Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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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 2.1, Problem 49E

(a)

To determine

To find: The value of limx→0−F(x) with the help of graph.

(a)

Expert Solution

Answer to Problem 49E

The value of limx→0−F(x) is equal to 4 .

Explanation of Solution

Given information:

The graph of the function is:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 2.1, Problem 49E , additional homework tip 1

Calculation:

As shown in graph, the function y=F(x) approaches to 4 as x approaches to 0 from the left side.

So, the left hand limit of y=F(x) at x=0 is equal to 4 .

Therefore, the value of limx→0−F(x) is equal to 4 .

(b)

To determine

To find: The value of limx→0+F(x) with the help of graph.

(b)

Expert Solution

Answer to Problem 49E

The value of limx→0+F(x) is equal to −3 .

Explanation of Solution

Given information:

The graph of the function is:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 2.1, Problem 49E , additional homework tip 2

Calculation:

As shown in graph, the function y=F(x) approaches to −3 as x approaches to 0 from the right side.

So, the right hand limit of y=F(x) at x=0 is equal to −3 .

Therefore, the value of limx→0+F(x) is equal to −3 .

(c)

To determine

To find: The value of limx→0F(x) with the help of graph.

(c)

Expert Solution

Answer to Problem 49E

The value of limx→0F(x) does not exist as limx→0−F(x)≠limx→0+F(x) .

Explanation of Solution

Given information:

The graph of the function is:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 2.1, Problem 49E , additional homework tip 3

Calculation:

As calculated in part (a) and part (b), the value of limx→0−F(x) is equal to 4 and the value of limx→0+F(x) is equal to −3 .

It can be seen that the left hand limit of the function is not equal to the right hand limit of the function.

Therefore, the limx→0F(x) does not exist.

(d)

To determine

To find: The value of F(0) with the help of graph.

(d)

Expert Solution

Answer to Problem 49E

The value of F(0) is equal to 4 .

Explanation of Solution

Given information:

The graph of the function is:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 2.1, Problem 49E , additional homework tip 4

Calculation:

As shown in graph, the value of function y=F(x) is 4 at the point x=0 .

Therefore, the value of F(0) is equal to 4 .

