How to calculate $\lim \limits_{(x,y)→(0,0)} \frac{(x^2+y^2)^2}{xy}$? The wolframalpha gives the answer $0$:
Wolframalpha culculation
I tried like this:
Let $x=r\cos(\theta)$ and $y=r\sin(\theta)$,then:
$\lim \limits_{(x,y)→(0,0)} \frac{(x^2+y^2)^2}{xy}$ = $\lim \limits_{r→0} \frac{r^4}{r^2 \sin(\theta)\cos(\theta)}$ = $\lim \limits_{r→0} \frac{2r^2}{\sin(2\theta)} =0$
but it seems wrong when  $\theta =0$ and the limit $\lim \limits_{r→0} \frac{2r^2}{\sin(2\theta)}$ may not exist.
So how to calculate the limit?

We have that for $x=y=t\to 0$
$$\frac{(x^2+y^2)^2}{xy}=\frac{4t^4}{t^2}=4t^2 \to 0 $$
but for $x=t\to 0$ and $y=t^3$
$$\frac{(x^2+y^2)^2}{xy}=\frac{(t^2+t^6)^2}{t^{4}}=1+2t^4+t^8 \to 1 $$
therefore the limit doesn't exist.
