   Chapter 2.4, Problem 29E ### 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 → 2 ( x 2 − 4 x + 5 ) = 1

To determine

To prove: The limit of a function limx2(x24x+5) is equal to 1 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, a=2, L=1 and f(x)=x24x+5.

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

if 0<|x2|<δ then |(x24x+5)1|<ε.

Consider |(x24x+5)1|

|(x24x+5)1|=|x24x+51|=|x24x+4|=|x22x2x+4|

=|x(x2)2(x2)|=|(x2)(x2)|=|(x2)2|=|(x2)|2

Therefore, find a number δ such that if 0<|x2|<δ then |(x2)|2<ε

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