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For an exercise in my analysis course, I have to show that the function $$\newcommand{\sgn}{\operatorname{sgn}}f: (x,y) \mapsto \begin{cases} \frac{(x \sin y)^2}{|x|+|y|},&(x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$$ is $C^1$. I tried to proceed by showing that the partial derivatives exist and are continuous.

I have shown that $$\frac{\partial f(x,y)}{\partial x} = \begin{cases} \sin^2 y\frac{2x(|x|+|y|)-x^2\sgn(x)}{(|x|+|y|)^2}, & (x,y)\neq (0,0)\\ 0, & (x,y) = (0,0)\end{cases} $$

and $$\frac{\partial f(x,y)}{\partial y} = \begin{cases} x^2\frac{\sin^2y\cdot \cos y(|x|+|y|)-\sin^2(y)\sgn(y)}{(|x|+|y|)^2}, &(x,y) \neq (0,0) \\ 0, &(x,y) = (0,0)\end{cases}$$

Now, I tried to show that those derivatives are continuous. To show that they are continuous in $(x,y) \neq (0,0)$ is easy since we can apply limit laws since de denominator is not $0$ close to $(x,y)$.

However, To show that they are continuous in $(0,0)$ I do not know how to proceed. I tried the following:

$$ \begin{align*}\lim_{(x,y)\rightarrow (0,0)} \sin^2 y\frac{2x(|x|+|y|)-x^2\sgn(x)}{(|x|+|y|)^2} &= \lim_{(x,y)\rightarrow (0,0)} \sin^2 y \cdot \lim_{(x,y)\rightarrow (0,0)}\frac{2x(|x|+|y|)-x^2sign(x)}{(|x|+|y|)^2}\\& = 0 \cdot \lim_{(x,y)\rightarrow (0,0)}\frac{2x(|x|+|y|)-x^2sign(x)}{(|x|+|y|)^2} \end{align*}$$

So if $$\begin{align*} \lim_{(x,y)\rightarrow (0,0)}\frac{2x(|x|+|y|)-x^2\sgn(x)}{(|x|+|y|)^2} &= \lim_{(x,y)\rightarrow (0,0)}\frac{2x(|x|+|y|)}{(|x|+|y|)^2}-\lim_{(x,y)\rightarrow (0,0)}\frac{x^2\sgn(x)}{(|x|+|y|)^2}\\ &= \lim_{(x,y)\rightarrow (0,0)}\frac{2x}{|x|+|y|}-\lim_{(x,y)\rightarrow (0,0)}\frac{x^2\sgn(x)}{(|x|+|y|)^2}\end{align*}$$ converges I'm done. I'm not sure how to proceed now though. Could anyone give me a pointer?

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up vote 2 down vote accepted

You only need that $\frac{2x}{|x|+|y|}$ and $\frac {x^2\operatorname{sgn}(x)}{(|x|+|y|)^2}$ are bounded.

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Why would that suffice? – sxd Nov 15 '12 at 18:06
Because then for $|(x,y)-(0,0)|<\epsilon$ you have $|f_x|\le \sin^2 y \cdot L< L \epsilon^2$ and $|f_y|\le x^2L<L\epsilon^2$. – Hagen von Eitzen Nov 15 '12 at 18:09
Oh I see what you mean, thanks ! – sxd Nov 15 '12 at 18:18

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