Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - Review - Exercises - Page 1022: 10

Answer

Does not exist.

Work Step by Step

Along the line $x=0$, the value of limit is zero. Along $x=0$: $\lim\limits_{(x,y) \to (0,0)}\frac{2xy}{x^2+2y^2}=0$ Along $y=x$: $\frac{2x^2}{x^2+2x^2}=\frac{2}{3}$ Since along two different lines the limit has two different values, it does not exist.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.