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

Question

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

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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.

Answer 1

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

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

Answer 2

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

Answer 3

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.

Answer 4

Just another approach

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

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

Answer 5

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

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

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

Choose $\delta = \epsilon$.