Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.7 The Precise Definition of a Limit - 1.7 Exercises: 2

Answer

$\delta=0.4$ (or any smaller positive number)

Work Step by Step

To reword the problem, given $\epsilon=0.5,$ we need to find a small interval around $3$, ($3-\delta,3+\delta)$ so that the function values of all x's from that interval fall within $(2-\epsilon,2+\epsilon).$ From the graph we read the following: When $x\in(2.6, 3.8)=(3-0.4,3+0.8)$, then $f(x)\in(1.5, 2.5)$, that is $|f(x)-2| < \epsilon=0.5$ The interval $(2.6, 3.8)$ around x=3, contains the interval $(2.6, 3.4) = (3-0.4,3+0.4)$, for which $x\in(3-0.4,3+0.4)\ \ \Rightarrow\ \ $ $0 < |x-3| < 0.4\ \ \Rightarrow\ \ |f(x)-2| < 0.5$. So, for the given $\epsilon=0.5$, we have found $\delta=0.4 > 0$ such that if $ 0 < |x-3| < \delta$, then $|f(x)-2| < \epsilon$. Note: we can take $\delta$ to be any number from $(0, 0.4)$
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