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Math
CalculusSingle Variable Calculus: Early Transcendentals, Volume I8th Edition

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Chapter 2.3, Problem 24E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

Solutions

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Problem 1E
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Problem 3E
Problem 4E
Problem 5E
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Problem 7E
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Problem 13E
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Problem 15E
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Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
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Problem 27E
Problem 28E
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Problem 31E
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Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
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Problem 44E
Problem 45E
Problem 46E
Problem 47E
Problem 48E
Problem 49E
Problem 50E
Problem 51E
Problem 52E
Problem 53E
Problem 54E
Problem 55E
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Problem 57E
Problem 58E
Problem 59E
Problem 60E
Problem 61E
Problem 62E
Problem 63E
Problem 64E
Problem 65E
Problem 66E
BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Evaluate the limit, if it exists.

lim h → 0 ( 3 + h ) − 1 − 3 − 1 h

To determine

To evaluate: The limit of the function limh→0(3+h)−1−3−1h.

Explanation

Direct substitution property:

If f is a polynomial or a rational function and a is in the domain of f, then limx→af(x)=f(a).

Fact 1:

If f(x)=g(x) when x≠a, then limx→af(x)=limx→ag(x), provided the limit exist.

Evaluation:

Let f(h)=(3+h)−1−3−1h.

Rewrite the terms of the equation,

f(h)=1(3+h)−13h (1)

Note 1:

The direct substitution method is not applicable for the function f(h) since the function f(0) is in indeterminate form when ting h=0. That is,

f(0)=1(3+0)−130=13−130=00

Note 2:

The Quotient rule is not applicable for the function f(h) as the limit of the denominator is zero.

limh→0(h)=0 (by limit law 8)

Note 3:

“The limit may be infinite or it may be some finite value when both the numerator and the denominator approach to 0.”

Calculation:

By note 3, take the limit h approaches 0 but h≠0

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