Chapter 12.2, Problem 5E

(a)..

To determine

To find: The limit of the given expression with the help of graph and then determine the other function that agrees with the given function at all but one point.

Answer to Problem 5E

The value of $\underset{x\to 1}{\lim }g\left(x\right)=2$ .

Explanation of Solution

Given:

The given graph is shown in Figure 1

  

Figure 1

The given limit is $\underset{x\to 1}{\lim }g\left(x\right)$ .

The given equation is $g\left(x\right)=\frac{x^3-x}{x-1}$ .

Calculation:

Consider the given equation is,

  $g\left(x\right)=\frac{x^3-x}{x-1}$ .

Consider $x=1$ , it is clear from the graph that $y=2$ .

For $g\left(x\right)=2$ the limit is calculated as,

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

Thus, the value of $\underset{x\to 1}{\lim }g\left(x\right)=2$ .

(b)..

To determine

To find: The limit of the given expression with the help of graph and then determine the other function that agrees with the given function at all but one point.

Answer to Problem 5E

The value of $\underset{x\to -1}{\lim }g\left(x\right)=0$ .

Explanation of Solution

Given:

The given limit is $\underset{x\to -1}{\lim }g\left(x\right)$ .

The given equation is $g\left(x\right)=\frac{x^3-x}{x-1}$ .

Calculation:

Consider the given equation is,

  $g\left(x\right)=\frac{x^3-x}{x-1}$ .

Consider $x=-1$ , it is clear from the graph that $g\left(x\right)=0$ .

For $\underset{x\to -1}{\lim }g\left(x\right)$ the limit is calculated as,

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

Thus, the value of $\underset{x\to -1}{\lim }g\left(x\right)=0$ .

(c)..

To determine

To find: The limit of the given expression with the help of graph and then determine the other function that agrees with the given function at all but one point.

Answer to Problem 5E

The value of $\underset{x\to -1}{\lim }g\left(x\right)=0$ .

Explanation of Solution

Given:

The given limit is $\underset{x\to 0}{\lim }g\left(x\right)$ .

The given equation is $g\left(x\right)=\frac{x^3-x}{x-1}$ .

Calculation:

Consider the given equation is,

  $g\left(x\right)=\frac{x^3-x}{x-1}$ .

Consider $x=0$ , it is clear from the graph that $g\left(x\right)=0$ .

For $\underset{x\to 0}{\lim }g\left(x\right)$ the limit is calculated as,

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

Thus, the value of $\underset{x\to -1}{\lim }g\left(x\right)=0$ .