# Trying to demonstrate a multivariable limit: $\lim\limits_{(x,y) \to (0,0)} \frac{xy}{|x|+|y|}$ [duplicate]

i'm having some trouble trying to prove the next multivariable limit:

$$\lim\limits_{(x,y) \to (0,0)} \frac{xy}{|x|+|y|}$$

I've already tried the different paths and the limit is 0, however i'm stuck trying to demonstrate it because those absolute values on the denominator, by using the theorem that our teacher taught us to do so.

The theorem 1.) 2.) goes as:

$$1.) |f(x,y) - L| < g(x,y)$$

Where L is the limit we calculated from the differents paths we got, 0 in this case, then we are supposed to calculate the $g(x,y)$ function by:

$$|f(x,y) - 0 | = |\frac{xy}{|x|+|y|} - 0 |$$

This is the part where i'm stuck cause we are supposed to calculate that $g(x,y)$ through that formula, however i'm getting stuck because i don't know how to operate with this absolute value on the denominator $||x|+|y||$ any help on this one would be highly appreciated!

$$2.) \lim\limits_{(x.y) \to (x_o,y_o)} g(x,y) = 0$$

This one is just to evaluate the $g(x,y)$ we got from the formula above and it should be 0.

## marked as duplicate by Workaholic, John B, Adam Hughes, Martin Sleziak, Daniel W. FarlowDec 4 '16 at 1:39

$|\frac {xy}{|x|+|y|}| \le \frac {|xy|}{|y|} = |x|$ Similarly $|\frac {xy}{|x|+|y|}| \le |y|$

However you want to define you distance metric, $d((x,y),(0,0)) \le \max(|x|,|y|)$

$d((x,y),(0,0)) < \delta \implies |\frac {xy}{|x|+|y|}| < \delta$

$\delta = \epsilon$

$\forall \epsilon > 0, \exists \delta > 0$ such that $d((x,y),(0,0))<\delta \implies |f(x,y) - 0| < \epsilon.$

• Excuse my ignorance, but on the first line you're doing the squeeze theorem right? if so, why do you define them like that? i'm not familiar when you have such absolute value, sorry if this is a silly question or something like that! just wanna learn about this – NeptaliD Sep 1 '16 at 1:33
• I wasn't thinking about the squeeze theorem. Although that could be a short cut. I was thinking more about my distance metric. Anyway, I am saying that $|x|+|y| > |x|$ therefore $\frac {1}{|x|+|y|} < \frac 1{|x|}$ – Doug M Sep 1 '16 at 1:46
• Ohh got it, thanks so much! – NeptaliD Sep 1 '16 at 1:54

One way forward is to transform to polar coordinates $(r,\theta)$ where $x=r\cos(\theta)$ and $y=r\sin(\theta)$. Then, we can write

\begin{align} \left|\frac{xy}{|x|+|y|}\right|&=\left|\frac{r^2\cos(\theta)\sin(\theta)}{r|\cos(\theta)|+r|\sin(\theta)|}\right|\\\\ &=\left|\frac{\frac12 r \sin(2\theta)}{|\cos(\theta)|+|\sin(\theta)|}\right|\\\\ &\le \frac12 r\\\\ &<\epsilon \end{align}

whenever $r=\sqrt{x^2+y^2}<\delta =2\epsilon$