Chapter 2.1, Problem 48E

Chapter 2.1, Problem 50E

Chapter 2 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

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 - Prob. 51E Ch. 2.1 - Prob. 52E Ch. 2.1 - Prob. 53E Ch. 2.1 - Prob. 54E Ch. 2.1 - Prob. 55E Ch. 2.1 - Prob. 56E Ch. 2.1 - Prob. 57E Ch. 2.1 - Prob. 58E Ch. 2.1 - Prob. 59E Ch. 2.1 - Prob. 60E Ch. 2.1 - Prob. 61E Ch. 2.1 - Prob. 62E Ch. 2.1 - Prob. 63E Ch. 2.1 - Prob. 64E Ch. 2.1 - Prob. 65E Ch. 2.1 - Prob. 66E Ch. 2.1 - Prob. 67E Ch. 2.1 - Prob. 68E Ch. 2.1 - Prob. 69E Ch. 2.1 - Prob. 70E Ch. 2.1 - Prob. 71E Ch. 2.1 - Prob. 72E Ch. 2.1 - Prob. 73E Ch. 2.1 - Prob. 74E Ch. 2.1 - Prob. 75E Ch. 2.1 - Prob. 76E Ch. 2.1 - Prob. 77E Ch. 2.1 - Prob. 78E Ch. 2.1 - Prob. 79E Ch. 2.1 - Prob. 80E Ch. 2.2 - Prob. 1QR Ch. 2.2 - Prob. 2QR Ch. 2.2 - Prob. 3QR Ch. 2.2 - Prob. 4QR Ch. 2.2 - Prob. 5QR Ch. 2.2 - Prob. 6QR Ch. 2.2 - Prob. 7QR Ch. 2.2 - Prob. 8QR Ch. 2.2 - Prob. 9QR Ch. 2.2 - Prob. 10QR Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E Ch. 2.2 - Prob. 11E Ch. 2.2 - 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Prob. 62E Ch. 2.2 - Prob. 63E Ch. 2.2 - Prob. 64E Ch. 2.2 - Prob. 65E Ch. 2.2 - Prob. 66E Ch. 2.2 - Prob. 67E Ch. 2.2 - Prob. 68E Ch. 2.2 - Prob. 69E Ch. 2.2 - Prob. 70E Ch. 2.2 - Prob. 71E Ch. 2.2 - Prob. 1QQ Ch. 2.2 - Prob. 2QQ Ch. 2.2 - Prob. 3QQ Ch. 2.2 - Prob. 4QQ Ch. 2.3 - Prob. 1QR Ch. 2.3 - Prob. 2QR Ch. 2.3 - Prob. 3QR Ch. 2.3 - Prob. 4QR Ch. 2.3 - Prob. 5QR Ch. 2.3 - Prob. 6QR Ch. 2.3 - Prob. 7QR Ch. 2.3 - Prob. 8QR Ch. 2.3 - Prob. 9QR Ch. 2.3 - Prob. 10QR Ch. 2.3 - Prob. 1E Ch. 2.3 - Prob. 2E Ch. 2.3 - Prob. 3E Ch. 2.3 - Prob. 4E Ch. 2.3 - Prob. 5E Ch. 2.3 - Prob. 6E Ch. 2.3 - Prob. 7E Ch. 2.3 - Prob. 8E Ch. 2.3 - Prob. 9E Ch. 2.3 - Prob. 10E Ch. 2.3 - Prob. 11E Ch. 2.3 - Prob. 12E Ch. 2.3 - Prob. 13E Ch. 2.3 - Prob. 14E Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 16E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - Prob. 21E Ch. 2.3 - Prob. 22E Ch. 2.3 - Prob. 23E Ch. 2.3 - Prob. 24E Ch. 2.3 - Prob. 25E Ch. 2.3 - Prob. 26E Ch. 2.3 - Prob. 27E Ch. 2.3 - Prob. 28E Ch. 2.3 - Prob. 29E Ch. 2.3 - Prob. 30E Ch. 2.3 - Prob. 31E Ch. 2.3 - Prob. 32E Ch. 2.3 - Prob. 33E Ch. 2.3 - Prob. 34E Ch. 2.3 - Prob. 35E Ch. 2.3 - Prob. 36E Ch. 2.3 - Prob. 37E Ch. 2.3 - Prob. 38E Ch. 2.3 - Prob. 39E Ch. 2.3 - Prob. 40E Ch. 2.3 - Prob. 41E Ch. 2.3 - Prob. 42E Ch. 2.3 - Prob. 43E Ch. 2.3 - Prob. 44E Ch. 2.3 - Prob. 45E Ch. 2.3 - Prob. 46E Ch. 2.3 - Prob. 47E Ch. 2.3 - Prob. 48E Ch. 2.3 - Prob. 49E Ch. 2.3 - Prob. 50E Ch. 2.3 - Prob. 51E Ch. 2.3 - Prob. 52E Ch. 2.3 - Prob. 53E Ch. 2.3 - Prob. 54E Ch. 2.3 - Prob. 55E Ch. 2.3 - Prob. 56E Ch. 2.3 - Prob. 57E Ch. 2.3 - Prob. 58E Ch. 2.3 - Prob. 59E Ch. 2.3 - Prob. 60E Ch. 2.3 - Prob. 61E Ch. 2.3 - Prob. 62E Ch. 2.3 - Prob. 63E Ch. 2.3 - Prob. 64E Ch. 2.4 - Prob. 1QR Ch. 2.4 - Prob. 2QR Ch. 2.4 - Prob. 3QR Ch. 2.4 - Prob. 