Show that the limit in $(0,0)$ of $\frac{xy}{x^3-y}$ doesn't exist. I'm trying to show that the limit below doesn't exist, I think it's quite simples but I couldn't until now.
$$\lim_{(x,y)\to (0,0)}\dfrac{xy}{x^3-y}$$
My approach is try to find two curves in the domain that have different limits in $(0,0)$
I couldn't find any curve that the limit isn't $0$.
Any tips? Thanks.
 A: Your idea is fine. Take the curve $y=x^3+x^4$. Then$$f(x,x^3+x^4)=-x-1$$and the fact that $\lim_{x\to0}-x-1=-1$. On the other hand, $\lim_{x\to0}f(x,0)=\lim_{x\to0}0=0$.
A: Generally in problems like this, you often want to homogenize the degrees of $x$ and $y$ on the denominator if possible.
In this case you can see that if we take $y=ux^3$ ($u$ not constant, see it as $u(x,y)=\frac y{x^3}$) then you can factor $x^3$ out.
Then $\dfrac{xy}{x^3-y}=\dfrac{x^4}{x^3(1-u)}=\dfrac{u}{1-u}\times x$
Now it is easier to see that to make it divergent, it suffices to take $\frac{u}{1-u}=\frac 1{x^2}$ to get an overall $\dfrac 1x$.
Substituting back into $y$ gives $f(x,\frac{x^3}{1+x^2})=\dfrac 1x$ is divergent.
Another possibility is to take $\frac u{1-u}=O(x)$ to cancel out with the numerator.
Notice that $\frac u{1-u}=\dfrac Cx$ leads to $f(x,\frac {Cx^3}{x+C})\to C$ for an arbitrary $C$, making the limit non-existent, because multiple paths lead to multiple limit values.
On the other hand when you arrive to $f(x,y)=g(u)\times x^\alpha$ and $g(u)$ is bounded then this time $|f(x,y)|<M|x|^\alpha\to 0$ has a limit in $(0,0)$.
For instance the problem $f(x,y)=\dfrac{xy^2}{x^4+y^2}$ with $y=ux^2$ leads to $g(u)=\frac{u^2}{1+u^2}<1$
