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: 53


The path limit along $x=0$ and $y=kx$ is equal to $0$.

Work Step by Step

We are given that $f(x,y)=\dfrac{2x^2y}{x^4+y^2}$ a) When $x=0$ Then, we have $\lim\limits_{(x,y) \to (0,0)}\dfrac{2x^2y}{x^4+y^2}=0$ b) Next, we have the second approach along $y=kx$ So, $\lim\limits_{(x,y) \to (0,0)}\dfrac{2x^2y}{x^4+y^2}=\dfrac{2x^2(kx)}{x^4+(kx)^2}=0$ This implies that the path limit along $x=0$ and $y=kx$ is equal to $0$.
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