For the function $$f = \begin {cases} \frac{xy}{x^2+y^2}, \text{if } (x,y) \neq (0,0) \\ 0, \text{if } (x,y) = (0,0) \end {cases}$$ show that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are not continuous at (0,0).

My attempt: I found $$\frac{\partial f}{\partial x}(x,y)=\frac{y^3-x^2y}{(x^2+y^2)^2}$$ At $(x,y)=(0,0)$ $$\frac{\partial f}{\partial x}(0,0)=\lim_{t\rightarrow 0}\frac{f(0+t,0)-f(0,0)}{t}=0.$$ Similarly $\frac{\partial f}{\partial y}(0,0)=0$

Then $$\lim_{(x,y)\rightarrow(0,0)}\frac{\partial f}{\partial x}(x,y)=\lim_{(x,y)\rightarrow(0,0)}\frac{y^3-x^2y}{(x^2+y^2)^2}= \text{ (consider restriction to the line } x = my) = \lim_{y\rightarrow 0}\frac{y^3-m^2y^3}{(m^2y^2+y^2)^2}=\lim_{y\rightarrow 0}\frac{y^3(1-m^2)}{(m^2+1)^2y^4}=\infty \neq \frac{\partial f}{\partial x}(0,0).$$ Thus $\frac{\partial f}{\partial x}$ is not continuous at (0,0). Similarly, for $\frac{\partial f}{\partial y}.$

Is it correct? Also is it sufficient to show the inequality of the limit and the value of the partial, considering only one restriction to the line (in this case $x=my$)? Or should we consider other restrictions as well?

  • $\begingroup$ Since the limit should be the same irrespective of the approach chosen, it is enough to consider limit along $x=my$ here. $\endgroup$ – Exp ikx Feb 1 '19 at 19:20

You could also argue like this: Obviously, the partial derivatives are continuous in $\mathbb R^2 \setminus \{(0,0)\}$. If they were also continuous in $(0,0)$, the function would be continuously partially differentiable, hence totally differentiable and hence continuous. But $f(1/n,1/n)=1/2$ does not converge to $f(0,0)=0$ for $n\to\infty$.

To be exact, this only shows that $\frac{\partial f}{\partial x}$ or $\frac{\partial f}{\partial y}$ is discontinuous at $(0,0)$, but for symmetry reasons it cannot be only one of them.

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