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Proof: Let $\varepsilon>0$ be given.

For our proof we need to find $\delta>0$ such that $0<\sqrt{x^2+y^2}<\delta\to|\frac{xy}{\sqrt{x^2+y^2}}-0|\lt\varepsilon$


$x\le\sqrt{x^2+y^2},\,\,\,\,\forall x,y\in\mathbb{R}\to\frac{x}{\sqrt{x^2+y^2}}\le1,$

which implies$\frac{x|y|}{\sqrt{x^2+y^2}}\le|y|\le\sqrt{y^2}\le\sqrt{x^2+y^2}$, which is our condition for $\delta$.

So if we let $\varepsilon=\delta$ and $0\lt\sqrt{x^2+y^2}\lt\delta$, we have that:

$|\frac{xy}{\sqrt{x^2+y^2}}-0|\le\sqrt{x^2+y^2}\lt\delta=\varepsilon.$

So by definition of the limit, $\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}=0.$

Anthony Anguiano
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Perhaps slightly simpler: $\;\sqrt{x^2+y^2}\ge\sqrt{x^2}=|x|\;$ , and then

$$0\le\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\le\frac{|x||y|}{|x|}=|y|\xrightarrow[(x,y)\to(0,0)]{}0$$

Or with polar coordinates: $\;x=r\cos \theta\;,\;\;y=r\sin\theta\implies$

$$\frac{xy}{\sqrt{x^2+y^2}}=r\cos\theta\sin\theta\xrightarrow[r\to0]{}0$$

DonAntonio
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    I believe you mean $y = r \sin \theta$ – Jonas Mar 04 '16 at 23:20
  • @Jonas ZOf course, thank you very much. Edited – DonAntonio Mar 04 '16 at 23:21
  • In the first paragraph you want to treat $x\ne0, x = 0$ separately. – zhw. Mar 05 '16 at 01:10
  • @zhw. Thank you. I think you're right, yet by the complete symmetry of the expression and since we can (and must) assume that while $;(x,y)\to (0,0);$ it is also true that $;(x,y)\neq (0,0)\iff x\neq 0;or;y\neq0;$ , so at least one of the coordinates isn't zero at any given point and thus we choose that one. – DonAntonio Mar 05 '16 at 08:55