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: 18



Work Step by Step

Re-arrange the given equation as: $\lim\limits_{(x,y) \to (2,2)} \dfrac{(x+y)-4}{\sqrt {x+y}-2}=\lim\limits_{(x,y) \to (2,2)} \dfrac{(\sqrt{x+y})^2-(2)^2}{\sqrt {x+y}-2}$ Plug the limits, then we get $\lim\limits_{(x,y) \to (2,2)} \dfrac{(\sqrt{x+y})-2)(\sqrt{x+y}+2)}{\sqrt {x+y}-2}=\lim\limits_{(x,y) \to (2,2)}(\sqrt{x+y}+2)=2+2=4$
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