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I consider the function defined by

$$ \left\{ \begin{array}{ll} \frac{x^2y^2}{x^2+y^2}\quad\text{if}\quad(x,y)\neq (0,0)\\ 0\quad\quad\;\;\;\text{if}\quad (x,y) = (0,0) \end{array} \right. $$

I would like to show that this function is differentiable at $(0,0)$.

My attempt :

Using the fact that if a function $f :\mathbb{R}^n\to\mathbb{R}$ is differentiable at a point $x_0$ if and only if $$ \varepsilon(h) = \frac{1}{\lVert h\rVert}\left(f(x_0+h) -f(x_0) -df(x_0)(h) \right)\xrightarrow[h \to 0]{} 0 $$ I consider the function

$$ \varepsilon(h_1,h_2) = \frac{1}{\sqrt{h_1^2+h_2^2}}\left(\frac{h_1^2h_2^2}{h_1^2+h_2^2 }\right) = \frac{h_1^2h_2^2}{(h_1^2+h_2^2)^{3/2}} $$

Clearly we have

$$ \lvert h_1\rvert \leq (h_1^2 + h_2^2)^{1/2}\quad\text{and}\quad h_2^2\leq h_1^2 + h_2^2\implies\lvert h_1\rvert h_2^2\leq(h_1^2 + h_2^2)^{3/2} $$

which implies

$$ 0\leq \lvert\varepsilon(h_1,h_2)\rvert = \frac{h_1^2h_2^2}{(h_1^2+h_2^2)^{3/2}}\leq\frac{h_1^2h_2^2}{\lvert h_1\rvert h_2^2} = \lvert h_1\rvert\xrightarrow[(h_1,h_2) \to (0,0)]{} 0 $$

Is this seems correct to you ?

Thank you a lot !

G2MWF
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    In last line, last inequality, seems, you forgot $3$ in power. Correct? – zkutch Feb 04 '23 at 23:58
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    Isn't it faster to use polar coordinates? – Sine of the Time Feb 05 '23 at 00:06
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    @SineoftheTime not really, and changing to polar coordinates yields incorrect conclusions if not careful (of which there are many examples on this site and in other online “notes”) – peek-a-boo Feb 05 '23 at 00:31
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    @peek-a-boo can you explain me why? – Sine of the Time Feb 05 '23 at 00:41
  • @zkutch You are totally right. Thank you ! – G2MWF Feb 05 '23 at 00:45
  • @SineoftheTime To be honest I did not think of that at all, in fact I am not familiar with polar coordinates – G2MWF Feb 05 '23 at 00:46
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    @coboy In general switching to polar coordinates it's preatty useful when dealing with 2 variable limits. By the way, your reasoning seems correct – Sine of the Time Feb 05 '23 at 00:48
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    @SineoftheTime I was not aware of the existence of a such method, I will definitively find out what it is. Thank you ! – G2MWF Feb 05 '23 at 00:50
  • @SineoftheTime See Why is this method wrong on calculating this multivariable limit? for just one instance of how naively using polar coordinates leads to wrong answers. Anyway, polar coordinates here do not simplify things in any way, what OP did is (almost) perfect; use simple and obvious estimates which are more easily generalizable. – peek-a-boo Feb 05 '23 at 01:01
  • @peek-a-boo In the question you attacched, the OP clearly applied polar coordinated not correctly. In fact he can't conclude anything because if $\theta=\pi$ the function is not defined. But in this case, using polar coordinates is the fastest way: the function become $\frac{r^4\cos^2\theta\sin^2\theta}{r^3}$ that approaches $0$ ad $r \to 0^+$ – Sine of the Time Feb 05 '23 at 11:52

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It’s almost perfect. The only issue I have is in the last two steps, particularly when you write $\leq \frac{h_1^2h_2^2}{|h_1|h_2^2}$. This is troublesome because your original assumption is only that $(h_1,h_2)\neq (0,0)$, meaning that $h_1,h_2$ cannot be zero simultaneously. But, it is very much possible that $h_1\neq 0$ and $h_2=0$ (or vice-versa). As such, writing $h_2^2$ in the denominator is wrong (sure, here you have a $h_2^2$ in the numerator, which cancels things out, but keep in mind that you can only ‘cancel’ when you have non-zero numbers).

Here’s how I’d present the inequalities. We have for all $(h_1,h_2)\neq (0,0)$, \begin{align} |\epsilon(h_1,h_2)|&=\frac{h_1^2h_2^2}{(h_1^2+h_2^2)^{3/2}}=|h_1|\cdot\frac{|h_1|}{(h_1^2+h_2^2)^{1/2}}\cdot\frac{h_2^2}{h_1^2+h_2^2}\leq |h_1|\cdot 1\cdot 1=|h_1|. \end{align} The fact that the fractions are at most $1$ is obvious since the numerators are smaller or equal to the denominators. Of course, you can do things slightly differently and prove that $|\epsilon(h_1,h_2)|\leq |h_2|$ if you wanted to. By presenting the inequalities this way, notice that I have only kept $h_1^2+h_2^2$ in the denominator, which by my assumption of $(h_1,h_2)\neq (0,0)$ is non-zero (in fact strictly positive), so I avoid the fatal mistake of division by zero.

peek-a-boo
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  • I think I see your point yes. With your decomposition you avoid the division by zero since if one of the $h_i$'s is $0$ the denominator is still defined because it implies the other $h_i$'s to be nonzero.. it's slightly subtle but totally right and it's an important remark, I hadn't thought about the fact of having one of the $h_i$ equal to $0$. Thanks a lot! – G2MWF Feb 05 '23 at 01:21
  • @coboy indeed, it’s a very simple, but necessary fix. – peek-a-boo Feb 05 '23 at 01:23
  • I would use $h_1^2h_2^2\le {1\over 2}(h_1^2+h_2^2)^2.$ – Ryszard Szwarc Feb 05 '23 at 08:55