4QR Ch. 2.4 - Prob. 5QR Ch. 2.4 - Prob. 6QR Ch. 2.4 - Prob. 7QR Ch. 2.4 - Prob. 8QR Ch. 2.4 - Prob. 9QR Ch. 2.4 - Prob. 10QR Ch. 2.4 - Prob. 1E Ch. 2.4 - Prob. 2E Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - Prob. 5E Ch. 2.4 - Prob. 6E Ch. 2.4 - Prob. 7E Ch. 2.4 - Prob. 8E Ch. 2.4 - Prob. 9E Ch. 2.4 - Prob. 10E Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - Prob. 14E Ch. 2.4 - Prob. 15E Ch. 2.4 - Prob. 16E Ch. 2.4 - Prob. 17E Ch. 2.4 - Prob. 18E Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - Prob. 21E Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - Prob. 24E Ch. 2.4 - Prob. 25E Ch. 2.4 - Prob. 26E Ch. 2.4 - Prob. 27E Ch. 2.4 - Prob. 28E Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Prob. 32E Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.4 - Prob. 37E Ch. 2.4 - Prob. 38E Ch. 2.4 - Prob. 39E Ch. 2.4 - Prob. 40E Ch. 2.4 - Prob. 41E Ch. 2.4 - Prob. 42E Ch. 2.4 - Prob. 43E Ch. 2.4 - Prob. 44E Ch. 2.4 - Prob. 45E Ch. 2.4 - Prob. 46E Ch. 2.4 - Prob. 47E Ch. 2.4 - Prob. 48E Ch. 2.4 - Prob. 49E Ch. 2.4 - Prob. 50E Ch. 2.4 - Prob. 51E Ch. 2.4 - Prob. 52E Ch. 2.4 - Prob. 53E Ch. 2.4 - Prob. 54E Ch. 2.4 - Prob. 55E Ch. 2.4 - Prob. 1QQ Ch. 2.4 - Prob. 2QQ Ch. 2.4 - Prob. 3QQ Ch. 2.4 - Prob. 4QQ Ch. 2 - Prob. 1RE Ch. 2 - Prob. 2RE Ch. 2 - Prob. 3RE Ch. 2 - Prob. 4RE Ch. 2 - Prob. 5RE Ch. 2 - Prob. 6RE Ch. 2 - Prob. 7RE Ch. 2 - Prob. 8RE Ch. 2 - Prob. 9RE Ch. 2 - Prob. 10RE Ch. 2 - Prob. 11RE Ch. 2 - Prob. 12RE Ch. 2 - Prob. 13RE Ch. 2 - Prob. 14RE Ch. 2 - Prob. 15RE Ch. 2 - Prob. 16RE Ch. 2 - Prob. 17RE Ch. 2 - Prob. 18RE Ch. 2 - Prob. 19RE Ch. 2 - Prob. 20RE Ch. 2 - Prob. 21RE Ch. 2 - Prob. 22RE Ch. 2 - Prob. 23RE Ch. 2 - Prob. 24RE Ch. 2 - Prob. 25RE Ch. 2 - Prob. 26RE Ch. 2 - Prob. 27RE Ch. 2 - Prob. 28RE Ch. 2 - Prob. 29RE Ch. 2 - Prob. 30RE Ch. 2 - Prob. 31RE Ch. 2 - Prob. 32RE Ch. 2 - Prob. 33RE Ch. 2 - Prob. 34RE Ch. 2 - Prob. 35RE Ch. 2 - Prob. 36RE Ch. 2 - Prob. 37RE Ch. 2 - Prob. 38RE Ch. 2 - Prob. 39RE Ch. 2 - Prob. 40RE Ch. 2 - Prob. 41RE Ch. 2 - Prob. 42RE Ch. 2 - Prob. 43RE Ch. 2 - Prob. 44RE Ch. 2 - Prob. 45RE Ch. 2 - Prob. 46RE Ch. 2 - Prob. 47RE Ch. 2 - Prob. 48RE Ch. 2 - Prob. 49RE Ch. 2 - Prob. 50RE Ch. 2 - Prob. 51RE Ch. 2 - Prob. 52RE Ch. 2 - Prob. 53RE Ch. 2 - Prob. 54RE Ch. 2 - Prob. 55RE

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The value of lim x → 0 − F ( x ) with the help of graph.

Concept explainers

Videos

To find: The value of limx→0−F(x) with the help of graph.

Answer to Problem 49E

Explanation of Solution

To find: The value of limx→0+F(x) with the help of graph.

Answer to Problem 49E

Explanation of Solution

To find: The value of limx→0F(x) with the help of graph.

Answer to Problem 49E

Explanation of Solution

To find: The value of F(0) with the help of graph.

Answer to Problem 49E

Explanation of Solution

Chapter 2 Solutions

Additional Math Textbook Solutions