limit $\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}}$ Let $$\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}}$$

How do I show it's limit is $0$? 

I thought about:

$$\frac{xy}{\|(x,y)\|} \le \frac{xy}{|x|} \le \frac{xy}{x} \le y \to 0$$

Is that correct?
It looks fine. 

As I wrote in another post, usually it is useful to set $x = \rho \cos \theta$, $y = \rho \sin \theta$ and taking the limit as $\rho \to 0$. More specifically, if you find that 
$|f(\rho \cos \theta, \rho \sin \theta)| \le g(\rho) \to 0$, then you showed that $f(x, y) \to 0$

In your case 

$$|f(\rho \cos \theta, \rho \sin \theta)| = \frac{|\rho^2 \sin \theta \cos \theta|}{|\rho|} \le \rho = g(\rho) \to 0$$

so your limit is indeed $0$
Wrong answer

Actually,it has no limits.
From different directions to (0,0),it's limit is different.
 For example:(x,0) to (0,0),the limit is 0;(t,t) to (0,0),the limit is 1/sqrt(2)