Htykuut

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

$$ frac{a^2b^2}{a^2 + b^2}=r^2cos^4thetasin^4theta to 0$$

otherwise as an alternative use that

$$0lefrac{a^2b^2}{a^2 + b^2} le frac{(a^2+b^2)^2}{a^2 + b^2}=a^2+b^2 to 0$$

First, your approach is correct. Second, try polar coordinates.

$a^2+b^2 ge 2|ab| ge |ab|.$

$0 le sqrt{|ab| dfrac{|ab|}{a^2+b^2}} le sqrt{ |ab| cdot 1} le$

$sqrt{ a^2+b^2}.$

Choose $delta = epsilon$.

One may solve the problem geometrically. As the signs of $a$ and $b$ don't matter, let's assume them to be non-negative. Now consider $a$ and $b$ as the legs of a right triangle, then the hypotenuse is $c=sqrt{a^2+b^2}$, hence
$$sqrt{frac{a^2b^2}{a^2+b^2}}=frac{ab}{c}.$$
If we call the height of the triangle $h$, we know that $ba=ch$, hence
$frac{ab}{c}=h$. Now if the legs approach zero, so does the triangle's height.

Just another approach

Let $(a, b) to (0, 0)$ along the line $b=am$, $m$ is a constant.

$displaystyle lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} =displaystyle lim_{(a, am) to (0, 0)} sqrt{frac{a^2(am)^2}{a^2 + (am)^2}}=displaystyle lim_{substack{ato 0\\text{along}: b=am}} sqrt{frac{a^2m^2}{1 + m^2}} =0$