For what values of $a$ and $b$ is the function $\frac{x^ay^b}{x^2+y^2}$ continuous at $(0,0)$?

Question

I have the function $$f(x,y)=\begin{cases}\dfrac{x^ay^b}{x^2+y^2} &(x,y)\neq(0,0)\\ 0 &(x,y)=(0,0) \end{cases}$$ I am trying to figure out what constants $a$ and $b$ will make the function continuous at $(0,0)$. I know that the limit has to be $0$ as $(x,y)\to(0,0)$ for it to be continuous.

Using polar coordinates, I think that $a+b \geq 3$ since $x^2+y^2= r^2$ and the limit of any polar function $r^x$ with $x > 0$ as $r\to0$ is $0$. Am I on the right track or am I missing something?

Answer 1

$0\leq (x-y)^2=x^2+y^2-2xy$

so

$xy\leq \frac{1}{2}(x^2+y^2)$

then

$|\frac{x^ay^b}{x^2+y^2}|\leq|\frac{x^{a-1}y^{b-1}\frac{1}{2}(x^2+y^2)}{x^2+y^2}|=\frac{1}{2}|x^{a-1}y^{b-1}|\to 0$

if $a>1 $ and $b>1$

with the polar coordinate you get a better condition on $a,b$ :

$a+b>2$

Answer 2

$\dfrac{x^ay^b}{x^2+y^2} \overset{POLAR \, \, COORD.}{\Longrightarrow} \dfrac{r^a\cos^a(\theta) r^b\sin^b(\theta)}{r^2} \Longleftrightarrow r^{a+b-2}\cos^a(\theta)\sin^b(\theta)$

What happens for different values of $a+b-2$ in the limit,

$\lim_{r\rightarrow 0}r^{a+b-2}\cos^a(\theta)\sin^b(\theta)=0$ ?