4
$\begingroup$

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.

$\endgroup$

4 Answers 4

12
$\begingroup$

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?

$\endgroup$
10
$\begingroup$

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 ?

$\endgroup$
5
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.