Where is the function differentiable

[Lee33]

Homework Statement

Where is the function $f:E^2\to\mathbb{R}$ given by $f(x,y)=\begin{cases}\frac{xy}{|x|+|y|} & , \ \text{if} \ (x,y)\ne(0,0)\\ 0 & , \ \text{if} \ (x,y)=(0,0) \end{cases}$ differentiable?

Homework Equations

None

The Attempt at a Solution

The function is continuous so the partials exists, thus we have $\frac{\partial f}{\partial x} = \begin{cases} \frac{y(|x|+|y|)\pm xy|y|}{(|x|+|y|)^2}, & x>0\\ \frac{y(|x|+|y|)\pm xy|y|}{(|x|+|y|)^2}, & x<0. \end{cases}$

$\frac{\partial f}{\partial y} = \begin{cases} \frac{x(|x|+|y|)\pm xy|x|}{(|x|+|y|)^2}, & y>0\\ \frac{x(|x|+|y|)\pm xy|x|}{(|x|+|y|)^2}, & y<0. \end{cases}$

For $(x,y)\ne (0,0),$ the partials $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ exists and are continuous so $F$ is differentiable at any $(x,y)\ne (0,0).$

For $(x,y)=(0,0),$ I am having trouble showing why the partials are not continuous at $(0,0).$

 

[pasmith]

Homework Statement

Where is the function $f:E^2\to\mathbb{R}$ given by $f(x,y)=\begin{cases}\frac{xy}{|x|+|y|} & , \ \text{if} \ (x,y)\ne(0,0)\\ 0 & , \ \text{if} \ (x,y)=(0,0) \end{cases}$ differentiable?

Homework Equations

None

The Attempt at a Solution

The function is continuous so the partials exists, thus we have $\frac{\partial f}{\partial x} = \begin{cases} \frac{y(|x|+|y|)\pm xy|y|}{(|x|+|y|)^2}, & x>0\\ \frac{y(|x|+|y|)\pm xy|y|}{(|x|+|y|)^2}, & x<0. \end{cases}$

$\frac{\partial f}{\partial y} = \begin{cases} \frac{x(|x|+|y|)\pm xy|x|}{(|x|+|y|)^2}, & y>0\\ \frac{x(|x|+|y|)\pm xy|x|}{(|x|+|y|)^2}, & y<0. \end{cases}$

For $(x,y)\ne (0,0),$ the partials $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ exists and are continuous so $F$ is differentiable at any $(x,y)\ne (0,0).$

For $(x,y)=(0,0),$ I am having trouble showing why the partials are not continuous at $(0,0).$

Given that f(0,0) = 0 and f(0,y) = 0, you have
<br /> \left.\frac{\partial f}{\partial x}\right|_{(0,0)} = \lim_{x \to 0} \frac{f(x,0)}{x}<br />
but for y \neq 0 you have
<br /> \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{x \to 0} \frac{f(x,y)}{x}<br />
Does
<br /> \lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)}
exist, and if it does is it equal to \left.\dfrac{\partial f}{\partial x}\right|_{(0,0)}?

 

[Lee33]

Given that f(0,0) = 0 and f(0,y) = 0, you have
<br /> \left.\frac{\partial f}{\partial x}\right|_{(0,0)} = \lim_{x \to 0} \frac{f(x,0)}{x}<br />
but for y \neq 0 you have
<br /> \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{x \to 0} \frac{f(x,y)}{x}<br />
Does
<br /> \lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)}
exist, and if it does is it equal to \left.\dfrac{\partial f}{\partial x}\right|_{(0,0)}?

Hm, let's see if I understand your reasoning because I am a bit confused.

$\lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{y \to 0} \frac{f(x,y)}{x}$ which does equal $\left.\dfrac{\partial f}{\partial x}\right|_{(0,0)}$.

Am I correct on that?

 

[pasmith]

Hm, let's see if I understand your reasoning because I am a bit confused.

$\lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{y \to 0} \frac{f(x,y)}{x}$ which does equal $\left.\dfrac{\partial f}{\partial x}\right|_{(0,0)}$.

Am I correct on that?

No; you have
<br /> \lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{y \to 0} \left(\lim_{x \to 0} \frac{xy}{x(|x| + |y|)}\right)
so you need to take the x limit first. What is
<br /> \lim_{x \to 0} \frac{xy}{x(|x| + |y|)}
when y \neq 0?

 

[Lee33]

No; you have
<br /> \lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{y \to 0} \left(\lim_{x \to 0} \frac{xy}{x(|x| + |y|)}\right)
so you need to take the x limit first. What is
<br /> \lim_{x \to 0} \frac{xy}{x(|x| + |y|)}
when y \neq 0?

By l'hopital we have that \lim_{x \to 0} \frac{xy}{x(|x| + |y|)} = \frac{y}{|y|}

Is that correct? The absolute value is throwing me off if I am wrong.