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 690: 3

Answer

$2 \sqrt 6$

Work Step by Step

Solve the limit $\lim\limits_{(x,y) \to (3,4)} \sqrt {x^2+y^2-1}$ $\lim\limits_{(x,y) \to (3,4)} \sqrt {x^2+y^2-1}=\sqrt {3^2+4^2-1}$ or, $\sqrt {24}=2 \sqrt 6$
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