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?

  • $\begingroup$ Have you tried Stirling's approximation? $\endgroup$ Oct 25, 2019 at 10:46
  • $\begingroup$ No, I haven't.... $\endgroup$
    – Melz
    Oct 25, 2019 at 10:47
  • $\begingroup$ @CadeReinberger Stirling's approximation holds as $x\to\infty$. Here we care about the behavior around $x=1$. $\endgroup$
    – Wojowu
    Oct 25, 2019 at 10:49
  • $\begingroup$ The limit is the Eulergamma constant $\endgroup$ Oct 25, 2019 at 10:52
  • $\begingroup$ @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. $\endgroup$
    – Wojowu
    Oct 25, 2019 at 10:54

2 Answers 2


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





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

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,

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.