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 693: 65

Answer

$\dfrac{\pi}{2}$

Work Step by Step

$ \lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } arctan (\dfrac{|x|+|y|}{x^2+y^2})$ Use polar-coordinates: $x= r \cos \theta , y = r \sin \theta \\ r^2=x^2+y^2$ Now, $\lim\limits_{(x,y) \to (0,0) } arctan (\dfrac{|x|+|y|}{x^2+y^2})=\lim\limits_{r \to 0} arctan (\dfrac{ |r \cos \theta| +|r \sin \theta|}{r^2})$ and $\lim\limits_{r \to 0} arctan (\dfrac{ |\cos \theta| +|\sin \theta|}{r}) =tan^{-1} (\infty) $ and $\lim\limits_{(x,y) \to (0,0) } f(x,y)=\dfrac{\pi}{2}$
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