## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises: 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 $f(0,0)=\frac{0}{0}$ We get intermediate form. Therefore, we will calculate limit of function in following way. To evaluate limit along x-axis; put $y=0$ $f(x,0)=\frac{xy^{2}cosy}{x^{2}+y^{4}}=\frac{x.0^{2}cos.0}{x^{2}+0^{4}}=0$ 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 let us say $x=y$ and $x\ne0$ $\lim\limits_{(x,y) \to (0,0)}f(y,y)=\lim\limits_{(x,y) \to (0,0)}f(y,y)\frac{y^{3}cosy}{y^{2}+x^{4}}=0$ 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. 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.

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.