Find $\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} $ I'm trying to find $$\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} .$$ 
After I tried couple of algebraic manipulation, I decided to use the polaric method.
I choose $x=r\cos \theta $ , $y=r\sin \theta$, and $r= \sqrt{x^2+y^2}$, so I get 
$$\lim\limits_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^4\cos^4 \theta+r^4 \sin^4 \theta } $$
What do I do from here? 
Thanks a lot!

Since $2(x^4+y^4)\geqslant(x^2+y^2)^2$, the ratio $r(x,y)=\dfrac{\mathrm e^{-1/(x^2+y^2)}}{x^4+y^4}$ is such that 
$$
0\lt r(x,y)\leqslant u\left(\frac1{x^2+y^2}\right),
\quad\text{where}\ u:z\mapsto2z^2\mathrm e^{-z}.
$$
Now, $\frac1{x^2+y^2}\to+\infty$ when $(x,y)\to(0,0)$ and $u(z)\to0$ when $z\to+\infty$ (as the saying goes, at infinity the exponential prevails on the powers) hence $r(x,y)\to0$ when $(x,y)\to(0,0)$.
