I'd like to show that $$ \lim_{(x,y)\to(0,0)} \frac{2xy^6 - 2x^5y^2}{(x^4+y^4)^2} $$ doesn't exist.

I've tried approaching along $x=0$, $y=0$, and $y=kx$, and haven't been able to get inconsistent values.


The original problem is: given $f: \mathbb{R}^2 \rightarrow \mathbb{R}$

$$ \begin{align} f(x,y) &= \frac{x^2y^2}{x^4+y^4},\quad if (x,y) \ne (0,0); & 0\quad , if (x,y)=(0,0) \end{align} $$ decide if it's differentiable at $(0,0)$; if not, verify it does not have continuous partials in a neighborhood of $(0,0)$.

Clearly $f$ is not differentiable as it's not continuous at $(0,0)$; I'm having trouble showing $\frac{\partial f}{\partial x} $ is not continous at $(0,0)$.

  • 1
    $\begingroup$ $y=kx$ should work for a suitably chosen $k$. The numerator is $2x^7(k^6-k^2)$ and the denominator $(1+k^4)^2x^8$, so that if $k^6-k^2\ne 0$, the quotient goes to plus or minus infinity as $x\to 0$. $\endgroup$ – ForgotALot Dec 10 '16 at 2:56
  • 1
    $\begingroup$ Or minus infinity, if you approach from $x<0$ side $\endgroup$ – Andrei Dec 10 '16 at 2:57
  • $\begingroup$ can't believe i didn't see that... thanks! $\endgroup$ – Yibo Yang Dec 10 '16 at 2:57
  • 1
    $\begingroup$ when numerator and denominator are homogeneous, try polar coordinates. If that shows a limit that depends on $\theta,$ you are done, no limit. $\endgroup$ – Will Jagy Dec 10 '16 at 3:13

Your Answer

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

Browse other questions tagged or ask your own question.