Does this go to 0 for large enough x?

[ChrisVer]

I have one question... In general I always thought that the exponential function was "dying" out faster than any other polynomial function, such that:
$e^{-x} x^a \rightarrow 0$ for $x \rightarrow \infty$.
[eg this is was used quiet commonly and so I got it as a rule-of-thumb, when deriving wavefunctions for a simple example for the Hydrogen atom]

However recently I read in a paper that this is not true, and as an illustration of how can that be, they logarithm-ized the function like:
$\ln (e^{-x} x^n) = -x + n \ln x$ which goes to infinity for $x,n\rightarrow \infty$.
This I read in here:
http://arxiv.org/pdf/1108.4270v5.pdf
in Sec4 (the new paragraph after Eq4.1)

This has confused me, can someone shred some light?

 

[micromass]

$n$ is variable. In your example, $a$ is fixed.

 

[ChrisVer]

well even if $n$ was not a variable, then the quantity I wrote $-x + n \ln x$ (fixed n), is not really going to zero for large x...

oops sorry... you want the logarithm to go to infinity.

 

[Stephen Tashi]

they logarithm-ized the function like:
$\ln (e^{-x} x^n) = -x + n \ln x$ which goes to infinity for $x,n\rightarrow \infty$.
This I read in here:
http://arxiv.org/pdf/1108.4270v5.pdf
in Sec4 (the new paragraph after Eq4.1)

which is on page 10 of the PDF.

Your rule-of-thumb is correct about limits involving single variables.

However, there is a distinction between limits taken with respect to one variable and limits taken with respect to two variables.

$\lim_{s\rightarrow \infty} g(s)$ is defined differently than $\lim_{s\rightarrow\infty,\ n\rightarrow\infty} g(s,n)$.

(The wording in the paper is "when both $n$ and $s$ go to $\infty$".)

There is a further distinction between the definition of a "double limit" $\lim_{s\rightarrow\infty,\ n\rightarrow\infty} g(s,n)$ and the definition of the two "iterated limits":
$\lim_{s\rightarrow\infty}( \lim_{n\rightarrow\infty} g(s,n))$
and
$\lim_{n\rightarrow\infty}( \lim_{s\rightarrow\infty} g(s,n))$.

 

[vanhees71]

The first limit can be evaluated by repeated application of de L'Hospital's rule,
$$\lim_{x \rightarrow \infty} \frac{x^{a}}{\exp x}=\lim_{x \rightarrow \infty} \frac{a x^{a-1}}{\exp x}=\ldots = \lim_{x \rightarrow \infty} \frac{a(a-1) \ldots (a-n+1) x^{a-n}}{\exp x}=0,$$
where I made $n>a-1$.

 

[Stephen Tashi]

where I made $n>a-1$.

L'hospitals rule doesn't apply once the numerator becomes constant, so we should stop when that happens.

 

[vanhees71]

If $a \in \mathbb{N}$ then for $n=a$ the numerator becomes constant, and then the limit is also shown to be 0. So there's nothing wrong arguing with de L'Hospital in all cases.

 

[Stephen Tashi]

So there's nothing wrong arguing with de L'Hospital in all cases.

I'm just saying L'Hospital's rule does not apply to a case like $lim_{x\rightarrow\infty} \frac{6}{f(x)}$ since L'Hospital's rule only applies when both the numerator and denominator both have a limit of zero or both have infinite limits. Once you get a constant like 6 in the numerator, you have to use a different justification for finding the limit.

So we shouldn't write a chain of equalities that implies L'Hospital's rule is being applied to the case where the numerator is constant because it is not, in general, true that $lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} = lim_{x\rightarrow\infty} \frac{f'(x)}{g'(x)} =lim_{x\rightarrow\infty} \frac{f"(x)}{g"(x)} = ... = lim_{x\rightarrow\infty} \frac{\frac{d^n}{dx} f(x)}{\frac{d^n}{dx}g(x)}$, running through an arbitary number $n$ of differentiations.

 

[vanhees71]

Of course it works only for limits of the type "$0/0$" or $\infty/\infty$. Perhaps I was not strict enough in my formulation. I did not mean to use it arbitrarily many times of course but only as long as it is applicable.