$\lim _{(x,y)\to(0,0)}\frac{x(x-y)}{x^4+y^4}$ does not exists Show that
$$\lim_{(x,y)\to(0,0)} \frac{xy(x-y)}{x^4+y^4}$$
does not exists.
I've tried the "traditional" paths, with $(x,x)$, $(0,x)$, $(0,-x)$, but I only get $0$ as answer. Any hint?
Thanks!
 A: Hint: Consider the paths you've already taken together with the path $t\mapsto (t,t^2)$.
A: Take the limit as $(x,y)\rightarrow(0,0)$ along the line $y=-x$:
this gives $\displaystyle \lim_{x\to0}\frac{-2x^3}{2x^4}=\lim_{x\to0}-\frac{1}{x}$, which does not exist.
A: Consider the path of the arbitrary straight line $y=mx$, $m \in \mathbb{R}$:
\begin{align*}
\lim_{(x,y)\to (0,0)} \frac{xy(x-y)}{x^4 + y^4} &= \lim_{x\to0}\frac{mx^2(x-mx)}{x^4 + m^4x^4} \\&= \lim_{x\to 0} \frac{mx^3(1-m)}{x^4(1+m)} \\&= \lim_{x\to 0} \frac{m(1-m)}{x(1+m)}
\end{align*}
Now, firstly notice that the limit is dependent on $m$, which means that the limit of the function does not exist for any value that $x$ approaches because the limit will differ for every value of $m$.
But let's look at this a bit further and manipulate what we have into the following form:
\begin{align*}\lim_{x\to 0} (\frac{1-m}{1+m}\times \frac{1}{x} ) &= \lim_{x\to 0}(\frac{1-m}{1+m})\times\lim_{x\to 0}(\frac{1}{x})\end{align*}
From which we can then clearly see the following:
In the case that $x \to 0$, the limit does not exist, since \begin{align*}\lim_{x\to 0^-} \frac{1}{x} \neq\lim_{x\to 0^+} \frac{1}{x}\end{align*}
