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

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!

• Use polar coordinates :)
– Anna
Commented Oct 28, 2013 at 1:33
• A general useful inequality is $|x|,|y|\leqslant \sqrt{x^2+y^2}$, i.e. $|x|,|y|\leqslant \lVert (x,y)\rVert$.
– Pedro
Commented Oct 28, 2013 at 1:58

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:

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

Convert into polar coordinates: so $x=r\cos\theta\space$ and $y=r\sin\theta.$

$\displaystyle\lim_{(x,y)\to(0,0)} \dfrac{xy}{\sqrt{x^2+y^2}}\\=\displaystyle\lim_{r\to 0}\dfrac{r\cos\theta\cdot r\sin\theta}{\sqrt{r^2\cos\theta+r^2\sin\theta}}\\=\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta}{\sqrt{r^2(\sin^2\theta+\cos^2\theta)}} \text{Recall the Pythagorean Identity,}\sin^2\theta+\cos^2\theta=1.\\=\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta}{\sqrt{r^2}}\\=\displaystyle\lim_{r\to 0}\space r\sin\theta\cos\theta\\=0$

• thx, I didn't expected the full answer. But thank you for your "latex" effort Commented Oct 28, 2013 at 2:06
• You're welcome. :) Commented Oct 28, 2013 at 2:07
• double thx, because that limit was in my exam this morning Commented Oct 28, 2013 at 16:37
• This doesn't show that the function is continuous at (0, 0). There is always a possibility that the limit as (x,y) goes to (0, 0) along a curve (a path that is not a straight line) is not 0. So this is not a complete proof. Commented Sep 6, 2015 at 23:14
• @Nick How would you fix the proof? It looks like they fixed $\theta$ and let $r$ goes to zero.
– john
Commented Oct 10, 2017 at 1:33

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 $x,y\to 0$.

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 $y = x$.

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...

It is correct to use polar. But don't forget to use the squeeze theorem as $$\theta$$ could depend on $$r$$ since it as any path: $$|r \sin(\theta (r))\cos (\theta(r))| \leq |r| \leq \sqrt{x^2 + y^2}$$. The squeeze theorem is essential to a proper argument.