University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.2 - Limits and Continuity in Higher Dimensions - Exercises - Page 692: 57



Work Step by Step

Consider $ |\sin \dfrac{1}{x}| \leq 1, |y \sin \dfrac{1}{x}| \leq y$ $\implies 0 \leq |y \sin \dfrac{1}{x}| \leq |y|$ Since, by the Squeeze Theorem $\lim\limits_{(x,y) \to (0,0) } |y \sin \dfrac{1}{x}| =0 $ and $\lim\limits_{(x,y) \to (0,0) } y \sin (1/x)=0$ Yes, by the Squeeze Theorem the limit for $ \lim\limits_{(x,y) \to (0,0) } y \sin \dfrac{1}{x} =0$ .
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