Chapter 2.3, Problem 38E

(a)

To determine

To find:

The value of given limit at that every point .

Answer to Problem 38E

  $\underset{x\to 1^{+}}{\lim }F\left(x\right)=2,\underset{x\to 1^{-}}{\lim }F\left(x\right)=-2$

Explanation of Solution

Given:

The function $f$ is

  $F\left(x\right)=\frac{x^2-1}{\left|x-1\right|}$

Concept used:

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$ , except possibly at $x=a$

Then

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

Calculation:

The existence of the limit

  $F\left(x\right)=\frac{x^2-1}{\left|x-1\right|}$

Apply by the limit and using the value of the given limit

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

(b)

To determine

To find:

The value of given limit at that every point .

Answer to Problem 38E

  $\underset{x\to 1^{+}}{\lim }F\left(x\right)=2,\underset{x\to 1^{-}}{\lim }F\left(x\right)=-2$

Explanation of Solution

Given:

The function $f$ is

  $F\left(x\right)=\frac{x^2-1}{\left|x-1\right|}$

Concept used:

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$ , except possibly at $x=a$

Then

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

Calculation:

The existence of the limit

  $F\left(x\right)=\frac{x^2-1}{\left|x-1\right|}$

Apply by the limit and using the value of the given limit

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

No, the limit does not exist .

The two different left and right hand limit are different .

(c)

To determine

To sketch:

The graph of the given function .

To find:

The value of given limit at that every point .

Answer to Problem 38E

  $\underset{x\to 1^{+}}{\lim }F\left(x\right)=2,\underset{x\to 1^{-}}{\lim }F\left(x\right)=-2$

Explanation of Solution

Given:

The function $f$ is

  $F\left(x\right)=\frac{x^2-1}{\left|x-1\right|}$

Concept used:

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$ , except possibly at $x=a$

Then

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

Calculation:

The existence of the limit

  $F\left(x\right)=\frac{x^2-1}{\left|x-1\right|}$

Test one point in each of the region formed by the graph

If the point satisfies the function then shade the entire region to denote that every point in the region satisfies the function .

To draw the table

  $F\left(x\right)=\frac{x^2-1}{\left|x-1\right|}$

$x-axis$	$0$	$-1$	$1$	$-1.8$
$y-axis$	$-1$	$0$	$2$	$0.8$

To draw a graph