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: 30


$\ln 7$

Work Step by Step

Solve $\lim\limits_{(x,y,z) \to (2,-3,6)} \ln \sqrt {x^2+y^2+z^2}$ Now, $\lim\limits_{(x,y,z) \to (2,-3,6)} \ln \sqrt {x^2+y^2+z^2}=\ln \sqrt {(2)^2+(-3)^2+(6)^2}$ Thus, we get $\ln \sqrt{49}=\ln 7$
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