How to prove $\lim_{(x,y)\to(0,0)} \frac{xy}{x+y} = 0$ I want to find $\displaystyle\lim_{(x,y)\to(0,0)} \frac{xy}{x+y} = 0$. After trying different ways to approach $(0,0)$, I am fairly convinced the limit is $0$, but I need to prove it by definition, and I seem to be stuck. I want to prove that $\forall \epsilon >0, \exists \delta > 0 $ such that $0<\parallel(x,y)\parallel<\delta \implies |\frac{xy}{x+y}| < \epsilon$.
I'm having trouble with the denominator. I know that to get $|\frac{xy}{x+y}|$ to be less than something, I have to show that $|x+y|$ can be made greater than something, but I don't know what. Any suggestions?
 A: If you approach $(0,0)$ along the line $x=0$ the function has constant value $0$ and the limit is $0$.
But now suppose that you approach along a curve like $y=x^2-x$. Then 
$$\frac{xy}{x+y} = \frac{x^3-x^2}{x^2} =x - 1$$ 
and the limit as $x\to 0$ is...
A: The sum and product of two numbers are independent, you can assign them arbitrary values at will.
So while you make them both converge to zero, you can follow any $(s,p)$ curve, with arbitrary ratios.
For example, following an hyperbolic spiral $\rho=s/\theta$,
$$\frac{xy}{x+y}=\frac{s^2}{\theta^2}\frac{\theta}{s}\frac{\cos\theta\sin\theta}{\cos\theta+\sin\theta}=s\frac{\sin\theta}\theta\frac{\cos\theta}{\cos\theta+\sin\theta}.$$
This expression alternates between minus and plus infinity on every turn.
A: Put $x=rcos\theta$ and $y=rsin\theta$ s.t. $$lim_{(r,\theta)\rightarrow (0,0)}\frac{rcos\theta.sin\theta}{cos\theta+sin\theta}=0$$ if and only if the denominator $cos\theta+sin\theta\ne 0$. However, approaching $r=0$ along $\theta=-\pi/4$ the given limit doesn't exist.
