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: 13



Work Step by Step

Here, $\lim\limits_{(x,y) \to (1,1)} \dfrac{(x-y)^2}{x-y}=\lim\limits_{(x,y) \to (1,1)}(x-y)$ Thus, $\lim\limits_{(x,y) \to (1,1)} \dfrac{(x-y)^2}{x-y}=1-1=0$
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.