What is the asymptotic behavior of factorial The question of the asymptotic behavior of $n!$ when $n\to\infty$ usually arises in calculation of time complexity for algorithms and relates to calculation of the following limit.
$$\lim_{n\to+\infty}\frac{\ln(n!)}{n\ln(n)}$$
At the first sight, I figured out that it is an $\frac{\infty}{\infty}$ indeterminate form. I found it hard to use the L'Hopital rule, so I decided to solve this via founding some bounds on $\ln(n!)$. First, I noted that
$$\ln(n!)=\ln(n\times(n-1)\times\dots\times 1)=\ln(n)+\ln(n-1)+\dots+\ln(1)=\sum_{i=1}^{n}\ln(i)\tag{1}$$
and consequently
$$\int_{1}^{n}\ln(x)dx<\sum_{i=1}^{n}\ln(i)\lt\sum_{i=1}^{n}\ln(n),\tag{2}$$
where I used the fact that $(1)$ is an upper Reimann sum for the shown integral. Simplifying $(2)$ yields
$$n\ln(n)-n+1<\sum_{i=1}^{n}\ln(i)\lt n\ln(n).\tag{3}$$
Dropping the $1$ on the left most inequality and dividing by $n\ln(n)$ leads to
$$1-\frac{1}{\ln(n)}<\frac{\ln(n!)}{n\ln(n)}\lt 1,\tag{4}$$
Finally, I used the sandwich theorem for limits to get
$$\lim_{n\to+\infty}\frac{\ln(n!)}{n\ln(n)} = 1$$
Is my approach OK? What are other ways to tackle this problem?
 A: To answer your titular question:
$$n!\sim n^ne^{-n}\sqrt{2\pi n}\left(1+\frac{1}{12n}+\cdots\right)$$
A nice truncation of the latter series (the "Stirling series") due to Ramanujan is:
$$n!\sim n^ne^{-n}\sqrt{\pi}\cdot\sqrt[6]{8n^3+4n^2+n+\frac{1}{30}}$$
But you can omit the tail series for an asymptotic approximation in the right-half complex plane if I remember correctly (it becomes a somewhat divergent series, losing accuracy as more terms beyond a certain number are given in the left-half plane) and the most commonly seen version is just $n!\sim n^ne^{-n}\sqrt{2\pi n}$. Certainly the series is good for the real axis.
So:
$$\frac{\ln(n!)}{n\ln n}\sim\frac{n\ln n-n+\ln(\sqrt{2\pi})+\frac{1}{2}\ln n}{n\ln n}=\frac{n+1/2}{n}+\frac{\ln(\sqrt{2\pi})-n}{n\ln n}$$
From which convergence to $1$ is clear.
Your approach was ok, and certainly more direct!
For further reading see "Stirling's approximation" which can be proved in general for the Gamma function. The full series can be derived from Binet's formulae for the log-gamma function.
For real positive $x$, and complex $z$ restricted off the negative real axis (although it does diverge...): $$\begin{align}\ln(\Gamma(x+1))=\ln(x!)&=(x+1/2)\ln x-x+\frac{1}{2}\ln2\pi+\sum_{n=1}^\infty\frac{B_{2n}}{2n(2n-1)}x^{1-2n}\\&=(x+1/2)\ln x-x+\frac{1}{2}\ln2\pi+\left(\frac{1}{12x}-\frac{1}{360x^3}+\frac{1}{1260x^5}-\cdots\right)\\\Gamma(x+1)=x!&=x^xe^{-x}\sqrt{2\pi x}\cdot\exp\left(\frac{1}{12x}-\frac{1}{360x^3}+\frac{1}{1260x^5}-\cdots\right)\\&=x^xe^{-x}\sqrt{2\pi x}\cdot\left(1+\frac{1}{12x}+\frac{1}{288x^2}-\frac{139}{51840x^3}+\cdots\right)\end{align}$$
Where $B_n$ are the Bernoulli numbers (positive or negative it doesn't matter).
