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

How should I approach this?

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  • $\begingroup$ i think the searched Limit is Zero, try to use polar coordinates $\endgroup$ May 21 '17 at 12:46
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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 $0$.

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    $\begingroup$ Nice answer. For the last limit, $t^2=u$ will make LHR quicker, I presume. $\endgroup$ May 21 '17 at 13:47
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Try taking $x^2 + y^2 = t$ and then write $x^6 + y^6$ in its term.

Will update after solving.

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