Is this a correct proof for $\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}=0$?

Question

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, $\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}=0.$

Answer 1

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: $\;x=r\cos \theta\;,\;\;y=r\sin\theta\implies$

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