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: 16

Answer

$-\dfrac{1}{10}$

Work Step by Step

$\lim\limits_{(x,y) \to (2,-4)} \dfrac{(y+4)}{x(xy-y+4x-4)}=\lim\limits_{(x,y) \to (2,-4)} \dfrac{(y+4)}{x(x-1)(y+4)}=\dfrac{1}{2(-4-1)}$ Plug the limits, then we get $\dfrac{1}{2(-5)}=-\dfrac{1}{10}$
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