c}{\lim }\sqrt{f\left(x\right)}$ using the given information.

  $\underset{x\to c}{\lim }f\left(x\right)=27$ and $\underset{x\to c}{\lim }g\left(x\right)=12$

Answer to Problem 10RE

  $\underset{x\to c}{\lim }\sqrt[3]{f\left(x\right)}=3$

Explanation of Solution

Given:

The limits $\underset{x\to c}{\lim }f\left(x\right)=27$ and $\underset{x\to c}{\lim }g\left(x\right)=12$

Concept Used:

For the functions $f\left(x\right)$ and $g\left(x\right)$ with the limits $\underset{x\to c}{\lim }f\left(x\right)=L$ and $\underset{x\to c}{\lim }g\left(x\right)=K$

  $\begin{array}{c}\underset{x\to c}{\lim }\sqrt[b]{f\left(x\right)}=\sqrt[b]{\underset{x\to c}{\lim }f\left(x\right)}\\ =\sqrt[b]{L}\end{array}$

Calculation:

Given the limits $\underset{x\to c}{\lim }f\left(x\right)=27$ and $\underset{x\to c}{\lim }g\left(x\right)=12$

Using the rule

For the functions $f\left(x\right)$ and $g\left(x\right)$ with the limits $\underset{x\to c}{\lim }f\left(x\right)=L$ and $\underset{x\to c}{\lim }g\left(x\right)=K$

  $\begin{array}{c}\underset{x\to c}{\lim }\sqrt[b]{f\left(x\right)}=\sqrt[b]{\underset{x\to c}{\lim }f\left(x\right)}\\ =\sqrt[b]{L}\end{array}$

  $\begin{array}{c}\underset{x\to c}{\lim }\sqrt[3]{f\left(x\right)}=\sqrt[3]{\underset{x\to c}{\lim }f\left(x\right)}\\ =\sqrt[3]{27}\\ =3\end{array}$

Therefore, $\underset{x\to c}{\lim }\sqrt[3]{f\left(x\right)}=3$

Conclusion:

  $\underset{x\to c}{\lim }\sqrt[3]{f\left(x\right)}=3$

b.

To determine

To Evaluate:The limit $\underset{x\to c}{\lim }\frac{f\left(x\right)}{18}$ using the given information.

  $\underset{x\to c}{\lim }f\left(x\right)=27$ and $\underset{x\to c}{\lim }g\left(x\right)=12$

Answer to Problem 10RE

  $\underset{x\to c}{\lim }\frac{f\left(x\right)}{18}=\frac{3}{2}$

Explanation of Solution

Given:

The limits $\underset{x\to c}{\lim }f\left(x\right)=27$ and $\underset{x\to c}{\lim }g\left(x\right)=12$

Concept Used:

For the functions $f\left(x\right)$ and $g\left(x\right)$ with the limits $\underset{x\to c}{\lim }f\left(x\right)=L$ and $\underset{x\to c}{\lim }g\left(x\right)=K$

  $\begin{array}{c}\underset{x\to c}{\lim }\frac{f\left(x\right)}{b}=\frac{\underset{x\to c}{\lim }f\left(x\right)}{b}\\ =\frac{L}{b}\end{array}$

Calculation:

Given the limits $\underset{x\to c}{\lim }f\left(x\right)=27$ and $\underset{x\to c}{\lim }g\left(x\right)=12$

Using the rule

For the functions $f\left(x\right)$ and $g\left(x\right)$ with the limits $\underset{x\to c}{\lim }f\left(x\right)=L$ and $\underset{x\to c}{\lim }g\left(x\right)=K$

  $\begin{array}{c}\underset{x\to c}{\lim }\frac{f\left(x\right)}{b}=\frac{\underset{x\to c}{\lim }f\left(x\right)}{b}\\ =\frac{L}{b}\end{array}$

  $\begin{array}{c}\underset{x\to c}{\lim }\frac{f\left(x\right)}{18}=\frac{\underset{x\to c}{\lim }f\left(x\right)}{18}\\ =\frac{27}{18}\\ =\frac{3}{2}\end{array}$

Therefore, $\underset{x\to c}{\lim }\frac{f\left(x\right)}{18}=\frac{3}{2}$

Conclusion:

  $\underset{x\to c}{\lim }\frac{f\left(x\right)}{18}=\frac{3}{2}$

c.

To determine

To Evaluate:The limit $\underset{x\to c}{\lim }\left[f\left(x\right)g\left(x\right)\right]$ using the given information.

