Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 2 - Section 2.4 - The Precise Definition of a Limit - 2.4 Exercises - Page 114: 26

Answer

$\lim\limits_{x \to 0} x^3 = 0$

Work Step by Step

Let $\epsilon \gt 0$ be given. Let $\delta = \sqrt[3]{e}$ If $\vert x-0 \vert \lt \delta$, then $\vert x^3 - 0\vert \lt \delta^3 = \epsilon$ Therefore, $\lim\limits_{x \to 0} x^3 = 0$
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