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 797: 59


a) $\sin 2 \theta$ b) $\sin 2 \theta$

Work Step by Step

a) Substitute $y=mx$ in the function, so we have: $f(x,y)=\dfrac{2m}{1+m^2}$ Now, set $ m =\tan \theta$ Thus, $f(x,y)=\dfrac{2m}{1+m^2}=\dfrac{2 \tan \theta}{1+(\tan \theta)^2}$ or, $f(x,y)=2 \sin \theta \cos \theta=\sin 2 \theta$ b) $\lim\limits_{(x,y) \to (0,0) } \dfrac{2 xy}{x^2+y^2}=\dfrac{2m}{1+m^2}$ Need to consider part (a), set $ m =\tan \theta$ $\dfrac{2 \tan \theta}{1+(\tan \theta)^2}=\sin 2 \theta$
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