How to calculate $\lim\limits_{x\to\infty} \frac{\log(x!)}{x\log(x)}$. Assume base $e$ (so $\ln)$.
My attempt:
$$\lim_{x\to\infty} \frac{\log(x!)}{x\log(x)}=\lim_{x\to\infty}\frac{\log(1\cdot 2\cdot 3\cdots x)}{x\log x}=\lim_{x\to\infty}\frac{\log(ax)}{x\log x}, a\gt 0$$
Applying LH rule:
$$\lim_{x\to\infty} \frac{\frac{1}{x}}{\log(x)+1}=\lim_{x\to\infty}\frac{1}{x(\log(x)+1)}=0$$
Wolfram tells me the answer is $1$. Where is my mistake?
On
By Stirling's Approximation: $\log n! = n\log n-n$ as $n \to \infty$. As pointed out in the commentsby @stochasticboy321 and @saulspatz you misjudged the order of the term $\log n!$.
$$\lim_{n\to \infty} \dfrac{\log n!}{n\log n}=\lim_{n\to \infty}\left(1-\dfrac{1}{\log n}\right)= 1$$
On
Hint (with the same caveat that @JustDroppedIn): write $$\log(n!) = \log(1) + \cdots + \log(n)$$ and apply Stolz–Cesàro.
First of all, if $x$ is a real parameter, the "factorial" doesn't make sense (unless it's some notation for the Gamma function). Anyway, I'll use $n$ instead of $x$ for an integer variable; So you want to compute $$\lim_{n\to\infty}\frac{\log(n!)}{n\log(n)}$$ Well, first of all, observe that $\log(n!)=\log(1)+\dots+\log(n)\leq\log(n)+\dots+\log(n)=n\log(n)$ so your limit (if existent) is $\leq 1$. On the other hand, let $n\in\mathbb{N}$ and $f(x)=\log(x)$. Let $P$ be the partition $\{1,\dots n\}$ of the interval $[0,n]$. Then We have that $\displaystyle{\log(n!)=\sum_{k=1}^{n}\log(k)=U(f,P)}$, i.e. the upper Darboux sum of $f$ for this partition; The Riemann integral $\displaystyle{\int_1^n\log(x)dx}$ is defined as the infimum of all upper Darboux sums of $f$ over all possible partitions of $[1,n]$, hence $\log(n!)\geq\int_1^n\log(t)dt=n\log(n)-n+1$. Deviding on both sides with $n\log(n)$ yields $$\frac{\log(n!)}{n\log(n)}\geq1-\frac{n-1}{n\log(n)}$$ Taking limits and observing that the RHS limit is $1$ gives the rest.