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: 69

Answer

$|f(x,y) -f(0,0) | \lt \epsilon $

Work Step by Step

We have $f(x,y)=x^2+y^2$ and $f(0,0)=0$ Now, $|f(x,y) -f(0,0) | \lt \epsilon $ or, $|x^2+y^2-0| \lt 0.01 $ This implies that $\sqrt {x^2+y^2 } \lt 0.1$ Suppose $\delta =0.1$ So $\sqrt {x^2+y^2 } \lt \delta$ Thus, $|f(x,y) -f(0,0) | \lt \epsilon $
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