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: 18

Answer

limit does not exist.

Work Step by Step

Given: $\lim\limits_{(x,y) \to (0,0)}f(x,y)=\frac{xy^{4}}{ {x^{2}+y^{8}}}$ We notice that if we directly substitute limits in the given function $f(x,y)=\frac{xy^{4}}{ {x^{2}+y^{8}}}$ 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 $y=0$ $f(x,0)=\frac{x.0^{4}}{ {x^{2}+0^{8}}}=0$ To evaluate limit along y-axis; put $x=0$ $f(0,y)=\frac{0.y^{4}}{ {0^{2}+y^{8}}}=0$ Approach (0,0) along another line let us say $x=y^{4}$ $f(y^{4},y)=\frac{y^{4}.y^{4}}{ {y^{8}+y^{8}}}$ Thus, $\lim\limits_{(x,y) \to (0,0)}f(x,y)=\lim\limits_{(x,y) \to (0,0)}\frac{y^{4}.y^{4}}{ {y^{8}+y^{8}}}$ $\lim\limits_{y \to 0}\frac{y^{8}}{ 2y^{8}}$ $=\frac{1}{2}$ For a limit to exist, all the paths must converge to the same point.Since, function $f(x,y)$ has two different values along two different paths, it follows that limit does not exist, Hence, the limit does not exist.
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