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I'd like your help with this:

I tried using L'Hôpital's Rule and all kinds of arithmetic to prove that $$\lim_{x \to 0 }\left(x^{-a}e^{\left(\frac{-1}{x^{2}}\right)}\right) = 0$$ for every $a$, and it didn't work.

($a=0$ is trivial)

Any hints?

Thank you.

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3 Answers 3

up vote 3 down vote accepted

Try setting $x = y^{-1}$. Then it should be easy. If you need more help, just ask.

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2  
No, It was enough. Thank you. –  user6163 May 20 '11 at 23:06
    
@FUZ Could you please elaborate on how you propose to solve the problem after changing variables. There are many possibilities so it's hard to guess. –  Bill Dubuque May 21 '11 at 0:08
    
@Bill: Quite similar to what you did. But I arrived at $\lim\limits_{y\to\infty}a\ln y - y^2=-\infty$... Maybe I calculated wrong. –  FUZxxl May 21 '11 at 13:36
    
@FUZ No, that's the same as I hinted. After factoring one gets $\ \log\ y\ (a - y^2/{\log\ y})\ $ so it reduces to calculating $\ y^2/{\log\ y}\ $ by L'Hôpital, as I said. But it's not the change of variable $y = 1/x$ that is essential here. Rather it's the taking of logs, since that allows L'Hôpital to succeed. It trades off an exponential for a log. And generally logs work better for L'Hôpital since they go away upon differentiation, but exp's do not. –  Bill Dubuque May 21 '11 at 15:29
    
@Bill Dubuque: I am not very good at limits (I am still in highschool, they will teach these things next year). To prove that, I just sad, that $a\ln y$ grows slower than $y^2$ and thus $a\ln y - y^2\to-\infty$ as $y\to\infty$. I guess that's not a good solution. –  FUZxxl May 21 '11 at 15:33

Actually, this result you hope for is false if $a$ is allowed to be an arbitrary real number (but it only fails for stupid reasons). If $a = -\frac{1}{2}$, for example, then

$$\lim_{x \to 0^{-}} \frac{\sqrt{x}}{e^{\frac{1}{x^2}}} $$

does not exist.

If one is restricted to integer values of $a$, then the result does as was previously discussed. Also, the limit $\displaystyle \lim_{x \to 0} |x|^{-a} e^{- \frac{1}{x^2}}$ always exists for all $a$.

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No, the above limit is 0. Why do you think it doesn't exist? –  Bill Dubuque May 21 '11 at 15:17
    
Because $\frac{\sqrt{x}}{e^{1/x^2}}$ is undefined for $x < 0$. –  JavaMan May 21 '11 at 15:44
1  
But that's a bit artifical since the limit will still be 0 if one allows complex values. –  Bill Dubuque May 21 '11 at 16:23
    
Why is it undefined? Then we have $\lim_{x\to0^-}{\sqrt x}/{\exp x^{-2}} = \lim_{x\to0^-}i\sqrt{|x|}/{\exp x^{-2}}$ which is the same except for an additional $i$. (Am I wrong?) –  FUZxxl May 21 '11 at 17:18

HINT $\ $ Take the $\rm\:log\:,\:$ put $\rm\: z = 1/x\:$ to reduce it to $\rm\: z^2/\:log(z)\to \infty\:$ as $\rm\: z\to\infty\:,\:$ by L'Hôpital.

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