Tough limit

chiph588@ · #1 $Old$ July 3rd, 2009, 09:59 AM

Show $\lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^3} = 0$ .

Twig · #2 $Old$ July 3rd, 2009, 10:17 AM

Let $t=\frac{1}{x^{2}}$ then $t \to \infty \mbox{ , as } x \to \ 0$ .

$\lim_{t \to \infty} \frac{e^{-t}}{\frac{1}{t^{\frac{3}{2}}}} =$ $\lim_{t \to \infty} t^{\frac{3}{2}}e^{-t} = \lim_{t \to \infty}\frac{t^{\frac{3}{2}}}{e^{t}}=0$

Because $\frac{x^{\alpha}}{e^{x}} \to 0 \mbox{ , as } x \to \infty$ , (standard limit).

Danny · #3 $Old$ July 3rd, 2009, 10:19 AM

Quote:

Originally Posted by chiph588@ $View Post$

Show $\lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^3} = 0$ .

Let $x = \frac{1}{t}$ and use L'Hopital's rule a few times.

Moo · #4 $Old$ July 4th, 2009, 09:58 AM

Hello,

Note that your limit is $\lim_{x\to 0} \frac{1}{x^3 e^{1/x^2}}$

Then you can apply l'Hopital's rule on the denominator to get $-3\cdot\frac{e^{1/x^2}}{x}$

And this obviously goes to $\infty$ as x goes to 0.

So the limit is 0.

NonCommAlg · #5 $Old$ July 4th, 2009, 12:15 PM

Quote:

Originally Posted by Twig $View Post$

Because $\frac{x^{\alpha}}{e^{x}} \to 0 \mbox{ , as } x \to \infty$ , (standard limit).

that's true for all $\alpha \in \mathbb{R}.$ a L'Hopital-free way to prove it: choose any positive integer $n > \alpha.$ then $e^x > \frac{x^n}{n!}$ because $x>0.$ thus $0 < \frac{x^{\alpha}}{e^x} < \frac{n!}{x^{n-\alpha}}.$ now apply the squeeze theorem.