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: 11


limits does not exist

Work Step by Step

We notice that if we directly substitute limits in the given function $f(x,y)=\frac{y^{2}sin^{2}x}{x^{4}+y^{4}}$ 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 $y=0$ $f(x,0)=\frac{y^{2}sin^{2}x}{x^{4}+y^{4}}=\frac{(0)^{2}sin^{2}x}{x^{4}+0}=0$ To evaluate limit along y-axis; put $x=0$ $f(0,y)=\frac{y^{2}sin^{2}0}{(0)^{4}+y^{4}}=0$ Although the obtained identical limits along the axes do not show that the given limit is 0. Then, approach (0,0) along another line let us say $y=x$ and $x\ne0$ $f(x,x)=\frac{x^{2}sin^{2}x}{x^{4}+x^{4}}=\frac{sin^{2}x}{2x^{4}}$ $f(x,x)=\frac{1}{2}(\frac{sinx}{x})^{2}$ Now, $\lim\limits_{(x,y) \to (0,0)}f(x,x)=\lim\limits_{(x,y) \to (0,0)}\frac{1}{2}(\frac{sinx}{x})^{2}$ $\lim\limits_{(x,y) \to (0,0)}f(x,x)=\frac{1}{2}$ For a limit to exist, all the paths must converge to the same point. Hence, both the limits are different and follow different paths, therefore, the limits does not exist.
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