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I am trying to find out if the partial derivatives of my function $$f(x,y)=\begin{cases} 2\frac{x^3y}{x^2+y^2}-xy & \text{if } (x,y)\neq 0 \space \\ 0 & \text{if } (x,y)= 0 \space\\ \end{cases}$$ are continuous at all points. I have already calculated the derivatives for $\mathbb{R}\setminus\{0\}$ which are continous because they are combinations of polynomials. For $\frac{\partial}{\partial x}(0,0)$ and $\frac{\partial}{\partial y}(0,0)$ I have gotten $0$ as a result. In order to prove that my derivatives are continuous in their origin $(0,0)$, do I unterstand correctly that it is enough to show that $\lim_{(x,y) \to (0,0)} \frac{\partial}{\partial x} f(x,y) = \frac{\partial}{\partial x}(0,0) = 0$? And considering that is true, do I look at $\lim_{(x,y) \to (0,0)} \frac{\partial}{\partial x} f(x,y)$ for $x$ and $y$ seperately by setting one of them to a constant or can I just say $x = my$ for my prove? Any help is highly appreciated.

Best regards.

zhw.
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psyph
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  • Adding the fractions makes this $$f(x,y)=xy\frac{x^2-y^2}{x^2+y^2},\quad x\neq0\neq y,$$ a slight simplification (but maybe irrelevant). – mr_e_man Jul 04 '19 at 03:39
  • I'm sorry that I don't understand, but adding which fractions? – psyph Jul 04 '19 at 03:42
  • Oh okay, you were talking about a simplification. I don't really know if that would help me with my problem, but thanks. – psyph Jul 04 '19 at 03:43
  • Try using the second technique (switching to polar) on this thread about multivariable limits. Using @mr_e_man 's simplification, you've got plenty of reason to switch to polar; and it seems like it might be easy to prove existence or nonexistence. – Vedvart1 Jul 04 '19 at 03:55
  • Thanks @Vedvart1, but I think we haven't covered polar in class yet. Are the assumptions that I'm making correct? Because they work for me, I'm just not sure whether or not the conclusion is right? – psyph Jul 04 '19 at 03:58
  • Your first assumption, namely that it's enough to show that $\frac{\partial f}{\partial x} → 0$ as $(x,y)→(0,0)$, is right. However, I'm a bit rusty on the second one, so I can't remember if showing the consistency of a linear path like $x=my$ as $(x,y)→(0,0)$ is sufficient or if it's necessary to show that all paths are consistent. – Vedvart1 Jul 04 '19 at 04:05

1 Answers1

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You need to show

$$\lim_{(x,y)\to (0,0)}\frac{\partial f}{\partial x}(x,y) = \frac{\partial f}{\partial x}(0,0)=0,$$

and the same for $\partial f/\partial y.$

For this it is not enough to consider limits on straight lines through $(0,0).$ There are examples where such limits along are all $0$ yet $f$ is not even continuous at $(0,0).$

Use your forumula for $\partial f/\partial x$ when $(x,y)\ne (0,0)$ to show that it has limit $0$ as $(x,y)\to 0.$ (Note that by definition, $(x,y)\to 0$ is the same as saying $(x^2+y^2)^{1/2}\to 0.$)

zhw.
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