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

Question

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?

Answer 1

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

Answer 2

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,

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