Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.2 - Limits and Continuity - 14.2 Exercise: 15

Answer

limit does not exist.

Work Step by Step

We notice that if we directly substitute limits in the given function $f(x,y)=\frac{xy^{2}cosy}{x^{2}+y^{4}}$ Then $f(0,0)=\frac{0}{0}$ Therefore, we will calculate limit of the function in the following way. To evaluate the limit along x-axis; put $y=0$ $f(x,0)=\frac{xy^{2}cosy}{x^{2}+y^{4}}=\frac{x.0^{2}cos.0}{x^{2}+0^{4}}=0$ To evaluate limit along y-axis; put $x=0$ $f(0,y)=\frac{xy^{2}cosy}{x^{2}+y^{4}}=\frac{0*y^{2}cosy}{0^{2}+y^{4}}=0$ Approach (0,0) along another line, $x=y$ and $x\ne0$ $\lim\limits_{(x,y) \to (0,0)}f(y,y)=\lim\limits_{(x,y) \to (0,0)}f(y,y)\frac{y^{3}cosy}{y^{2}+y^{4}}=0$ Approach (0,0) along another line let us say $x=y^{2}$ $\lim\limits_{(x,y) \to (0,0)}f(y^{2},y)=\lim\limits_{(x,y) \to (0,0)}\frac{y^{2}.y^{2}cosy}{y^{4}+x^{4}}$ $=\lim\limits_{y \to 0}\frac{cosy}{2}$ $=\frac{cos0}{2}$ $=\frac{1}{2}$ For a limit to exist, all the paths must converge to the same point. Hence, the limit does not exist.
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