Chapter 11.3, Problem 35E

Solution

a)

To find: the limit $\underset{x\to 4}{\lim }\left(g\left(x\right)+2\right)$ assuming that $\underset{x\to 4}{\lim }f\left(x\right)=-1$ and $\underset{x\to 4}{\lim }g\left(x\right)=4$ .

  $\underset{x\to 4}{\lim }\left(g\left(x\right)+2\right)=6$

Given information:

  $\underset{x\to 4}{\lim }f\left(x\right)=-1$ and $\underset{x\to 4}{\lim }g\left(x\right)=4$ .

Formula used:

Addition Rule: $\underset{x\to c}{\lim }\left(f\left(x\right)+g\left(x\right)\right)=\underset{x\to c}{\lim }f\left(x\right)+\underset{x\to c}{\lim }g\left(x\right)$

Calculation:

Substitute $g\left(x\right)=4$ and $2$ in the addition rule formula.

  $\begin{array}{c}\underset{x\to 4}{\lim }\left(g\left(x\right)+2\right)=\underset{x\to 4}{\lim }\left(g\left(x\right)\right)+\underset{x\to 4}{\lim }\left(2\right)\\ =4+2\\ =6\end{array}$

Hence, $\underset{x\to 4}{\lim }\left(g\left(x\right)+2\right)=6$ .

b)

To find: the limit $\underset{x\to 4}{\lim }xf\left(x\right)$ assuming that $\underset{x\to 4}{\lim }f\left(x\right)=-1$ and $\underset{x\to 4}{\lim }g\left(x\right)=4$ .

  $\underset{x\to 4}{\lim }xf\left(x\right)=-4$

Given information:

  $\underset{x\to 4}{\lim }f\left(x\right)=-1$ and $\underset{x\to 4}{\lim }g\left(x\right)=4$ .

Formula used:

Product rule: $\underset{x\to c}{\lim }\left(f\left(x\right)\cdot g\left(x\right)\right)=\underset{x\to c}{\lim }f\left(x\right)\cdot \underset{x\to c}{\lim }g\left(x\right)$

Calculation:

  $xf\left(x\right)$

  $\underset{x\to 4}{\lim }xf\left(x\right)=\left(x\right)\underset{x\to 4}{\lim }f\left(x\right)$

Substitute $x$ and $f\left(x\right)$ in the product rule formula.

  $\underset{x\to 4}{\lim }xf\left(x\right)=\underset{x\to 4}{\lim }\left(x\right)\cdot \underset{x\to 4}{\lim }f\left(x\right)$

Substitute $-1$ for $\underset{x\to 4}{\lim }f\left(x\right)$ .

  $\begin{array}{l}\underset{x\to 4}{\lim }xf\left(x\right)=\underset{x\to 4}{\lim }\left(x\right)\cdot \left(-1\right)\\ =\left(4\right)\cdot \left(-1\right)\\ =-4\end{array}$

Hence, $\underset{x\to 4}{\lim }xf\left(x\right)=-4$ .

c)

To find: the limit $\underset{x\to 4}{\lim }{g}^2\left(x\right)$ assuming that $\underset{x\to 4}{\lim }f\left(x\right)=-1$ and $\underset{x\to 4}{\lim }g\left(x\right)=4$ .

  $\underset{x\to 4}{\lim }{g}^2\left(x\right)=16$

Given information:

  $\underset{x\to 4}{\lim }f\left(x\right)=-1$ and $\underset{x\to 4}{\lim }g\left(x\right)=4$ .

Formula used:

Power Rule: $\underset{x\to c}{\lim }{\left(f\left(x\right)\right)}^n={\left(\underset{x\to c}{\lim }f\left(x\right)\right)}^n$

Calculation:

Apply the power rule.

  $\underset{x\to 4}{\lim }{g}^2\left(x\right)={\left(\underset{x\to 4}{\lim }\left(g\left(x\right)\right)\right)}^2$

Substitute $\underset{x\to 4}{\lim }g\left(x\right)=4$ in $\underset{x\to 4}{\lim }\left(g\left(x\right)\right)$ .

  $\begin{array}{c}\underset{x\to 4}{\lim }{\left(g\left(x\right)\right)}^2=\underset{x\to 4}{\lim }\left(4^2\right)\\ =16\end{array}$

Hence, $\underset{x\to 4}{\lim }{g}^2\left(x\right)=16$ .

d)

To find: the limit $\underset{x\to 4}{\lim }\frac{g\left(x\right)}{f\left(x\right)-1}$ assuming that $\underset{x\to 4}{\lim }f\left(x\right)=-1$ and $\underset{x\to 4}{\lim }g\left(x\right)=4$ .

  $\underset{x\to 4}{\lim }\frac{g\left(x\right)}{f\left(x\right)-1}=-2$

Given information:

  $\underset{x\to 4}{\lim }f\left(x\right)=-1$ and $\underset{x\to 4}{\lim }g\left(x\right)=4$ .

Formula used:

Quotient Rule: $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to c}{\lim }f\left(x\right)}{\underset{x\to c}{\lim }g\left(x\right)}$

Calculation:

Apply the quotient rule by substituting $\underset{x\to 4}{\lim }g\left(x\right)=4$ and $\underset{x\to 4}{\lim }f\left(x\right)=-1$ .

  $\underset{x\to 4}{\lim }\frac{g\left(x\right)}{f\left(x\right)-1}=\frac{\underset{x\to 4}{\lim }g\left(x\right)}{\underset{x\to 4}{\lim }f\left(x\right)-1}$

Substitute $\underset{x\to 4}{\lim }f\left(x\right)=-1$ in $\frac{\underset{x\to 4}{\lim }g\left(x\right)}{\underset{x\to 4}{\lim }f\left(x\right)-1}$ .

  $\begin{array}{c}\frac{\underset{x\to 4}{\lim }g\left(x\right)}{\underset{x\to 4}{\lim }f\left(x\right)-1}=\frac{4}{-1-1}\\ =\frac{4}{-2}\\ =-2\end{array}$

Hence, $\underset{x\to 4}{\lim }\frac{g\left(x\right)}{f\left(x\right)-1}=-2$ .