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I need to calculate this limit: $$\lim\limits_{(x,y)\to(0,0)} \cfrac{1}{x^6 + y^6} exp \Biggr(\cfrac{-1}{x^2 + y^2} \Biggl) $$

It should be zero. I thought about the sandwich theorem but I'm not sure what to do from here:

$$0\leq \Biggr|\cfrac{1}{x^6 + y^6} exp \Biggr(\cfrac{-1}{x^2 + y^2} \Biggl) \Biggr|$$

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Our goal is to show $$\lim\limits_{(x,y)\to(0,0)}\;\left(\frac{e^{-\frac{1}{x^2+y^2}}}{x^6+y^6}\right)\!=0$$ Changing to polar coordinates, let \begin{cases} x=r\cos(\theta)\\[4pt] y=r\cos(\theta)\\ \end{cases} Noting that for $\theta\in [0,2\pi]$, the function $$\theta \mapsto \cos^6(\theta)+\sin^6(\theta)$$ has positive minimum value, $a$ say, we get $$ 0 < \frac {e^{-\frac{1}{x^2+y^2}}} {x^6+y^6} = \frac {e^{-\frac{1}{r^2}}} {r^6\bigl(\cos^6(\theta)+\sin^6(\theta)\bigr)} \le \frac {e^{-\frac{1}{r^2}}} {ar^6} $$ for all $(x,y)\in\mathbb{R}^2{\setminus}\{(0,0)\}$.

Hence, to prove the desired limit, it suffices to prove $$ \lim_{r\to {0^{+}}} \frac {e^{-\frac{1}{r^2}}} {r^6} = 0 $$ Then, letting $u=\frac{1}{r^2}$, we get $$ \lim_{r\to {0^{+}}} \frac {e^{-\frac{1}{r^2}}} {r^6} = \lim_{u\to \infty} \frac{u^6}{e^u} = 0 $$ as was to be shown.

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