Chapter 1.1, Problem 56E

(a)

To determine

To calculate: The value of $\underset{x\to b}{\lim }f(x)+\underset{x\to b}{\lim }g(x)$

Answer to Problem 56E

The value is 4.

Explanation of Solution

Given information: The limit $\underset{x\to b}{\lim }f(x)=7$ and $\underset{x\to b}{\lim }g(x)=-3$

Calculation:

Apply the Sum rule and put the given limits:

  $\begin{array}{c}=\underset{x\to b}{\lim }f(x)+\underset{x\to b}{\lim }g(x)\\ =7+(-3)\\ =4\end{array}$

Therefore the required value of $\underset{x\to b}{\lim }f(x)+\underset{x\to b}{\lim }g(x)$ is 4.

(b)

To determine

To calculate: The value of $\underset{x\to b}{\lim }f(x)\cdot \underset{x\to b}{\lim }g(x)$

Answer to Problem 56E

The value is $-21$

Explanation of Solution

Given information: The limit $\underset{x\to b}{\lim }f(x)=7$ and $\underset{x\to b}{\lim }g(x)=-3$

Calculation:

The product rule can be used to set limits.

  $\underset{x\to b}{\lim }(f(x)\cdot g(x))=\underset{x\to b}{\lim }f(x)\cdot \underset{x\to b}{\lim }g(x)$

Since know that:

  $\begin{array}{c}\underset{x\to b}{\lim }f(x)=7\\ \underset{x\to b}{\lim }g(x)=-3\end{array}$

These values can be substituted to obtain the desired outcome:

  $\begin{array}{c}\underset{x\to b}{\lim }f(x)\cdot \underset{x\to b}{\lim }g(x)=7\cdot (-3)\\ =-21\end{array}$

Therefore the required value of $\underset{x\to b}{\lim }f(x)\cdot \underset{x\to b}{\lim }g(x)$ is $-21$ .

(c)

To determine

To calculate: The value of $\underset{x\to b}{\lim }4\cdot g(x)$

Answer to Problem 56E

The value is $-12$

Explanation of Solution

Given information: The limit $\underset{x\to b}{\lim }f(x)=7$ and $\underset{x\to b}{\lim }g(x)=-3$

Calculation:

The constant multiple rule must be applied for limitations in this exercise.

  $\underset{x\to b}{\lim }c\cdot g(x)=c\cdot \underset{x\to b}{\lim }g(x)$

Where $c$ is constant

In example $c=4$ so put it into the rule.

  $\underset{x\to b}{\lim }g(x)=-3$

Therefore,

  $\begin{array}{c}\underset{x\to b}{\lim }4\cdot g(x)=4\cdot \underset{x\to b}{\lim }g(x)\\ =4\cdot (-3)\\ =-12\end{array}$

Hence the required value of $\underset{x\to b}{\lim }f(x)\cdot \underset{x\to b}{\lim }g(x)$ is $-12$ .

(d)

To determine

To calculate: The value of $\frac{\underset{x\to b}{\lim }f(x)}{\underset{x\to b}{\lim }g(x)}$

Answer to Problem 56E

The value is $-\frac{7}{3}$

Explanation of Solution

Given information: The limit $\underset{x\to b}{\lim }f(x)=7$ and $\underset{x\to b}{\lim }g(x)=-3$

Calculation:

The quotient rule for the limits must be applied in this exercise and it states that:

  $\underset{x\to b}{\lim }\frac{f(x)}{g(x)}=\frac{\underset{x\to b}{\lim }f(x)}{\underset{x\to b}{\lim }g(x)}$

Suppose that,

  $\begin{array}{c}\underset{x\to b}{\lim }f(x)=7\\ \underset{x\to b}{\lim }g(x)=-3\end{array}$

Put this into the rule and get,

  $\frac{\underset{x\to b}{\lim }f(x)}{\underset{x\to b}{\lim }g(x)}=\frac{7}{-3}$

Therefore the required value of $\frac{\underset{x\to b}{\lim }f(x)}{\underset{x\to b}{\lim }g(x)}$ is $-\frac{7}{3}$