Compute $\lim\limits_{x \to \infty} \frac{\ln (x!)}{x \ln (x)}$ I was just playing with logarithms and factorials, and then realized that $f(x) = \frac{\ln (x!)}{\ln(x)}$ is slower than $f(x) = x$. From there I got $\frac{\ln x!}{(x \ln (x))}$. In most calculators, I can't find the limit, and when I put a ridiculously large number it doesn't go below 0.6. So is the limit 0?
If the limit is considered along $\mathbb{N}$ it is enough to invoke Cesàro-Stolz:
$$\lim_{n\to +\infty}\frac{\log(n!)}{n\log n}=\lim_{n\to +\infty}\frac{\log(n+1)}{(n+1)\log(n+1)-n\log n}=\lim_{n\to +\infty}\frac{\log(n+1)}{\log(n+1)+\underbrace{n\log\left(1+\frac{1}{n}\right)}_{\to 1}}=1. $$
If the limit is considered along $\mathbb{R}$ it is enough to invoke De l'Hospital:
$$ \lim_{x\to +\infty}\frac{\log\Gamma(x+1)}{x\log x} = \lim_{x \to +\infty}\frac{\psi(x+1)}{1+\log x}=\lim_{x\to +\infty} x\cdot\psi'(x+1)=\lim_{x\to +\infty}x\sum_{n\geq 1}\frac{1}{(n+x)^2} $$
where
$$\sum_{n\geq 1}\frac{1}{(n+x)^2}&gt;\sum_{n\geq 1}\frac{1}{(n+x)(n+x+1)}=\frac{1}{x+1}, $$
$$\sum_{n\geq 1}\frac{1}{(n+x)^2}&lt;\frac{1}{(x+1)^2}+\sum_{n\geq 1}\frac{1}{(n+x)(n+x+1)}=\frac{x+2}{(x+1)^2}.$$
Stirling's approximation:
$$\frac{\ln (x!)}{x \ln (x)}\sim \frac{\ln (x!)}{\ln (x^x)}\sim \frac{x!}{x^x}\sim \frac{(x/e)^x\sqrt{2\pi x}}{x^x}\sim\frac{\sqrt{2\pi x}}{e^x}\to 0.$$