calculate $ \lim_{(x,y)\to (0,0)} \frac{e^{\frac{-1}{x^2+y^2}}}{x^4+y^4} $ calculate the limit:
$$\lim_{(x,y)\to (0,0)} \frac{e^{\frac{-1}{x^2+y^2}}}{x^4+y^4} $$
I have tried to use polar coordinates,
I also tried to show there is no limit, but I'm pretty sure now the limit is $0$,
but I can't prove it.
 A: Hint:
Since $\|(x,y)\|_2$ and $\|(x,y)\|_4$ are equivalents, there is a $D>0$ such that $\|(x,y)\|_4\geq D\|(x,y)\|_2$ and thus
$$0\leq \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}\leq \frac{1}{D}\cdot \frac{e^{-\frac{1}{\|(x,y)\|_2^2}}}{\|(x,y)\|_2^4}.$$
Then you can use polar coordinate to conclude.
Added :
With polar coordinate, you get
$$\frac{1}{D}\cdot  \frac{e^{-\frac{1}{\|(x,y)\|_2^2}}}{\|(x,y)\|_2^4}=\frac{1}{D}\cdot  \underbrace{\frac{e^{-\frac{1}{r^2}}}{r^4}}_{\underset{r\to 0}{\longrightarrow } 0}{\longrightarrow }0$$
The fact that $$\lim_{r\to 0}\frac{e^{-\frac{1}{r^2}}}{r^4}=0$$ come from the fact that $$\lim_{u\to +\infty }u^2e^u=+\infty.$$
Indeed,
$$e^u\geq u$$ for all $u\in\mathbb R$ and thus $$u^2e^u\geq u^3\underset{u\to+\infty }{\longrightarrow }+\infty.$$
A: Let  $(r,\theta)$ be the polar coordinates of the point $(x,y)$. Set $u=\dfrac 1{r^2}$. We can write 
$$\frac{\mathrm e^{-\tfrac{1}{x^2+y^2}}}{x^4+y^4}=\mathrm e^{-u}\cdot\frac1{r^4(1-2\sin^2\theta\cos^2\theta)}=_{+\infty}\mathrm e^{-u}O(u^2)\xrightarrow[u\to+\infty]{}0$$
since $\;1-2\sin^2\theta\cos^2\theta=1-\dfrac12\sin^2 2\theta\ge\dfrac12. $
