Limit of one exponential function

[mnb96]

Hello,
How can I find the limit of $$x^{k}\mathrm{e}^{-ax^n}$$ for $x\to +\infty$.
a,k,n are positive reals.
I even have troubles solving it for k=1, a=1, n=2.
Any hint?

 

[Char. Limit]

e^x grows faster than any power of x. That's the really important thing here.

 

[mnb96]

Thanks, I've heard that explanation somewhere but:

- Can you please formally specify what you mean by "grows faster".

- And also, is there a theorem that proves something about the fact that if one function grows faster than the other, it somehow "wins".
Here x goes to infinity, and exp goes to zero. It's not trivial to deduce that the limit is zero.

 

[Char. Limit]

Now that's a good question... I don't know the specific theorem (although I know there is one), but I mean to say that, for example, and this is actually not exactly formal anyway, that e^-x will fall to zero much faster than x^n will climb to infinity. I'm not sure how to put that formally...

 

[CompuChip]

The formal statement would be something like:

"Let two functions f(x) and g(x) be given, and let a be some common boundary point of their domain. We say that f(x) tends to zero faster than g(x) if the limit of f(x) / g(x) as x tends to a is zero."

If you want to get really formal, you may want to check out the "Big oh" (and related little o) notation.

 

[Cyosis]

This theorem is helpful when dealing with indeterminate forms.

http://en.wikipedia.org/wiki/L'Hôpital's_rule

 

[mnb96]

Now I am thinking, could it be this way?
We consider $f(x)=x^k$ and $g(x)=\mathrm{e}^{ax^n}$.

Consider $f(x)/g(x)$ and apply the De L'Hopital's rule?

 

[Cyosis]

Yes that will work.

 

[Landau]

Now that's a good question... I don't know the specific theorem (although I know there is one), but I mean to say that, for example, and this is actually not exactly formal anyway, that e^-x will fall to zero much faster than x^n will climb to infinity. I'm not sure how to put that formally...

It's very easy to prove
$$\lim_{x\to\infty}e^{-x}x^n=0.$$
Just consider the power series expansion of exp.
$$\frac{e^x}{x^n}=\frac{1+x+\frac{x^2}{2!}+...}{x^n}=\underbrace{\frac{1}{x^n}}_{\to 0}+\underbrace{\frac{1}{x^{n-1}}}_{\to 0}+\underbrace{...}_{\to 0}+\frac{1}{n!}+\frac{x}{(n+1)!}+...\ \ \to\infty.$$

 

[Char. Limit]

I thought the Taylor series would unlock it... thanks for the help.