## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.2 - Limits and Continuity in Higher Dimensions - Exercises 14.2 - Page 798: 74

#### Answer

$\delta$ exists when $\delta =0.02$

#### Work Step by Step

We have $f(x,y)=\dfrac{x^3+y^4}{x^2+y^2}$ and $f(0,0)=0; \epsilon =0.02$ The polar-coordinates are defined as: $x= r \cos \theta , y = r \sin \theta$ and $r^2=x^2+y^2$ Consider $\sqrt {x^2+y^2} \lt \delta$ This implies that $\dfrac{|x+y|}{x^2+1} \lt 0.01$ Since,$|\dfrac{(r \cos \theta)^3+(r \sin \theta)^4}{x^2+y^2}| =| r \cos ^3 \theta +r^2 \sin^4 \theta| \lt 0.02$ When $\theta = \dfrac{\pi}{2} \implies |r^2| \lt 0.02$ [let us say $|r^2| =\delta$] Thus, we can conclude that $\delta$ exists when $\delta =0.02$

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