Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Practice Exercises - Page 101: 32

Answer

See explanations.
1575131256

Work Step by Step

As $x\to0$, the function $f(x)=sin\frac{1}{x}$ will oscillate in a range of $[-1,1]$. Thus the limit $\lim_{x\to0}sin\frac{1}{x}$ does not exist, which means that the discontinuity at $x=0$ can not be extended.
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