Explanation Given: The function is lim h → 0 g ( h ) = L . Formula used: The limit law is lim x → c x = c . The power law: If lim x → c f ( x ) and lim x → c g ( x ) both exists, then lim x → c ( f ( x ) ) n = ( lim x → c f ( x ) ) n Calculation: Part a: The function is lim h → 0 g ( h ) = L . We have to prove that lim h → 0 g ( a h ) = L . As you know that h → 0 so a h → 0 Let v = a h . Now, find the value of function. lim h → 0 g ( a h ) = lim h → 0 g ( w ) = L Hence, lim h → 0 g ( a h ) = L . Part b: The function is lim h → 1 g ( h ) = L

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Question

Chapter 2.3, Problem 41E

To determine

a. Explain why limh→0g(ah)=L for any constant a≠0.

b. Assume that limh→1g(h)=L, is it still necessarily true that limh→1g(ah)=L ?

c. Illustrate (a) and (b) with the function f(x)=x2.

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Chapter 2.3, Problem 40E

Chapter 2.3, Problem 42E

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Chapter 2 Solutions

Calculus - Standalone book

Ch. 2.1 - Prob. 1PQ Ch. 2.1 - Prob. 2PQ Ch. 2.1 - Prob. 3PQ Ch. 2.1 - Prob. 4PQ Ch. 2.1 - Prob. 5PQ 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. 2.2 - Prob. 1PQ Ch. 2.2 - Prob. 2PQ Ch. 2.2 - Prob. 3PQ Ch. 2.2 - Prob. 4PQ Ch. 2.2 - Prob. 5PQ Ch. 2.2 - Prob. 6PQ Ch. 2.2 - Prob. 7PQ 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 - Prob. 12E Ch. 2.2 - Prob. 13E Ch. 2.2 - 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Prob. 31E Ch. 2.7 - Prob. 32E Ch. 2.7 - Prob. 33E Ch. 2.7 - Prob. 34E Ch. 2.7 - Prob. 35E Ch. 2.7 - Prob. 36E Ch. 2.7 - Prob. 37E Ch. 2.7 - Prob. 38E Ch. 2.7 - Prob. 39E Ch. 2.7 - Prob. 40E Ch. 2.7 - Prob. 41E Ch. 2.7 - Prob. 42E Ch. 2.7 - Prob. 43E Ch. 2.7 - Prob. 44E Ch. 2.7 - Prob. 45E Ch. 2.8 - Prob. 1PQ Ch. 2.8 - Prob. 2PQ Ch. 2.8 - Prob. 3PQ Ch. 2.8 - Prob. 4PQ Ch. 2.8 - Prob. 5PQ Ch. 2.8 - Prob. 1E Ch. 2.8 - Prob. 2E Ch. 2.8 - Prob. 3E Ch. 2.8 - Prob. 4E Ch. 2.8 - Prob. 5E Ch. 2.8 - Prob. 6E Ch. 2.8 - Prob. 7E Ch. 2.8 - Prob. 8E Ch. 2.8 - Prob. 9E Ch. 2.8 - Prob. 10E Ch. 2.8 - Prob. 11E Ch. 2.8 - Prob. 12E Ch. 2.8 - Prob. 13E Ch. 2.8 - Prob. 14E Ch. 2.8 - Prob. 15E Ch. 2.8 - Prob. 16E Ch. 2.8 - Prob. 17E Ch. 2.8 - Prob. 18E Ch. 2.8 - Prob. 19E Ch. 2.8 - Prob. 20E Ch. 2.8 - Prob. 21E Ch. 2.8 - Prob. 22E Ch. 2.8 - Prob. 23E Ch. 2.8 - Prob. 24E Ch. 2.8 - Prob. 25E Ch. 2.8 - Prob. 26E Ch. 2.8 - Prob. 27E Ch. 2.8 - Prob. 28E Ch. 2.8 - Prob. 29E Ch. 2.8 - Prob. 30E Ch. 2.8 - Prob. 31E Ch. 2.8 - 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a. Explain why lim h → 0 g ( a h ) = L for any constant a ≠ 0 . b. Assume that lim h → 1 g ( h ) = L , is it still necessarily true that lim h → 1 g ( a h ) = L ? c. Illustrate (a) and (b) with the function f ( x ) = x 2 .

Solution Summary: The author explains how to determine why undersethto 0mathrmlimg(ah-)=L for any constant.

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a. Explain why limh→0g(ah)=L for any constant a≠0.

b. Assume that limh→1g(h)=L, is it still necessarily true that limh→1g(ah)=L ?

c. Illustrate (a) and (b) with the function f(x)=x2.

Want to see the full answer?

Chapter 2 Solutions

a. Explain why lim h → 0 g ( a h ) = L for any constant a ≠ 0 . b. Assume that lim h → 1 g ( h ) = L , is it still necessarily true that lim h → 1 g ( a h ) = L ? c. Illustrate (a) and (b) with the function f ( x ) = x 2 .

Solution Summary: The author explains how to determine why undersethto 0mathrmlimg(ah-)=L for any constant.

Concept explainers

Videos

a. Explain why limh→0g(ah)=L for any constant a≠0. b. Assume that limh→1g(h)=L, is it still necessarily true that limh→1g(ah)=L ? c. Illustrate (a) and (b) with the function f(x)=x2.

Want to see the full answer?

Chapter 2 Solutions

a. Explain why limh→0g(ah)=L for any constant a≠0.

b. Assume that limh→1g(h)=L, is it still necessarily true that limh→1g(ah)=L ?

c. Illustrate (a) and (b) with the function f(x)=x2.