## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

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

Yes

#### 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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