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Math
CalculusSingle Variable Calculus: Early Transcendentals, Volume I8th Edition

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Chapter 2.4, Problem 22E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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Chapter
Section
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
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Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Prove the statement using the ε, δ definition of a limit.

lim x → − 1.5 9 − 4 x 2 3 + 2 x = 6

To determine

To prove: The limit of a function limx→−1.5(9−4x23+2x) is equal to 6 by using the ε,δ definition of a limit.

Explanation

Definition used:

“Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then, the limit of f(x) as x approaches a is L, limx→af(x)=L if for every number ε>0 there is a number δ>0 such that if 0<|x−a|<δ then |f(x)−L|<ε”.

Difference of squared formula:(a2−b2)=(a+b)(a−b)

To guess: The number δ.

Let ε be a given positive integer. Here, f(x)=9−4x23+2x, a=−1.5 and L=6.

By the definition of ε and δ, it is enough to find a number δ such that if 0<|x−(−1.5)|<δ, then |(9−4x23+2x)−6|<ε.

Consider |(9−4x23+2x)−6|.

|(9−4x23+2x)−6|=|(32)−(2x)23+2x−6|

Apply difference of squared formula,

|(9−4x23+2x)−6|=|(3+2x)(3−2x)3+2x−6|

Since the limit x approaches −1.5 but not equal to −1.5, cancel the common term 3+2x(≠0) from both the numerator and the denominator,

|(9−4x23+2x)−6|=|(3−2x)−6|=|−2x−3|=|−2||x+1

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