Does $\lim_{(a, b) \to (0, 0)} \sqrt{\frac{a^2b^2}{a^2 + b^2}} = 0$? 
Does $$\lim_{(a, b) \to (0, 0)} \ \sqrt{\frac{a^2b^2}{a^2 + b^2}} = 0 \ ?$$

I'm trying to show that $$\lim_{(a, b) \to (0, 0)} \ \sqrt{\frac{a^2b^2}{a^2 + b^2}} = 0 $$
but I am getting stuck. I was thinking that as a starting point I could show that $$\lim_{(a, b) \to (0, 0)} \ \frac{a^2b^2}{a^2 + b^2} = 0 $$
and then conclude that since $$\lim_{x \to 0} \sqrt{x} = 0$$ and $\frac{a^2b^2}{a^2 + b^2} \to 0$ as $(a, b) \to (0, 0)$ we arrive at $$\lim_{(a, b) \to (0, 0)} \ \sqrt{\frac{a^2b^2}{a^2 + b^2}} = 0.$$

Firstly is my approach above a correct one. Secondly how can show that $$\lim_{(a, b) \to (0, 0)} \ \frac{a^2b^2}{a^2 + b^2} = 0. $$
Because I don't see any way to show the above (apart from perhaps proving it from the definition directly, which I would like to avoid if there is an easier way to do it). Also it could be the case that the initial limit doesn't even exist.

By polar coordinates we have that
$$ \frac{a^2b^2}{a^2 + b^2}=r^2\cos^4\theta\sin^4\theta \to 0$$
otherwise as an alternative use that
$$0\le\frac{a^2b^2}{a^2 + b^2} \le \frac{(a^2+b^2)^2}{a^2 + b^2}=a^2+b^2 \to 0$$
