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 796: 45


Limit does not exist

Work Step by Step

Consider our approach : $(x,y) \to (0,0)$ along $y=kx; m\ne -10$ Then, we get $\lim\limits_{x \to 0} \dfrac{x-kx}{x+kx}=\lim\limits_{x \to 0} \dfrac{x(1-k)}{x(1+k)}=\dfrac{1-k}{1+k}$ This shows that there are multiple limit values when the approach is different, so, the limit does not exist at the point (0,0) for the function $f(x,y)=\dfrac{x-y}{x+y}$.
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.