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: 28

Answer

$\arctan (-\dfrac{\pi}{4})$

Work Step by Step

Here, we have $\lim\limits_{(x,y,z) \to (\frac{-1}{4},\frac{\pi}{2}, 2)} \tan^{-1} (xyz)=\tan^{-1} [(\dfrac{-1}{4})(\dfrac{\pi}{2})(2)]$ Plug the limits $\tan^{-1}(-\dfrac{\pi}{4})=\arctan (-\dfrac{\pi}{4})$
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