# Find \lim_{(x,y)\to(0,0)} \frac{2xy^6 - 2x^5y^2}{(x^4+y^4)^2}

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.

[Background]

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 $(0,0)$.

• $y=kx$ 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 $x\to 0$. – ForgotALot Dec 10 '16 at 2:56
• Or minus infinity, if you approach from $x<0$ side – Andrei Dec 10 '16 at 2:57
• can't believe i didn't see that... thanks! – Yibo Yang Dec 10 '16 at 2:57
• when numerator and denominator are homogeneous, try polar coordinates. If that shows a limit that depends on $\theta,$ you are done, no limit. – Will Jagy Dec 10 '16 at 3:13