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Chapter 2 | Limits
Let ε >0. The first part of the definition begins “For every ε >0.” This means we must prove that whatever follows is true no matter what positive value of ε is chosen. By stating “Let ε >0,” we signal our intent to do so. Choose δ = ε 2 . The definition continues with “there exists a δ >0. ” The phrase “there exists” in a mathematical statement is always a signal for a scavenger hunt. In other words, we must go and find δ . So, where exactly did δ = ε /2 come from? There are two basic approaches to tracking down δ . One method is purely algebraic and the other is geometric. We begin by tackling the problem from an algebraic point of view. Since ultimately we want | (2 x +1)−3 | < ε , we begin by manipulating this expression: | (2 x +1)−3 | < ε is equivalent to |2 x −2| < ε , which in turn is equivalent to |2|| x −1| < ε . Last, this is equivalent to | x −1| < ε /2. Thus, it would seem that δ = ε /2 is appropriate. We may also find δ through geometric methods. Figure 2.40 demonstrates how this is done.
Figure 2.40 This graph shows how we find δ geometrically.
Assume 0< | x −1| < δ . When δ has been chosen, our goal is to show that if 0< | x −1| < δ , then | (2 x +1)−3 | < ε . To prove any statement of the form “If this, then that,” we begin by assuming “this” and trying to get “that.” Thus, | (2 x +1)−3 | = |2 x −2| property of absolute value
= | 2( x −1) | = |2|| x −1| =2| x −1| <2· δ
| 2 | =2
here’s where we use the assumption that 0 < | x −1| < δ
=2· ε
ε
here’s where we use our choice of δ = ε /2
2 =
Analysis
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