Thomas' Calculus 13th Edition

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

Chapter 14: Partial Derivatives - Section 14.2 - Limits and Continuity in Higher Dimensions - Exercises 14.2 - Page 797: 58


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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.