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I'm trying to find $$\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} .$$ After I tried couple of algebraic manipulation, I decided to use the polaric method. I choose $x=r\cos \theta $ , $y=r\sin \theta$, and $r= \sqrt{x^2+y^2}$, so I get

$$\lim\limits_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^4\cos^4 \theta+r^4 \sin^4 \theta } $$

What do I do from here?

Thanks a lot!

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If there was a tug of war between an exponential function and an algebraic function, who would win? –  Fabian Feb 8 '12 at 19:20
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3 Answers 3

up vote 2 down vote accepted

You are almost done!

$$\lim_{r\to 0}\frac{e^{-\frac{1}{r^2}}}{r^4(\cos^4\theta+\sin^4\theta)}=\frac 1{\cos^4\theta+\sin^4\theta}\lim_{r\to 0}\frac{e^{-\frac{1}{r^2}}}{r^4}.$$ For the last limit you can use L'Hopital's rule.

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Thanks! Do I treat $\theta$ as a fixed one and assume that $cos^4\theta+sin^4\theta$ is never 0? –  Jozef Feb 8 '12 at 19:46
... and show the $\theta$ factor is bounded. –  GEdgar Feb 8 '12 at 19:48
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We have for $t>0$ that $e^t\geq\frac{t^3}3$ so $$\frac{\exp\left(-\frac 1{x^2+y^2}\right)}{x^4+y^4}\leq \frac 3{\frac 1{(x^2+y^2)^3}}\frac 1{x^4+y^4}=\frac{3(x^2+y^2)^3}{x^4+y^4}=3(x^2+3y^2+3x^2+y^2)=12(x^2+y^2),$$ and you are done.

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Since $2(x^4+y^4)\geqslant(x^2+y^2)^2$, the ratio $r(x,y)=\dfrac{\mathrm e^{-1/(x^2+y^2)}}{x^4+y^4}$ is such that $$ 0\lt r(x,y)\leqslant u\left(\frac1{x^2+y^2}\right), \quad\text{where}\ u:z\mapsto2z^2\mathrm e^{-z}. $$ Now, $\frac1{x^2+y^2}\to+\infty$ when $(x,y)\to(0,0)$ and $u(z)\to0$ when $z\to+\infty$ (as the saying goes, at infinity the exponential prevails on the powers) hence $r(x,y)\to0$ when $(x,y)\to(0,0)$.

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