# Continuity of $\frac{xy}{(x^2+y^2)^\alpha}$

I am asked to study the continuity of the function $$f_\alpha(x,y)=\begin{cases}\frac{xy}{(x^2+y^2)^\alpha} \text{ if } (x,y)\neq(0,0)\\0 \text{ if }(x,y)=(0,0)\end{cases}$$ depending on the value of $$\alpha$$. I have seen that for $$\alpha\leq 0$$ the function is continuous and for $$\alpha= 1$$ it is not continous (taking the limit at $$x=y$$). What do I do to determine the continuity of the function for other values of $$\alpha$$?

Hint: Put $$x=r\cos{\theta}$$ and $$y=r\sin{\theta}$$ then rewrite $$f_{\alpha}$$ in terms of $$r$$ and $$\theta$$
• I already tried that since similar exercices were done this way. However I got to the point where $f(rcos(\theta),rsin(\theta)=r^{2-2\alpha}sin(\theta)cos(\theta)=r^{2-2\alpha}sin(2\theta)/2$ and I don't know what to do from here. Nov 6 '19 at 0:59
• @JohnKeeper what is the limit when $\alpha>1$ and $r$ going to $0$? What if $\alpha<1$. In the case $\alpha=1$ what if $\theta=0$? $\theta=\dfrac{\pi}{4}$? Nov 6 '19 at 1:03
• So, I assume we take the limit of $r$ going to $0$ because thats equivalent to the limit $(x,y)$ approaches $(0,0)$. Then, if $\alpha>1$ and $r$ goes to $0$, $r^{2-2\alpha}sin(2\theta)/2$ goes to infinity (therefore not continuous), for $\alpha<1$ the limit goes to 0 and therefore is continuous. For $\alpha=1$, taking the $\theta=0$ and $\theta=\pi/4$ yields two different values for the limit, therefore we don't have continuity at $\alpha=1$. Is this it? Nov 6 '19 at 1:11