# Continuity of partial derivatives at (0,0)

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?

• Since the limit should be the same irrespective of the approach chosen, it is enough to consider limit along $x=my$ here. – 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.