Finding $\lim_{(x,y)\to(0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^6+y^6}$ 
$$\lim_{(x,y)\to(0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^6+y^6}$$

How should I approach this?
 A: I would suggest using polar coordinates:
\begin{align}
x&=r\cos(\theta)\\
y&=r\sin(\theta)\\
r^2&=x^2+y^2.
\end{align}
Since $x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)$, we could also compute:
$$
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{r^6(\cos^4(\theta)-\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta))}.
$$
Since $\cos^6(\theta)+\sin^6(\theta)=\cos^4(\theta)-\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta))$ is always positive (on the unit circle), it is, therefore, bounded and bounded away from zero.  Hence, there exist constants $C_1$ and $C_2$ so that
$$
0<C_1\leq \cos^4(\theta)-\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta))\leq C_2.
$$
Then, by the squeeze theorem,
$$
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{C_2r^6}\leq
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{r^6(\cos^4(\theta)-\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta))}\leq
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{C_1r^6}
$$
Therefore, it make sense to look at 
$$
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{r^6}
$$
This is an indeterminate form of the form $\frac{0}{0}$.  Using the substitution $t=\frac{1}{r}$, this limit becomes
$$
\lim_{t\rightarrow \infty}\frac{t^6}{e^{t^2}}.
$$
Using l'Hopital's rule a few times, you get that the limit is $0$.  Hence, by the squeeze theorem, the desired limit is $0$.
A: Try taking $x^2 + y^2 = t$ and then write $x^6 + y^6$ in its term.
Will update after solving.
