How to show that $f(x,y)$ is continuous. How to show that $f(x,y)$ is continuous. 
$$f(x,y)=\frac{4y^3(x^2+y^2)-(x^4+y^4)2x\alpha}{(x^2+y^2)^{\alpha +1}}$$  for $\alpha <3/2$.
Please show me   Thanks :) 
 A: Notice that $|x|,|y|\leq (x^2+y^2)^{1/2}=||(x,y)||$ so we have
\begin{align}|f(x,y)|&=\frac{|4y^3(x^2+y^2)-(x^4+y^4)2x\alpha|}{(x^2+y^2)^{\alpha +1}}\leq \frac{4|y|^3(x^2+y^2)+(x^4+y^4)2|x||\alpha|}{(x^2+y^2)^{\alpha +1}}\\
&\leq \frac{4||(x,y)||^3||(x,y)||^2+(||(x,y)||^4+||(x,y)||^4)2||(x,y)|||\alpha|}{||(x,y)||^{2\alpha +2}}\\
&= \frac{4(|\alpha|+1)||(x,y)||^5}{||(x,y)||^{2\alpha +2}}=4(|\alpha|+1)||(x,y)||^{3-2\alpha}\rightarrow 0\end{align}
So if $f(0,0)=0$ then $f$ is continuous at $(0,0)$ and hence everywhere.
A: The only point where we can have trouble is $(x,y)=0$. To figure out what happens there, use the polar coordinates, as suggested in the comment of davin. We have $x=r\cos\theta$, $y=r\sin\theta$ and therefore 
$$x^2+y^2=r^2,\qquad x^4+y^4=r^4\left(\cos^4\theta+\sin^4\theta\right)$$
so that
$$f=r^{3-2\alpha}\left(4\sin^3\theta-2\alpha\cos^5\theta-2\alpha\cos\theta\sin^4\theta\right).\tag{1}$$
Obviously, as $\alpha<3/2$ and $(x,y)\rightarrow(0,0)$, the limit of $f$ in (1) exists and is equal to the value of $f$ at $(0,0)$ (equal to zero). 
