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 - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 59: 81

Answer

a. 0, b. see explanations.

Work Step by Step

(a) See graph, $\lim\limits_{x\to 0}g(x)=0$ (b) $\lim\limits_{x\to 0}g(x)=\lim\limits_{x\to 0}(x\cdot sin(\frac{1}{x}))=\lim\limits_{x\to 0}x\cdot \lim\limits_{x\to 0}sin(\frac{1}{x})$. We know that $ \lim\limits_{x\to 0}sin(\frac{1}{x})$ does not exit but the value oscillates within $[-1,1]$. Thus $\lim\limits_{x\to 0}g(x)=\lim\limits_{x\to 0}x\cdot \lim\limits_{x\to 0}sin(\frac{1}{x})=0\cdot \lim\limits_{x\to 0}sin(\frac{1}{x})=0$
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