Calculus 8th Edition

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

Chapter 1 - Functions and Limits - Review - Exercises: 42

Answer

please see step-by-step

Work Step by Step

By definition , $\displaystyle \lim_{x\rightarrow 0}\sqrt[3]{x}=0$ is true if for any $\epsilon > 0$, there is a $\delta > 0$ such that if $ 0 < |x-0| < \delta$, then $|\sqrt[3]{x}-0| < \epsilon$. analyzing the last expression, $|\sqrt[3]{x}-0|=|\sqrt[3]{x} | < \epsilon$ $|x|=|\sqrt[3]{x}|^{3} < \epsilon^{3}$ So, for any given $\epsilon,$ if we choose $\delta=\epsilon^{3}$,we will have $0 < |x-0| < \delta\ \ \Rightarrow\ \ |\sqrt[3]{x}-0| < \epsilon$. Thus, $\displaystyle \lim_{x\rightarrow 0}\sqrt[3]{x}=0$ by the definition of a limit.
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