Proving the continuity of a function with two variables using partial derivatives Let 
$$f(x,y) = \left\{\begin{array}{rcl}\frac{x^{\alpha}y^{\beta}}{x^2+y^2} & \mathrm{if} & (x,y) \ne (0,0)\\ 0 & \mathrm{if} & (x,y)=(0,0)\end{array}\right.$$
The question is: for what values of $\alpha$ and $\beta$ is $f$ continuous at $(0,0)?$
My (partial) answer is below.
 A: I found the answer.
We have $|x^\alpha|\le \(sqrt{(x^2+y^2)})^alpha$ and similarly for $|y^\beta|$.
So $$|x^\alpha y^\beta|\le (x^2+y^2)^{(\alpha+\beta)/2}$$
$$|f(x,y)|\le (x^2+y^2)^{(\alpha+\beta)/2-1}$$
so $\lim_{(x,y)\to (0,0)}  f(x,y)$ is smaller than $\lim_{(x,y)\to (0,0)} (x^2+y^2)^{(\alpha+\beta)/2-1}$. 
So we make now discussion of the limit according to the values of $\alpha$ and $\beta$
i.e. if $\alpha +\beta <2$ so $(x^2+y^2)^{(\alpha+\beta)/2-1}$ tends to infinity so take $y=x$
then $\lim f(x,x) =1/2$ as $x$ tends to $0$. So it is different from $0$... so f is not cont. At 0 in this case 
And so on for $\alpha +\beta >2$ and for $\alpha +\beta =2$ similarly we solve it.
A: Using polar coordinates ( $x=r\cos \theta,\, y = r\sin \theta $ ) we have

$$\lim_{r\to 0} \frac{r^{\alpha+\beta} \cos^{\alpha} \theta \sin^{\beta} \theta }{r^2} = \lim_{r\to 0 }\, r^{\alpha+\beta-2} \cos^{\alpha} \theta \sin^{\beta} \theta = 0 $$

if 

$$ \alpha+\beta - 2> 0 \implies \alpha + \beta > 2. $$

Note:

$$ |\cos^{\alpha} \theta \sin^{\beta} \theta| \leq 1 . $$

