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

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### 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 limx1.5(94x23+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, limxaf(x)=L if for every number ε>0 there is a number δ>0 such that if 0<|xa|<δ then |f(x)L|<ε”.

Difference of squared formula:(a2b2)=(a+b)(ab)

To guess: The number δ.

Let ε be a given positive integer. Here, f(x)=94x23+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 |(94x23+2x)6|<ε.

Consider |(94x23+2x)6|.

|(94x23+2x)6|=|(32)(2x)23+2x6|

Apply difference of squared formula,

|(94x23+2x)6|=|(3+2x)(32x)3+2x6|

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,

|(94x23+2x)6|=|(32x)6|=|2x3|=|2||x+1

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 