# Show that $\lim\limits_{(x,y)\to(0,0)}\frac{x^3y-xy^3}{x^4+2y^4}$ does not exist.

Show that $$\lim_{(x,y)\to(0,0)}\frac{x^3y-xy^3}{x^4+2y^4}$$ does not exist.

I'm not even sure how to approach this. I tried factoring out $$xy$$ in the numerator to get $$xy(x^2 - y^2)$$, but I don't think that gets me anywhere with the denominator.

HINT:

What happens if the limit is taken along $$y=2x$$? What happens when the limit is taken along $$y=0$$? Are these equal? If not, what can one conclude?

Let's approach the limit along the line $$y=mx.$$

\begin{align} &\lim_{(x,y)\to (0,0)}\dfrac{x^3y-xy^3}{x^4+2y^4}\\ &=\lim_{x\to 0}\dfrac{x^3mx-x(mx)^3}{x^4+2(mx)^4}\\ &=\lim_{x\to 0}\dfrac{x^4m-m^3x^4}{x^4+2m^4x^4}\\ &=\lim_{x\to 0}\dfrac{m-m^3}{1+2m^4}\\ &=\dfrac{m-m^3}{1+2m^4}\\ \end{align}

So what can you conclude about the limit ?

Putting $$x=y$$ your expression vanishes, and for$$x=2y$$, the limit will be $$1/3$$. Therefore the limit does not exist

Doing the change to polar coordinates: $$\frac{x^3y - xy^3}{x^4 + 2y^4} = \frac {r^3\cos^3\theta\,r\sin\theta - r\cos\theta\,r^3\sin^3\theta} {r^4\cos^4\theta + 2r^4\sin^4\theta} = \frac {\cos^3\theta\sin\theta - \cos\theta\sin^3\theta} {\cos^4\theta + 2\sin^4\theta},$$ dependent of $$\theta$$ (and independent of $$r$$), so the limit does not exists.