Limit of a factorial over an exponential

It's been a while since I've taken Calc II, so I'm having trouble remember the exact steps to solve a problem like this:

limit as x goes to infinity of x!/(x^x).

I know that the bottom grows faster than the top and that the limit is nonzero, but I don't remember how to get to that answer:

There are n factors on the top: n * (n-1) * (n-2) * ....* (n-(n-1)) * 1

And there are n on the bottom: n * n * n .... * n

Every factor on the top is >= n, so if you rewrite the division you get:

(n/n) * ((n-1)/n) * ((n-2)/n) ... * (1/n)

Half of the terms are less than 1/2 and half of the terms are greater than 1/2. Intuitively I want to say that as N goes to infinity, it's a wash and the limit is 1/2, but I don't know how to show that mathematically.

Thanks.

Re: Limit of a factorial over an exponential

Quote:

Originally Posted by

**Open that Hampster!** It's been a while since I've taken Calc II, so I'm having trouble remember the exact steps to solve a problem like this:

limit as x goes to infinity of x!/(x^x).

I know that the bottom grows faster than the top Mr F says: Therefore it should be clear that the limit is equal to zero.

and that the limit is nonzero, Mr F says: NO it's not. See above.

but I don't remember how to get to that answer: Mr F says: Perhaps you should consider the gamma function ..... Furthemore, when x is an integer the analysis is not too difficult.

There are n factors on the top: n * (n-1) * (n-2) * ....* (n-(n-1)) * 1

And there are n on the bottom: n * n * n .... * n

Every factor on the top is >= n, so if you rewrite the division you get:

(n/n) * ((n-1)/n) * ((n-2)/n) ... * (1/n)

Half of the terms are less than 1/2 and half of the terms are greater than 1/2. Intuitively I want to say that as N goes to infinity, it's a wash and the limit is 1/2, but I don't know how to show that mathematically.

Thanks.

..

Re: Limit of a factorial over an exponential

Alright, I'm glad I haven't forgotten as much math as I thought I had.

However, it turns out I was trying to solve a problem that didn't need to be solved. The reason the limit wasn't 0 is because I'm actually supposed to be dealing with this limit:

lim (n->inf) lg(n!)/lg(n^n)

So far I've gotten:

lg(n!) - lg(n^n)

lim lg(n!) - lim n*lg(n)

Which is just infinity - infinity. But since n! doesn't have a derivative, L'Hospitals rule is useless, and I'm not sure I've ever learned a technique to deal with such a limit. Unless I wasn't supposed to break the fraction apart to begin with.

EDIT: Nevermind. You can use Stirling's approximation to show that the limit is equal to 1. This thread can be closed/marked as solved/etc