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: 71


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

Work Step by Step

We have $f(x,y)=\dfrac{x+y}{x^2+1}$ and $f(0,0)=0$ Now, $|f(x,y) -f(0,0) | \lt \epsilon \implies |\dfrac{x+y}{x^2+1}-0| \lt 0.01 $ This implies that $\dfrac{|x+y|}{x^2+1} \lt 0.01$ Since,$|x+y| \leq 2 \sqrt {x^2+y^2 }; x^2+1 \geq 1$ So, $\dfrac{|x+y|}{x^2+1} \leq 2 \sqrt {x^2+y^2}$ This implies that $2 \sqrt {x^2+y^2} \lt 0.01 \implies \sqrt {x^2+y^2} \lt 0.005$ Suppose $\delta =0.005 \implies \sqrt {x^2+y^2 } \lt \delta$ Thus, $|f(x,y) -f(0,0) | \lt \epsilon $
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