## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises - Page 950: 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.

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