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: 66

Answer

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{x^2-y^2}{x^2+y^2}$ or, $=\lim\limits_{r \to 0} \cos (\dfrac{r^2 \cos^2 \theta- r^2 \sin ^2 \theta}{r^2})$ or, $=\lim\limits_{r \to 0} \cos (\cos^2 \theta-\sin ^2 \theta)$ or, $=\cos^2 \theta-\sin ^2 \theta$ We see that $\cos^2 \theta-\sin ^2 \theta$ is not a unique value. Thus, $ \lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } \dfrac{x^2-y^2}{x^2+y^2}$ =Limit does not exist.
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