$\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}}$ Let me be honest, when some teacher ask me to calculate such limit, I always try to found two paths with different results to show that such limit don't exist.
But, I try four paths all fall at zero... and wolframalpha (internet boss (after math.stackexchange.com)) tells me this limit don't exit. Stewart's (book boss) calculus book tells me this limit is zero. But I don't know how to show this...
I hate limit calculus, because it's full of tricks... if you're not aware enough the day of your exam, you lost quickly some points...
Could you light my path?
thx to everybody!
 A: Convert into polar coordinates: so $x=r\cos\theta\space$ and $y=r\sin\theta.$
$\displaystyle\lim_{(x,y)\to(0,0)} \dfrac{xy}{\sqrt{x^2+y^2}}\\=\displaystyle\lim_{r\to 0}\dfrac{r\cos\theta\cdot r\sin\theta}{\sqrt{r^2\cos\theta+r^2\sin\theta}}\\=\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta}{\sqrt{r^2(\sin^2\theta+\cos^2\theta)}} \text{Recall the Pythagorean Identity,}\sin^2\theta+\cos^2\theta=1.\\=\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta}{\sqrt{r^2}}\\=\displaystyle\lim_{r\to 0}\space r\sin\theta\cos\theta\\=0$
A: You could do the following: $x^2+y^2\geq 2|xy|$, so $\sqrt{x^2+y^2}\geq \sqrt{2|xy|}$, therefore $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq\frac{|xy|}{\sqrt{2|xy|}}=\frac{\sqrt{|xy|}}{\sqrt{2}},$$ which converges to $0$ as $x,y\to 0$.
A: The hint is to use the standard trick of the polar coordinates thing:
Put $x = r \cos \theta $, $y = r \sin \theta $. 
Alternatively, you can see what happens if you approach the limit through the the line $y = x $.
A: Related problems: I, II. Here is how you advance
$$ \Bigg| \frac{xy}{\sqrt{x^2+y^2}}-0  \Bigg |\leq   \frac{|x||y|}{\sqrt{x^2+y^2}} \leq \frac{ \sqrt{x^2+y^2} \sqrt{x^2+y^2}  }{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}< \epsilon =\delta .$$
Note:

$$ |x| \leq \sqrt{x^2+y^2},\quad |y| \leq \sqrt{x^2+y^2}. $$

A: There really isn't any "trick" here; just applying standard absolute value estimates and polar coordinates. As @Citizen mentioned, polar coordinates turns the inside of your limit to
$$\dfrac{r^2\cos\theta\sin\theta}{r}.$$
Try it from there...
