Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises: 9

Answer

Limit does not exist.

Work Step by Step

We notice that if we directly substitute limits in the given function $f(x,y)=\frac{x^{4}-4y^{2}}{x^{2}+2y^{2}}$ Then $f(0,0)=\frac{0}{0}$ We get intermediate form. Therefore, we will calculate limit of function in following way. To evaluate limit along x-axis; put $y=0$ $f(x,0)=\frac{x^{4}-4y^{2}}{x^{2}+2y^{2}}=\frac{x^{4}-0}{x^{2}+0}=x^{2}$ To evaluate limit along y-axis; put $x=0$ $f(0,y)=\frac{x^{4}-4y^{2}}{x^{2}+2y^{2}}=-2$ Thus, both limits are different along different paths therefore, limit 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.