Calculus: Graphical, Numerical, Algebraic

3rd Edition

ISBN: 9780133688399

Author: Finney, Ross L.

Publisher: Pearson/Prentice Hall

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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topic…

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Question

Chapter 2, Problem 26RE

(a)

To determine

To find: The value of limx→1−k(x) .

(a)

Expert Solution

Answer to Problem 26RE

The value of limx→1−k(x) is 1.5.

Explanation of Solution

Given information: The domain of the function is −1≤x≤3 . The graph is given below:

Calculus: Graphical, Numerical, Algebraic, Chapter 2, Problem 26RE , additional homework tip 1

Calculation:

From the given graph, it can be observed that as x approaches 1 from left then the function approaches to 1.5.

limx→1−k(x)=1.5

Therefore, the value of limx→1−k(x) is 1.5.

(b)

To determine

To find: The value of limx→1+k(x) .

(b)

Expert Solution

Answer to Problem 26RE

The value of limx→1+k(x) is 0.

Explanation of Solution

Given information: The domain of the function is −1≤x≤3 . The graph is given below:

Calculus: Graphical, Numerical, Algebraic, Chapter 2, Problem 26RE , additional homework tip 2

Calculation:

From the given graph, it can be observed that as x approaches 1 from right then the function approaches to 0.

limx→1+k(x)=0

Therefore, the value of limx→1+k(x) is 0.

(c)

To determine

To find: The value of k(1) .

(c)

Expert Solution

Answer to Problem 26RE

The value is k(1) is 0.

Explanation of Solution

Given information: The domain of the function is −1≤x≤3 . The graph is given below:

Calculus: Graphical, Numerical, Algebraic, Chapter 2, Problem 26RE , additional homework tip 3

Calculation:

From the graph, it can be observed that the value of the function y=f(x) at x=1 is 0.

k(1)=0

Therefore, the value is k(1) is 0.

(d)

To determine

To find: Whether k(x) is continuous at x=1 or not.

(d)

Expert Solution

Answer to Problem 26RE

The function k(x) is discontinuous at x=1 .

Explanation of Solution

Given information: The domain of the function is −1≤x≤3 . The graph is given below:

Calculus: Graphical, Numerical, Algebraic, Chapter 2, Problem 26RE , additional homework tip 4

Calculation:

From the graph, it can be observed that, at x=1 there is a jump on the graph of the function. The jump shows the discontinuity as the left hand limit is not equal to the right hand limit of the function.

limx→1−k(x)≠limx→1+k(x)

Therefore, the function is discontinuous at x=1 .

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 26RE

The point of discontinuity of the given function is x=1 .

Explanation of Solution

Given information: The domain of the function is −1≤x≤3 . The graph is given below:

Calculus: Graphical, Numerical, Algebraic, Chapter 2, Problem 26RE , additional homework tip 5

Calculation:

A function y=f(x) is said to be continuous if f(c)=limx→c+f(x)=limx→c−f(x) .

It is already known that at x=1 left hand limit is not equal to right hand limit of the function. So, function is discontinuous at x=1 .

Therefore, the point of discontinuity of the given function is x=1 .

(f)

To determine

To find: Whether any point of discontinuity can be removed, if so describe new function and if not explain.

(f)

Expert Solution

Answer to Problem 26RE

There is no point for removable discontinuity.

Explanation of Solution

Given information: The domain of the function is −1≤x≤3 . The graph is given below:

Calculus: Graphical, Numerical, Algebraic, Chapter 2, Problem 26RE , additional homework tip 6

Calculation:

The discontinuity can be removable if left hand limit and right hand limit are equal. The right hand limit and left hand limit at x=1 are not equal.

The discontinuity is not removable because the function cannot be made continuous by filling in a point at x=1 .

Therefore, there is no point for removable discontinuity.

Chapter 2, Problem 25RE

Chapter 2, Problem 27RE

Chapter 2 Solutions

Calculus: Graphical, Numerical, Algebraic

Ch. 2.1 - Prob. 1QR Ch. 2.1 - Prob. 2QR Ch. 2.1 - Prob. 3QR Ch. 2.1 - Prob. 4QR Ch. 2.1 - Prob. 5QR Ch. 2.1 - Prob. 6QR Ch. 2.1 - Prob. 7QR Ch. 2.1 - Prob. 8QR Ch. 2.1 - Prob. 9QR Ch. 2.1 - Prob. 10QR

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.1 - Prob. 47E Ch. 2.1 - Prob. 48E Ch. 2.1 - Prob. 49E Ch. 2.1 - Prob. 50E Ch. 2.1 - 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