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, $\sqrt {24}=2 \sqrt 6$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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
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