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: 19



Work Step by Step

$\lim\limits_{(x,y) \to (2,0)} \dfrac{\sqrt {2x-y}-2}{(2x-y)-4}=\lim\limits_{(x,y) \to (2,0)}\dfrac{\sqrt {2x-y}-2}{(\sqrt {2x-y}+2)(\sqrt {2x-y}-2)}$ This implies that$\lim\limits_{(x,y) \to (2,0)}\dfrac{\sqrt {2x-y}-2}{(\sqrt {2x-y}+2)(\sqrt {2x-y}-2)}=\lim\limits_{(x,y) \to (2,0)}\dfrac{1}{(\sqrt {2x-y}+2)}$ Thus, we get $\dfrac{1}{(\sqrt {2(2)-0}+2)}=\dfrac{1}{4}$
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