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[?] Calculate and interpret Lim (x->0) ln(x!)/x ?

Posted 7 years ago
POSTED BY: Lucas Campos
4 Replies
POSTED BY: Daniel Lichtblau

The other answers are correct, but it's just worth noting that Mathematica uses x! as an abbreviation for Gamma[x+1]. (When x is a nonnegative integer, then the value of Gamma[x+1] is that of Factorial[x].

POSTED BY: Murray Eisenberg
In[2]:= Limit[Log[x!]/x, x -> 0]

Out[2]= -EulerGamma

In[3]:= N[%]

Out[3]= -0.577216
POSTED BY: Frank Kampas

Expanding with series and then taking a limit is simpler to solve:

func = Series[Log[x!]/x, {x, 0, 5}] // Normal
(*-EulerGamma + (?^2 x)/12 + (?^4 x^3)/360 + (?^6 x^5)/5670 + 1/6 x^2 PolyGamma[2, 1] + 1/120 x^4 PolyGamma[4, 1] *)
Limit[func, x -> 0]
(* -EulerGamma *)
% // N
(* -0.577216 *)

Another way is use L'Hôpital's rule. We have [0/0] then:

D[Log[x!], x]/D[x, x] // FullSimplify
(* PolyGamma[0, 1 + x] *)
% /. x -> 0
(* -EulerGamma *)
POSTED BY: Mariusz Iwaniuk
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