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: 60

Answer

$0$

Work Step by Step

$ \lim\limits_{(x,y) \to (0,0) } |f(x,y)|=\lim\limits_{(x,y) \to (0,0) }|xy \times \dfrac{x^2-y^2}{x^2+y^2}|$ or, $=\lim\limits_{(x,y) \to (0,0) } \dfrac{|xy|}{x^2+y^2}|x^2-y^2|$ or, $\lim\limits_{(x,y) \to (0,0) } \dfrac{|xy|}{x^2+y^2}|x^2-y^2| \leq \lim\limits_{(x,y) \to (0,0) } \dfrac{1}{2} \times |x^2-y^2|$ or, $=0$ So, by the Sandwich Theorem$ \lim\limits_{(x,y) \to (0,0) } |f(x,y)|=0$
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