Chapter 2.1, Problem 55E

(a)

To determine

To find: The value of $\underset{x\to 4}{\lim }\left(g\left(x\right)+3\right)$ .

Answer to Problem 55E

The value of $\underset{x\to 4}{\lim }\left(g\left(x\right)+3\right)$ is $6$ .

Explanation of Solution

Given information: The values $\underset{x\to 4}{\lim }f\left(x\right)=0$ and $\underset{x\to 4}{\lim }g\left(x\right)=3$ .

Calculation:

Use the sum property for limit.

  $\underset{x\to a}{\lim }\left(f\left(x\right)+g\left(x\right)\right)=\underset{x\to a}{\lim }\left(f\left(x\right)\right)+\underset{x\to a}{\lim }\left(g\left(x\right)\right)$

Simplify the given limit.

  $\begin{array}{c}\underset{x\to 4}{\lim }\left(g\left(x\right)+3\right)=\underset{x\to 4}{\lim }g\left(x\right)+\underset{x\to 4}{\lim }3\\ =\underset{x\to 4}{\lim }g\left(x\right)+3\end{array}$

Substitute $3$ for $\underset{x\to 4}{\lim }g\left(x\right)$ in the above expression.

  $\begin{array}{c}\underset{x\to 4}{\lim }\left(g\left(x\right)+3\right)=3+3\\ =6\end{array}$

Therefore, the value of $\underset{x\to 4}{\lim }\left(g\left(x\right)+3\right)$ is $6$ .

(b)

To determine

To find: The value of $\underset{x\to 4}{\lim }\left(xf\left(x\right)\right)$ .

Answer to Problem 55E

The value of $\underset{x\to 4}{\lim }\left(xf\left(x\right)\right)$ is $0$ .

Explanation of Solution

Given information: The values $\underset{x\to 4}{\lim }f\left(x\right)=0$ and $\underset{x\to 4}{\lim }g\left(x\right)=3$ .

Calculation:

Use the product rule for limit.

  $\underset{x\to a}{\lim }\left(f\left(x\right)\cdot g\left(x\right)\right)=\underset{x\to a}{\lim }\left(f\left(x\right)\right)\cdot \underset{x\to a}{\lim }\left(g\left(x\right)\right)$

Simplify the given limit.

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

Substitute $4$ for $x$ and $0$ for $\underset{x\to 4}{\lim }\left(f\left(x\right)\right)$ in the above expression.

  $\begin{array}{c}\underset{x\to 4}{\lim }\left(xf\left(x\right)\right)=4\times 0\\ =0\end{array}$

Therefore, the value of $\underset{x\to 4}{\lim }\left(xf\left(x\right)\right)$ is $0$ .

(c)

To determine

To find: The value of $\underset{x\to 4}{\lim }\left(g^2\left(x\right)\right)$ .

Answer to Problem 55E

The value of $\underset{x\to 4}{\lim }\left(g^2\left(x\right)\right)$ is $9$ .

Explanation of Solution

Given information: The values $\underset{x\to 4}{\lim }f\left(x\right)=0$ and $\underset{x\to 4}{\lim }g\left(x\right)=3$ .

Calculation:

Use the power rule for limit.

  $\underset{x\to a}{\lim }\left({\left(f\left(x\right)\right)}^c\right)={\left(\underset{x\to a}{\lim }\left(f\left(x\right)\right)\right)}^c$

Simplify the given limit.

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

Substitute $3$ for $\underset{x\to 4}{\lim }\left(g\left(x\right)\right)$ in the above expression.

  $\begin{array}{c}\underset{x\to 4}{\lim }\left(g^2\left(x\right)\right)={\left(3\right)}^2\\ =9\end{array}$

Therefore, the value of $\underset{x\to 4}{\lim }\left(g^2\left(x\right)\right)$ is $9$ .

(d)

To determine

To find: The value of $\underset{x\to 4}{\lim }\left(\frac{g\left(x\right)}{f\left(x\right)-1}\right)$ .

Answer to Problem 55E

The value of $\underset{x\to 4}{\lim }\left(\frac{g\left(x\right)}{f\left(x\right)-1}\right)$ is $-3$ .

Explanation of Solution

Given information: The values $\underset{x\to 4}{\lim }f\left(x\right)=0$ and $\underset{x\to 4}{\lim }g\left(x\right)=3$ .

Calculation:

Use the quotient rule for limit.

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

Simplify the given limit.

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

Substitute $0$ for $\underset{x\to 4}{\lim }\left(f\left(x\right)\right)$ and $3$ for $\underset{x\to 4}{\lim }\left(g\left(x\right)\right)$ in the above expression.

  $\begin{array}{c}\underset{x\to 4}{\lim }\left(\frac{g\left(x\right)}{f\left(x\right)-1}\right)=\frac{3}{0-1}\\ =\frac{3}{-1}\\ =-3\end{array}$

Therefore, the value of $\underset{x\to 4}{\lim }\left(\frac{g\left(x\right)}{f\left(x\right)-1}\right)$ is $-3$ .