Prove that the limit doesn't exist I have to prove that 
$$\lim_{(x,y)->(0,0)} \frac{xy}{2x-y}$$ doesn't exist.
I have tried to use these restrictions:


*

*$x=0;  y=0; y=x;  y=mx; y=mx+q; y=ax^2;y=ax^2+bx+c; y=1/x, y=1/x^2,..$


and for each of them, I have done the related limit. But I have always obtained 0. 
I know that I if want to prove that the limit doesn't exist, I have to find two restrictions that have different values for their limits.
Any suggestions? Many thanks
 A: $$
\lim_{(x,y)->(0,0)} \frac{xy}{2x-y}
$$
Let $y=ax^2+bx+c$ then we get
$$
\lim_{(x,y)->(0,0)} \frac{ax^3+bx^2+cx}{2x-ax^2-bx-c}
$$
We can see that for it to have a nonzero numerator, we must cancel an $x$. To do that we must set $c=0$ giving.
$$
\begin{align}
&\lim_{(x,y)->(0,0)} \frac{ax^3+bx^2}{2x-ax^2-bx}\\
=\:&\lim_{(x,y)->(0,0)} \frac{ax^2+bx}{2-ax-b}
\end{align}
$$
We now obtain the same problem as before, we must cancel an $x$, to do that we must set $b=2$.
$$
\begin{align}
&\lim_{(x,y)->(0,0)} \frac{ax^3+2x^2}{2-ax^2-2}\\
=\:&\lim_{(x,y)->(0,0)} \frac{ax^3+2x^2}{-ax^2}\\
=\:&\lim_{(x,y)->(0,0)} \frac{ax+2}{-a}
\end{align}
$$
Take the limit and obtain.
$$
\lim_{(x,y)->(0,0)} \frac{ax+2}{-a}=\frac{-2}a
$$
The limit is different for every $a\in\mathbb R$, so the original limit is undefined.
A: $$\lim\limits_{(x,y)\to (0,0)} \frac{xy}{2x-y}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+} \frac{r^2\cos\phi \sin\phi}{2r\cos\phi-r\sin\phi}$$
$$=\lim\limits_{r\to 0^+} \frac{r\cos\phi \sin\phi}{2\cos\phi-\sin\phi}$$
Now let's attempt to find a bound that is independent of $\phi$,
$$\left|\frac{\cos\phi \sin\phi}{2\cos\phi-\sin\phi}\right|\leq \left|\frac{\frac12}{2\cos\phi-\sin\phi}\right| $$
$$\left|\frac{\cos\phi \sin\phi}{2\cos\phi-\sin\phi}\right|\leq \left|\frac{1}{4\cos\phi-2\sin\phi}\right| $$
The right hand side is unbounded which implies that the limit is dependent on $\phi$. Therefore
$$\lim\limits_{(x,y)\to (0,0)} \frac{xy}{2x-y}\Rightarrow \mbox{does not exist}$$
