Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.4 - The Precise Definition of a Limit - 2.4 Exercises - Page 114: 28


$\lim\limits_{x \to -6^+} \sqrt[8] {6+x} = 0$

Work Step by Step

Let $\epsilon \gt 0$ be given. Let $\delta = \epsilon^8$ Suppose that $\vert x-(-6) \vert \lt \delta$. Then: $\vert \sqrt[8] {6+x}-0 ~\vert \lt \sqrt[8] \delta = \sqrt[8] \epsilon^8 = \epsilon$ Therefore, $\lim\limits_{x \to -6^+} \sqrt[8] {6+x} = 0$
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