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

Answer

$0$

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