# Compute $\lim_{x\rightarrow 0 }\biggr ( \dfrac{1}{x}\ln (x!)\biggr )$

I want to compute the following limit

$$\lim_{x\rightarrow 0 }\biggr ( \dfrac{1}{x}\ln (x!)\biggr )$$

Since factorial is only defined for integers, we must use the gamma function.

$$\lim_{x\rightarrow 0} \dfrac{\ln (\Gamma (x+1))}{x} = \lim_{x\rightarrow 0}\dfrac{\dfrac{d}{dx}(\ln(\Gamma(x+1))}{\dfrac{d}{dx}(x)} = \lim_{x\rightarrow 0} \dfrac{\Gamma'(x+1)}{\Gamma(x+1)} = \psi(1)$$

Where $$\psi$$ is the digamma function.

$$\psi(x+1) = -\gamma +\int^{1}_{0}\dfrac{1-t^{x}}{1-t}dt$$

What we want is

$$\psi(1) = -\gamma +\int^{1}_{0}\dfrac{1-t}{1-t}dt = -\gamma$$

where $$\gamma$$ is Euler-Mascheroni constant.

Is there a way to compute this limit without using digamma function?

• Have you tried Stirling's approximation? Oct 25, 2019 at 10:46
• No, I haven't....
– Melz
Oct 25, 2019 at 10:47
• @CadeReinberger Stirling's approximation holds as $x\to\infty$. Here we care about the behavior around $x=1$. Oct 25, 2019 at 10:49
• The limit is the Eulergamma constant Oct 25, 2019 at 10:52
• @Dr.SonnhardGraubner If you read the question, you will find out it's actually $-\gamma$ and OP knows that. They are interested in a proof not using the digamma function. Oct 25, 2019 at 10:54

If you know the series expansion for $$\log\Gamma(1+x)$$, then $$\lim_{x\to 0}\frac{1}{x}\log\Gamma(x+1)=\lim_{x\to 0}\frac{1}{x}\left(-\gamma x +O(x^2)\right)=-\gamma.$$

Hint:

$$\log((x+n)!)=\log(x!)+\sum_{k=1}^n\log(x+k)$$

and

$$(\log(x+n)!)'=(\log x!)'+\sum_{k=1}^n\frac1{x+k}$$

and by letting $$x$$ tend to $$0$$,

$$(\log(n)!)'=(\log 0!)'+\sum_{k=1}^n\frac1{k}$$

Then using Stirling and the asymptotic formula for the Harmonic numbers,

$$\log n\sim(\log 0!)'+\log n-\gamma.$$

• Interesting approach, but I don't see how you get an asymptotic on the derivative from Stirling. Oct 25, 2019 at 13:37
• @Wojowu: derivative of $\log((x/e)^x)$. The other terms can be dropped.
– user65203
Oct 25, 2019 at 14:06