   Chapter 2.4, Problem 20E ### 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 → 10 ( 3 − 4 5 x ) = − 5

To determine

To prove: The limit of a function limx10(345x) is equal to −5 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”.

To guess: The number δ.

Let ε be a given positive integer. Here, f(x)=345x, a=10 and L=5.

By the definition of ε and δ, it is enough to find a number δ such that,

if 0<|x10|<δ, then |(345x)(5)|<ε.

Consider |(345x)(5)|.

|(345x)(5)|=|345x+5|=|845x|=|404x5|=|4(x10)5|

Take out the constant term,

|(345x)(5)|=|45||x10|=45|x10|

That is, if 0<|x10|<δ then |x10|<54ε

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