limits $\displaystyle \lim_{(x,y)\rightarrow (0,0)}\frac{xy}{2x^2-y^2}$ Does this limit exist? 
$$\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{2x^2-y^2}$$
Also is there a general method that may be used to answer such questions?
Thanks.
 A: No it doesn't. Take the test sequence $(0,1/n)$ and the limit is $0$, taking a different test sequence $(1/n,1/n)$ gives the limit $1$.
The general method is to test different, non-trivial directions, usually a sequence along $x=0$ or $y=0$ then a sequence along $y = \pm x$ suffices. 
A: Choose $\,y=mx\,$ , so that $\,(x,y)=(x,mx)\,$ , and then
$$\frac{xy}{2x^2-y^2}=\frac{mx^2}{x^2(2-m^2)}=\frac{m}{2-m^2}\xrightarrow [(x,y)\to (0,0)\Longleftrightarrow x\to 0]{} \frac{m}{2-m^2}$$
and the limit is not independent of the way the variable approaches the origin, thus it doesn't exist.
A: No it does not exist. I will the function in the limit be f(x,y), we find that the sequence: f(0,1/1),f(0,1/2),f(0,1/3),... does not converge, thus we conclude that there is no L such that for all epsilon>0 there exists an open ball B around (0,0) such that for all (x,y) in B |f(x,y)-L|
A: Along $y=x$ then $xy/(2x^2-y^2)=x^2/2x^2$ has limit 1/2. Along the line
$y=2x$ then $xy/(2x^2-y^2)=2x^2/(-2x^2)$ has limit -1.
Thus the limit does not exist.
