University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.2 - Limits and Continuity in Higher Dimensions - Exercises - Page 691: 20



Work Step by Step

$\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(x-y)-1}=\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(\sqrt x+\sqrt {y+1})(\sqrt x-\sqrt {y+1})}$ This implies that$\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(\sqrt x+\sqrt {y+1})(\sqrt x-\sqrt {y+1})}=\lim\limits_{(x,y) \to (4,3)} \dfrac{1}{(\sqrt x-\sqrt {y+1})}$ Thus, we get $\dfrac{1}{(\sqrt 4-\sqrt {3+1})}=\dfrac{1}{4}$
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