  $\underset{x\to c}{\lim }f\left(x\right)=27$ and $\underset{x\to c}{\lim }g\left(x\right)=12$

Answer to Problem 10RE

  $\underset{x\to c}{\lim }\left[f\left(x\right)g\left(x\right)\right]=324$

Explanation of Solution

Given:

The limits $\underset{x\to c}{\lim }f\left(x\right)=27$ and $\underset{x\to c}{\lim }g\left(x\right)=12$

Concept Used:

For the functions $f\left(x\right)$ and $g\left(x\right)$ with the limits $\underset{x\to c}{\lim }f\left(x\right)=L$ and $\underset{x\to c}{\lim }g\left(x\right)=K$

  $\begin{array}{c}\underset{x\to c}{\lim }\left[f\left(x\right)g\left(x\right)\right]=\underset{x\to c}{\lim }f\left(x\right)\underset{x\to c}{\lim }g\left(x\right)\\ =LK\end{array}$

Calculation:

Given the limits $\underset{x\to c}{\lim }f\left(x\right)=27$ and $\underset{x\to c}{\lim }g\left(x\right)=12$

Using the rule

For the functions $f\left(x\right)$ and $g\left(x\right)$ with the limits $\underset{x\to c}{\lim }f\left(x\right)=L$ and $\underset{x\to c}{\lim }g\left(x\right)=K$

  $\begin{array}{c}\underset{x\to c}{\lim }\left[f\left(x\right)g\left(x\right)\right]=\underset{x\to c}{\lim }f\left(x\right)\underset{x\to c}{\lim }g\left(x\right)\\ =LK\end{array}$

  $\begin{array}{c}\underset{x\to c}{\lim }\left[f\left(x\right)g\left(x\right)\right]=\underset{x\to c}{\lim }f\left(x\right)\underset{x\to c}{\lim }g\left(x\right)\\ =\left(27\right)\left(12\right)\\ =324\end{array}$

Therefore, $\underset{x\to c}{\lim }\left[f\left(x\right)g\left(x\right)\right]=324$

Conclusion:

  $\underset{x\to c}{\lim }\left[f\left(x\right)g\left(x\right)\right]=324$

d.

To determine

To Evaluate:The limit $\underset{x\to c}{\lim }\left[f\left(x\right)-2g\left(x\right)\right]$ using the given information.

  $\underset{x\to c}{\lim }f\left(x\right)=4$ and $\underset{x\to c}{\lim }g\left(x\right)=5$

Answer to Problem 10RE

  $\underset{x\to c}{\lim }\left[f\left(x\right)-2g\left(x\right)\right]=3$

Explanation of Solution

Given:

The limits $\underset{x\to c}{\lim }f\left(x\right)=4$ and $\underset{x\to c}{\lim }g\left(x\right)=5$

Concept Used:

For the functions $f\left(x\right)$ and $g\left(x\right)$ with the limits $\underset{x\to c}{\lim }f\left(x\right)=L$ and $\underset{x\to c}{\lim }g\left(x\right)=K$

  $\begin{array}{c}\underset{x\to c}{\lim }\left[f\left(x\right)-bg\left(x\right)\right]=\underset{x\to c}{\lim }f\left(x\right)-b\underset{x\to c}{\lim }g\left(x\right)\\ =L-bK\end{array}$

Calculation:

Given the limits $\underset{x\to c}{\lim }f\left(x\right)=4$ and $\underset{x\to c}{\lim }g\left(x\right)=5$

Using the rule

For the functions $f\left(x\right)$ and $g\left(x\right)$ with the limits $\underset{x\to c}{\lim }f\left(x\right)=L$ and $\underset{x\to c}{\lim }g\left(x\right)=K$

  $\begin{array}{c}\underset{x\to c}{\lim }\left[f\left(x\right)-bg\left(x\right)\right]=\underset{x\to c}{\lim }f\left(x\right)-b\underset{x\to c}{\lim }g\left(x\right)\\ =L-bK\end{array}$

  $\begin{array}{c}\underset{x\to c}{\lim }\left[f\left(x\right)-2g\left(x\right)\right]=\underset{x\to c}{\lim }f\left(x\right)-2\underset{x\to c}{\lim }g\left(x\right)\\ =27-2\left(12\right)\\ =27-24\\ =3\end{array}$

Therefore, $\underset{x\to c}{\lim }\left[f\left(x\right)-2g\left(x\right)\right]=3$

Conclusion:

  $\underset{x\to c}{\lim }\left[f\left(x\right)-2g\left(x\right)\right]=3$