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: 25

Answer

please see step-by-step

Work Step by Step

$\displaystyle \lim_{x\rightarrow a}f(x)=L$ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that the following is valid: $($if $ 0 < |x-a| < \delta$ then $|f(x)-L| < \epsilon)$ ------------- $f(x)=x^{2}.$ Given any $\epsilon > 0$, we want to find a $\delta > 0$ such that $ 0 < |x-0| < \delta\ \ \Rightarrow\ \ |x^{2}-0| < \epsilon$. The conclusion of this implication... $|x^{2}-0| < \epsilon$ $| x^{2}| < \epsilon$ $ x^{2} < \epsilon$ $|x| < \sqrt{\epsilon}$ ... is equivalent to $|x| < \sqrt{\epsilon}$. So, for $\epsilon > 0$, we take $\delta=\sqrt{\epsilon}$, for which $ 0 < |x-0| < \delta \ \ \Rightarrow\ \ |x^{2}-0| < \epsilon$ By the definition, $\displaystyle \lim_{x\ \rightarrow 0}x^{2}=0$
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