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This is the problem I am trying to solve:

$$\lim_{x \to \infty} \frac{e^x x!}{x^x\sqrt{x}}.$$

I believe this is an indefinite form, thus use L'Hospitals's rule.

But the problem I am having is how to find the derivative of $x!$

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    I assume you mean $\lim_{x \to \infty}$...2012-12-05
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    @Jebruho yes I did.2012-12-05
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    http://www.wolframalpha.com/input/?i=derivative+of+x%212012-12-05
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    Look up the Gamma function.2012-12-05
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    @amzoti, I don't understand much of what wolfmanalpha returns. I just look up the terms and then get more confused. hehe2012-12-05
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    Well then, you had better resort to Stirling's Asymptotic Formula, which really simplifies your problem.2012-12-05
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    @Amzoti give me time, I just started calc 12012-12-05
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    Finding the limit is *equivalent* to proving Stirling's Formula. This is not at all easy.2012-12-05

2 Answers 2

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Use the Stirling asymptotic formula, $$x!\approx x^x e^{-x}\sqrt{2\pi x}$$ for $x\gg 1$.

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    is that the only method?2012-12-05
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    surely not, but you will need some specific results about factorials or about the gamma function.2012-12-05
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    Well if you want to use L'Hospitals rule you will have to replace $x!$ with a differentiable function which happens to equal $n!$ at non-negative integers $n$, the natural way to do this is via the Gamma function. As stated, $x!$ is only defined for integers $x$, so your expression is in fact a sequence, not a function on the positive reals.2012-12-05
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    @SeanGomes You seem to imply that there is an alternative method of solving this that not need L`Hospitals rule.2012-12-05
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    The above answer provides an example of such a method: Stirling's formula. This formula can be proven without any mention of the Gamma function.2012-12-05
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    The limit is indeed $\sqrt{2\pi}$. Stirling's approximation doesn't need the gamma function. And if you use Stirling's approximation, then you don't need L'Hopital's rule either. In fact, its almost "trivial" after you use Stirling's approximation. Everything just cancels.2012-12-05
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    Isn't this taking the answer for granted? (This query is addressed to the OP, not the answerer.)2012-12-05
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Let $f(x) = \frac{e^x x!}{x^x\sqrt{x}}$. Then $\frac{f(x+1)}{f(x)} = \frac{e^{x+1} (x+1)!}{(x+1)^{x+1}\sqrt{x+1}}/ \frac{e^x x!}{x^x\sqrt{x}} = \frac{e^{x+1} (x+1)!}{(x+1)^{x+1}\sqrt{x+1}} \frac{x^x\sqrt{x}}{e^x x!} = \frac{e (x+1) x^x \sqrt{x}}{(x+1)^{x+1} \sqrt{x+1}} = \frac{e x^x }{(x+1)^x \sqrt{1+1/x}} = \frac{e }{(1+1/x)^x \sqrt{1+1/x}} = \frac{e }{(1+1/x)^{x+1/2}} $.

This and what follows is just replicating what Stirling and those who preceded him did to get "Stirling's formula".

At this point, we need to show that $(1+1/x)^{x+1/2}$ is very close to $e$ for large $x$.

Tak the logs, so we are looking at $(x+1/2) \ln(1+1/x) = (x+1/2)(1/x - 1/(2x^2) + 1/(3x^3) + ... = (1 - 1/(2x) + 1/(3x^2) + ...) + (1/(2x) - 1/(4x^2) + ...) =1 + 1/(12x^2) + ...$.

Note that the $1/2$ in $x+1/2$ allows the $1/x$ term to cancel. That is why $(1+1/x)^{x+c}$ is closest to $e$ for $c = 1/2$. For any other real $c$, $\lim \frac{e}{(1+1/x)^{x+c}} = 1$, But the product of these terms does not cenverge unless $c = 1/2$.

Exponentiating this, $(1+1/x)^{x+1/2} = e e^{1/(12x^2) + ...} = e(1+1/(12x^2) + ...) $ so $\frac{e}{(1+1/x)^{x+1/2}} = \frac{1}{1+1/(12x^2) + ...} = 1-1/(12x^2) + ... $.

In all this, the "..." represents terms of higher order in $x$.

We now know that the ratio of consecutive terms is (1) less than one (though we would have to prove that the higher order terms are smaller than the $1/(12x^2)$ term), and (2) is close enough to one that the product of $1-1/(12x^2)$ converges.

If we look at $\prod_{x=n}^m (1-1/(12x^2))$ and take the log, we find, by an analysis similar to that above, the sum of the logs converges as $m \to \infty$, so the product converges.

Since the product $\frac{f(x+1)}{f(x)}$ converges, $f(x)$ must tend to a limit. To show this, let $P_n = \prod_{x=n}^{\infty} \frac{f(x+1)}{f(x)}$. Then $P_n = 1/f(n)$ and $\lim_{n \to \infty} P_n$ exists.

This is not completely rigorous, but it can be made so, and it was. It does show why the limit exists.

Actually, Stirling's contribution was to explicitly evaluate the limit - it had previously been shown that the limit exists.

This rambling essay was done off the top of my head at 11:30 at night. I hope it helps.

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    Which shows that the limit *exists*, but does not provide its value.2012-12-05