Chapter 1, Problem 26RE

(a)

To determine

To find: The limit $\underset{x\to 1^{-}}{\lim }k(x)$

Answer to Problem 26RE

The limit is 1.5

Explanation of Solution

Given information: The limit is $\underset{x\to 1^{-}}{\lim }k(x)$

Calculation:

The given limit is

  $\underset{x\to 1^{-}}{\lim }k(x)$

The graph shows that the value of $k(x)$ approaches 1.5 as the value of $x=1$ from the left side. Hence,

  $\underset{x\to 1^{-}}{\lim }k(x)=1.5$

Therefore the required limit is $\underset{x\to 1^{-}}{\lim }k(x)=1.5$ .

(b)

To determine

To find: The value of $\underset{x\to 1^{+}}{\lim }k(x)$

Answer to Problem 26RE

The value is $\underset{x\to 1^{+}}{\lim }k(x)=0$

Explanation of Solution

Given information: The limit is $\underset{x\to 1^{+}}{\lim }k(x)$

Calculation:

The given limit is $\underset{x\to 1^{+}}{\lim }k(x)$

See from the graph that when the right side of $x$ approaches 1, the value of $k$ approaches the value of 0. So

  $\underset{x\to 1^{+}}{\lim }k(x)=0$

Therefore the required limit is $\underset{x\to 1^{+}}{\lim }k(x)=0$ .

(c)

To determine

To find: The value of $k(1)$

Answer to Problem 26RE

The value is $k(1)=0$

Explanation of Solution

Given information: The function is $k(1)$

Calculation:

The given function is $k(1)$

From the graph that $k(x)$ is defined as $k(1)=0$ when $x$ equals 1. as opposed to point $(1,1.5)$ which is designated as a blank point, because point $(1,0)$ is colored inside.

  $k(1)=0$

Therefore the required value is $k(1)=0$ .

(d)

To determine

To find: determine the value of $k(x)$ is continuous $x=1$ or not.

Answer to Problem 26RE

  $k(x)$ is not continuous $x=1$ .

Explanation of Solution

Given information: From part a) $\underset{x\to 1^{-}}{\lim }k(x)=1.5$

And part b) $\underset{x\to 1^{+}}{\lim }k(x)=0$

Calculation:

Examine the graph and observe the jump discontinuity at $x=1$ . In terms of mathematics,

  $\underset{x\to 1^{-}}{\lim }k(x)\ne \underset{x\to 1^{+}}{\lim }k(x)$

Therefore, by definition $k(x)$ is not continuous $x=1$ .

(e)

To determine

To find: The points of discontinuity of. $k(x)$

Answer to Problem 26RE

  $x<-1$ or $x=1$ or $x>3$

Explanation of Solution

Given information: From part d) $\underset{x\to 1^{-}}{\lim }k(x)\ne \underset{x\to 1^{+}}{\lim }k(x)$

Calculation:

  $x=1$ is a point of discontinuity for $k$ , as seen in 26d. Additionally, there will be discontinuity points in regions where k is not defined, i.e., for $x<-1$ and $x>3$

Therefore, the points of discontinuity of. $k(x)$ : $x<-1$ or $x=1$ or $x>3$

(f)

To determine

To find: The removal points on discontinuity.

Answer to Problem 26RE

There are no removable points of discontinuity.

Explanation of Solution

Given information: The graph is:

  

Calculation:

The left and right sides must be equal for the point of discontinuity to be removed. However, at $x=1$ , there is a jump discontinuity, thus the left and right-hand bounds are not equal. Because cannot make $f(x)$ continuous by filling in a point at $x=1$ , the discontinuity is then irremovable.

Therefore, there are no removable points of discontinuity.