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 798: 64


Limit does not exist.

Work Step by Step

The polar-coordinates are defined as: $x= r \cos \theta , y = r \sin \theta$ and $r^2=x^2+y^2$ $ \lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } \dfrac{2x}{x^2+x+y^2}$ or, $=\lim\limits_{r \to 0} \dfrac{ 2 r \cos \theta}{r^2+r\cos \theta}$ or, $=\lim\limits_{r \to 0} \dfrac{ 2 \cos \theta}{r+\cos \theta}$ Suppose $ a \cos \theta \to 0$; $\lim\limits_{a \cos \theta \to 0} \dfrac{ 2 \cos \theta}{ cos \theta(a+1) }=\dfrac{2}{a+1}$ Then we can see that for this case, the limit is not a unique value. Thus, $ \lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } \dfrac{2x}{x^2+x+y^2}$ =Limit does not exist.
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