Prove that $\lim_{(x,y) \to (0,0)} \frac{xy(x-y)}{x^3 + y^3}$ does not exist I am trying to prove that the following limit does not exist.
$$\lim\limits_{(x,y) \to (0,0)} \frac{xy(x-y)}{x^3 + y^3}$$
I have tried several paths such as:

*

*$\operatorname{\gamma}(t) = (t,0)$


*$\operatorname{\gamma}(t) = (0,t)$


*$\operatorname{\gamma}(t) = (t,t)$


*$\operatorname{\gamma}(t) = (t,t^2)$


*$\operatorname{\gamma}(t) = (t^2,t)$
but all these paths equal $0$. I have started think that the limit is in fact $0$ and I expended a quite long time trying to prove it by the squeeze theorem and them I gave up and looked over WolframAlpha and learned that the limit does not exist. I do not know how to prove it.
Can someone, please:

*

*Show for this particular limit a path that is different than $0$?


*Explain the thought processes I should apply to this kind of problem? How does one get the feeling that this limit does not exist after trying so many paths?
 A: So I will explain how we can come up with a path for where the limit becomes something else than $0$. Expanding the expression a bit we get that
$$\frac{xy(x-y)}{x^3+y^3}=\frac{x^2y-xy^2}{x^3+y^3}.$$
Immediately we should suspect that there is some path for which we get something else than $0$, as the numerator and denominator both consist of third order terms, and so this is something we can try to capitalize on. Indeed what we can try is to make the numerator and denominator become some non-zero third order terms. This means that we want some $(x,y)=(\alpha t,\beta t)$ for appropriate $\alpha,\beta$. Now by the nature of the problem, we can simply set $\beta=1$, and just consider curves on the form $\gamma_\alpha(t)=(\alpha t,t)$. If we set $(x,y)=(\alpha t,t)$, we get that
$$\frac{x^2y-xy^2}{x^3+y^3}=\frac{\alpha^2t^3-\alpha t^3}{\alpha^3t^3+t^3}=\frac{\alpha(\alpha-1)t^3}{(\alpha^3+1)t^3}=\frac{\alpha(\alpha-1)}{(\alpha^3+1)}.$$
It is now easy to choose $\alpha$ so that this is non-zero.
A: Let, $(x,y)\to(0,0)$ such that $\frac yx\to k,\,k\not\in\left\{0,1\right\}$ then we have
$$\begin{align}\lim_{(x,y)\to (0,0)}\frac{xy(x-y)}{x^3 + y^3}&=\lim_{(x,y)\to (0,0)}\frac{\frac yx-\left(\frac yx\right)^2}{1+\left (\frac yx\right)^3}\\
&=\frac{k(1-k)}{k^3+1}\\
&≠0.\end{align}$$

Addendum
You can choose infinitely many pair  $(x,y)\to(0,0)$ such that $\frac yx\to k,\,k\not\in\left\{0,1\right\}.$
Algebraic path:

Let $a_i,b_i\in\mathbb R$ with $a_0≠b_0 $ and $a_0≠0,b_0≠0$, then $(a_0x^n+a_1x^{n-1}+\cdots +a_n; b_0x^n+b_1x^{n-1}+\cdots +b_n)\to (0,0)$, where $n\in\mathbb Z_+$, which gives $k=\frac ba$.

Trigonometric path:

Let $a,b\in\mathbb R\setminus \left\{0\right\}$ with $a≠b$ then $\left(\frac{x^n\sin ax}{bx},x^n\right)\to (0,0)$, where $n\in\mathbb Z_+$, then we have $k=\frac ba$.

A: As suggested in the comments, $(2t, t)$ should work. Here, you get a numerator of $2t^3$ and a denominator of $9t^3$, which clearly does not go to $0$.
In general, you can see the numerator is “like” a cubic polynomial as is the denominator. So you wouldn’t expect it to always go to $0$. So you want to find a path like this, where you get a constant times a term like $t^3$ in both the numerator and denominator. Hope this helps.
A: Consider these paths:
1)$x=y \rightarrow 0$;
2)$x=2y \rightarrow 0$.
What can you say?
