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: 17

Answer

$2$

Work Step by Step

Here, we have $\lim\limits_{(x,y) \to (0,0)} \dfrac{(\sqrt x-\sqrt y)(\sqrt x+\sqrt y+2)}{\sqrt x-\sqrt y}=\lim\limits_{(x,y) \to (0,0)} (\sqrt x+\sqrt y+2)$ Plug the limits, then we get $(0+0+2)=2$
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