Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Practice Exercises - Page 864: 10



Work Step by Step

Consider $l=\lim\limits_{(x,y) \to (0,0)} \dfrac{2+y}{x+\cos y}$ or, $l=\lim\limits_{(x,y) \to (0,0)} \dfrac{2+y}{x+\cos y}=\dfrac{2+0}{0+\cos (0)}$ Hence, $l=\dfrac{2}{0+1}=2$
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