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: 18



Work Step by Step

$\lim\limits_{(x,y) \to (2,2)} \dfrac{(x+y)-4}{\sqrt {x+y}-2}=\lim\limits_{(x,y) \to (2,2)} \dfrac{(\sqrt{x+y})^2-(2)^2}{\sqrt {x+y}-2}$ This implies that$\lim\limits_{(x,y) \to (2,2)} \dfrac{(\sqrt{x+y})-2)(\sqrt{x+y}+2)}{\sqrt {x+y}-2}=\lim\limits_{(x,y) \to (2,2)}(\sqrt{x+y}+2)$ Thus, we get $2+2=4$
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