<s>[INST] 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.
 [/INST] 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
$$0\le\frac{a^2b^2}{a^2 + b^2} \le \frac{(a^2+b^2)^2}{a^2 + b^2}=a^2+b^2 \to 0$$
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