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!
 A: Hint. For now, suppose that the limit exists and is equal to $L$. Then,
$$L = \lim\limits_{r\to0}{\frac{e^{-\frac{1}{r^2}}}{r^4(1-2\cos^2\theta \sin^2\theta)}}$$
Take $\theta = \frac\pi 2$ first. Then, $$L = \lim\limits_{r\to0}{\frac{e^{-\frac{1}{r^2}}}{r^4}}$$
Now, take $\theta = \frac\pi 4$. Then,
$$L = \lim\limits_{r\to0}{\frac{2e^{-\frac{1}{r^2}}}{r^4}} = 2L \implies L =  0$$
So, if the limit exists, it must be equal to $0$.

Proof of Existence.

Observe that $$\frac{e^{-1/r^2}}{r^4(\sin^4\theta+\cos^4\theta)} \leq \frac{2}{r^4e^{1/r^2}}$$


Now, since $r^5e^{1/r^2}\to \infty$ as $r\to 0^+$, there is $\delta'>0$ such that $$0<r<\delta' \implies r^5e^{1/r^2} > 2 \implies \frac{2}{r^4e^{1/r^2}} < r$$


Finally, choose $\delta = \min\{\delta',\epsilon\}$ to conclude that $$0<\sqrt{x^2+y^2}<\delta \implies 0 < r < \delta \implies \left|{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}\right|<\epsilon$$

A: 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.
A: Since $1-2\cos^2 \theta \sin^2 \theta \ne 0$ the limit in polar coordinates becomes
$$
\frac{1}{1-2\cos^2 \theta \sin^2 \theta}\cdot \lim_{r\to 0}\dfrac{e^{-1/r^2}}{r^4}=0, \quad \forall \theta \in[0,2\pi)
$$
This shows that the original limit is zero.
