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For 2 variable functions we define $ f\left(h\right)=o\left(h\right) $ if $ |f\left(h\right)|=o\left(||h||\right) $ for $h=(h_1,h_2) $.

consider the function $ x\sin\left(\frac{y}{\sqrt{|x|}}\right) $.

I have to determine if this function is differntiable at $ (0,0)$. So I already found that the function is continious at (0,0), and that the partial derivatives are both $ 0 $ at (0,0).

Thus Jacobi matrix would be $ (0,0) $ and by the definition of differntiability that I have, if $ f $ is differntiable at $(0,0)$ there exists a linear transformation $ L_0 $ such that $ f\left(h_{1},h_{2}\right)-f\left(0,0\right)=L_{0}h+o\left(h\right) $

Where $ L_0 $ is exactly tha Jacobi matrix composed of the partial deriviatives at $ (0,0) $. So if I want to check the definition, all I need to to is to check if $ f\left(h_{1},h_{2}\right)-f\left(0,0\right)=o\left(h\right) $

Because $L_0 $ is the zero matrix due to my calculations. Now that brings us to the main question, is $ h_{1}\sin\left(\frac{h_{2}}{\sqrt{|h_{1}|}}\right)=o\left(h\right) $ ?

Or in other words, How can I calculate the limit :

$ \lim_{h\to0}\frac{h_{1}\sin\left(\frac{h_{2}}{\sqrt{|h_{1}|}}\right)}{||h||}=\lim_{\left(h_{1},h_{2}\right)\to\left(0,0\right)}\frac{h_{1}\sin\left(\frac{h_{2}}{\sqrt{|h_{1}|}}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}}} $

(Also is this correct way of using the definitions of o(h) ? Im not sure about it and Im not sure if this is the right way to check differntiability).

Thanks in advance

FreeZe
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  • Use that the function $\sin(x) / x$ is bounded. With this you only need to establish that $\frac{\sqrt{h_1}h_2}{h} \to 0$ as $h\to 0$ to determine the limit. – Winther Nov 17 '20 at 16:57
  • How is this the same ? the argument of the sine here not necesseraly tends to 0 when h tends to zero – FreeZe Nov 17 '20 at 17:01
  • I actually think that this function is not o(h) – FreeZe Nov 17 '20 at 17:02
  • If you have a product $AB$ where $A$ is bounded and $B$ goes to zero then... btw if you don't think this function is $o(h)$ then that would imply that the function is not differentiable (and your other argument would have to be wrong). – Winther Nov 17 '20 at 17:10
  • What argument exactly? I didnt claim anything, I was just trying to figure out if this function is o(h) or not in order to know if its differntiable or not. Yes - I think it is not o(h) and thus not differntiable. Difference here is that you do not have such product AB with bounded A and B goes to zero – FreeZe Nov 17 '20 at 17:42
  • Thought you said you proved it and just wanted to double check, but I now see I misread it. Anyway the argument I was referring to was to write $\frac{h_1\sin(h_2/\sqrt{h_1})}{h} = \frac{\sqrt{h_1}h_2}{h}\cdot \frac{\sin(h_2/\sqrt{h_1})}{h_2/\sqrt{h_1}}$. – Winther Nov 17 '20 at 17:52
  • @Winther How can we prove that $ \lim_{\left(h_{1},h_{2}\right)\to\left(0,0\right)}\frac{h_{2}}{\sqrt{|h_{1}|}}=0 $ ? – FreeZe Nov 17 '20 at 18:05
  • You can't, because its not true. You can however show that $\frac{\sqrt{h_1}h_2}{h}$ goes to zero as its just $\sqrt{h_2} \cdot \sqrt{\frac{h_1h_2}{h_1^2+h_2^2}}$ and the latter factor is bounded (as is easily seen using polar coordinates). – Winther Nov 17 '20 at 18:07
  • @Winther So how can you conclude that $ \frac{\sin\left(\frac{h_{2}}{\sqrt{|h_{1}|}}\right)}{\frac{h_{2}}{\sqrt{|h_{1}|}}} $ is bounded around $ 0,0 $ if you dont know that the limit exists? – FreeZe Nov 17 '20 at 18:09
  • https://math.stackexchange.com/questions/2433051/sine-inequality-how-to-prove-that-sinx-le-x-for-x-in-mathbbr And note that the function you have there is just a normal function $\sin(x)/x$ where $x = h_2/\sqrt{h_1}$. – Winther Nov 17 '20 at 18:11
  • @Winther I see. so this function is actually differntiable, cool. One more question, when I try to calculate the limit using WolframAlpha, I get that $ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{y}{\sqrt{|x|}}=0$. Is this wrong? (cant we trust wolfram on this ? ) – FreeZe Nov 17 '20 at 18:27
  • Yes its wrong (try to take the limit along curves $y = a\sqrt{x}$). Its far from perfect on multidimensional limits, see e.g. https://math.stackexchange.com/questions/1526058/how-to-calculate-lim-x-y-to-0-0-frac-x2y3-sin-x8y4/ ; https://math.stackexchange.com/questions/1972398/the-limit-of-a-two-variable-function-lim-x-y-to-0-0-frac-xy-sin-y/1972416 ; https://math.stackexchange.com/questions/2053544/wrong-wolfram-alpha-limit-lim-limits-x-y-to0-0-fracxy3x2y6 for examples where it get things wrong. – Winther Nov 17 '20 at 18:33

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