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: 19



Work Step by Step

Re-arrange the given equation as: $\lim\limits_{(x,y) \to (2,0)} \dfrac{\sqrt {2x-y}-2}{(2x-y)-4}=\lim\limits_{(x,y) \to (2,0)}\dfrac{\sqrt {2x-y}-2}{(\sqrt {2x-y}+2)(\sqrt {2x-y}-2)}$ This implies that$\lim\limits_{(x,y) \to (2,0)}\dfrac{\sqrt {2x-y}-2}{(\sqrt {2x-y}+2)(\sqrt {2x-y}-2)}=\lim\limits_{(x,y) \to (2,0)}\dfrac{1}{(\sqrt {2x-y}+2)}$ Plug the limits, then we get $\dfrac{1}{(\sqrt {2(2)-0}+2)}=\dfrac{1}{(\sqrt {4}+2)}=\dfrac{1}{4}$
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