I Does this go to 0 for large enough x?

1. Aug 3, 2016

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 $x,n\rightarrow \infty$.
http://arxiv.org/pdf/1108.4270v5.pdf
in Sec4 (the new paragraph after Eq4.1)

This has confused me, can someone shred some light?

2. Aug 3, 2016

micromass

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

3. Aug 3, 2016

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.

4. Aug 3, 2016

Stephen Tashi

which is on page 10 of the PDF.

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
$\lim_{n\rightarrow\infty}( \lim_{s\rightarrow\infty} g(s,n))$.

5. Aug 4, 2016

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 $n>a-1$.

6. Aug 4, 2016

Stephen Tashi

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

7. Aug 5, 2016

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.

8. Aug 5, 2016

Stephen Tashi

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.

9. Aug 5, 2016

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.