## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.2 - Limits and Continuity in Higher Dimensions - Exercises 14.2 - Page 797: 57

#### Answer

Yes ; $\lim\limits_{(x,y) \to (0,0) } y \sin \dfrac{1}{x} =0$

#### Work Step by Step

Given : $|\sin \dfrac{1}{x}| \leq 1, |y \sin \dfrac{1}{x}| \leq y$ Re-arrange as: $0 \leq |y \sin \dfrac{1}{x}| \leq |y|$ Since, by the Squeeze Theorem $\lim\limits_{(x,y) \to (0,0) } |y \sin \dfrac{1}{x}| =0$ and $\lim\limits_{(x,y) \to (0,0) } y \sin (1/x)=0$ This implies that the limit for $\lim\limits_{(x,y) \to (0,0) } y \sin \dfrac{1}{x} =0$ by the Squeeze Theorem. So, our answer is Yes.

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