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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 !

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    $\begingroup$ In last line, last inequality, seems, you forgot $3$ in power. Correct? $\endgroup$ Commented Feb 4, 2023 at 23:58
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    $\begingroup$ Isn't it faster to use polar coordinates? $\endgroup$ Commented Feb 5, 2023 at 0:06
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    $\begingroup$ @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”) $\endgroup$ Commented Feb 5, 2023 at 0:31
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    $\begingroup$ @peek-a-boo can you explain me why? $\endgroup$ Commented Feb 5, 2023 at 0:41
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    $\begingroup$ @coboy In general switching to polar coordinates it's preatty useful when dealing with 2 variable limits. By the way, your reasoning seems correct $\endgroup$ Commented Feb 5, 2023 at 0:48

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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.

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  • $\begingroup$ 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! $\endgroup$ Commented Feb 5, 2023 at 1:21
  • $\begingroup$ @coboy indeed, it’s a very simple, but necessary fix. $\endgroup$ Commented Feb 5, 2023 at 1:23
  • $\begingroup$ I would use $h_1^2h_2^2\le {1\over 2}(h_1^2+h_2^2)^2.$ $\endgroup$ Commented Feb 5, 2023 at 8:55

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