University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 16


$-2.01\lt x \lt -1.99$ $\delta = 0.01$

Work Step by Step

We solve the inequality: $|2x-2+6|\lt 0.02$ $-0.02\lt|2x+4|\lt 0.02$ $-0.01\lt|x+2|\lt 0.01$ $-2.01\lt x \lt -1.99$ Finding $\delta$: $|x--2|\lt \delta$ $-\delta-2\lt x \lt \delta-2$ $\delta=0.01$
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