Solution Verification: using polar coordinates to solve this limit (as variables approach infinity). 
$\lim_{(x,y)\to \infty}\frac{2x+3y}{x^2-xy+y^2}$

Here's what I did: 
$\lim_{r\to \infty} \frac{r(2sin(a)+3cos(a))}{r^2(-sin(a)cos(a))}=\lim_{r\to \infty} \frac{2sin(a)+3cos(a)}{r(-sin(a)cos(a))}=0$

Am I allowed to use polar coordinates when the variables go to infinity? (would appreciate to know why incase it's a yes). 
I've never seen it in any example or question before and I'm a little skeptical.

If it's not allowed, I would appreciate a push in the right direction. 
Thanks in advance!
 A: Here's another method.  Let $x. y \gt 0.$
$$x^2-xy+y^2=\left(x-{y\over 2}\right)^2+{3\over 4}y^2\ge {3\over 4}y^2 \tag 1$$
By symmetry,
$$x^2-xy+y^2\ge {3\over 8}\left(x^2+y^2\right).$$
The rest is elementary. For example,
$$0\lt \frac{2x}{x^2-xy+y^2}\le {16x\over 3x^2},$$
and the right side goes to zero as $x\to\infty.$
This method will work whenever the denominator is a positive quadratic homogeneous polynomial and the numerator is linear.
(Explaining "symmetry" above: You can get (1) with $x^2$ instead of $y^2$ on the right. Bigger than two numbers implies bigger than their average.)
A: Your idea of using polar coordinates is a good one, but you have to be a bit more careful.  Suppose for some positive integer $N$, we have $x>N$ and $y>N$.  Then $(x,y)$ is a point in the first in the first quadrant, so we can write $$x=r\cos\theta\\y=r\sin\theta$$ for some $0<\theta<\frac\pi2$ and some $r>N$.  Now $$\frac{2x+3y}{x^2-xy+y^2}=\frac{2r\cos\theta+3r\sin\theta}{r^2-r^2\sin\theta\cos\theta}=\frac1r\frac{2\cos\theta+3\sin\theta}{1-\cos\theta\sin\theta}$$ (I notice you made a mistake in the denominator.)
Now the problem is that you can't $\theta$ fixed and let $r$ go to $\infty$, because the value of $\theta$ depends on $x$ and and $y$.  If we can show that $f(\theta)=\frac{2\cos\theta+3\sin\theta}{1-\cos\theta\sin\theta}$ is bounded, however, then we can say that the fraction goes to $0$.  This is true because the denominator never vanishes, so that $f$ is a continuous function on a closed interval, and is therefore bounded.
