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

Question

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.

Answer 1

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?

Answer 2

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 ?

Answer 3

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

Answer 4

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.