Calculate the limit: $ \lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}$ 
Calculate the limit:
$$\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}
$$

I tried to change to polar coordinates like that:
\begin{align*}
\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}} &=
\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{(x^2+y^2)^2-2x^2y^2}}\\
&=\lim\limits_{r\to0}{\frac{e^{-\frac{1}{r^2}}}{r^4(1-2\cos^2\theta \sin^2\theta)}}\\
\end{align*}
and I'm not sure how to continue from this point.
Thank you!

Hint: $x^{4}+y^{4} \geq \frac  1 2 ({x^{2}+y^{2}})^{2}$. Now use polar coordinates to see that the limit is $0$.
This allows you to get rid of $\theta$. You will also need the fact that $t^{4}e^{-t} \to 0$ as $ t \to \infty$ (Take $t=\frac  1r$). This fact is proved using L'Hopital's Rule.
