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: 20



Work Step by Step

Re-arrange the equation as follows: $\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(x-y)-1}=\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(\sqrt x+\sqrt {y+1})(\sqrt x-\sqrt {y+1})}$ and $\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(\sqrt x+\sqrt {y+1})(\sqrt x-\sqrt {y+1})}=\lim\limits_{(x,y) \to (4,3)} \dfrac{1}{(\sqrt x-\sqrt {y+1})}$ Plug the limits, then we get $\dfrac{1}{(\sqrt 4+\sqrt {3+1})}=\dfrac{1}{(\sqrt 4+\sqrt {4})}=\dfrac{1}{4}$
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