Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.2 Exercises - Page 923: 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 the limit of the function in the 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 $x=0$ $f(0,y)=\frac{x^{4}-4y^{2}}{x^{2}+2y^{2}}=-2$ Hence, both the limits are different along different paths, therefore, the limit does not exist.
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