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

Posted 5 months ago
4 Replies
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Ploting this in geogebra gives an answer close to -0.5772, but I don't get why. Pls help me understand it. Also, it seems to be the same value for the derivative of ln(x!) when x->0, but I'm not sure about it. I came to this problem when trying to figure out what lim (x->0) x!^-x was, and I then found out that this is the same as lim (x->0) e^(ln(x!)/x), meaning that if i understand the "power" part, then I'll get the whole problem. I know it's another case where you have 0/0, but I still want to understand where does that answer from geogebra comes from. Plss help me!!

4 Replies

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

func = Series[Log[x!]/x, {x, 0, 5}] // Normal
(*-EulerGamma + (π^2 x)/12 + (π^4 x^3)/360 + (π^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 *)
In[2]:= Limit[Log[x!]/x, x -> 0]

Out[2]= -EulerGamma

In[3]:= N[%]

Out[3]= -0.577216

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].

To add to the response by @MurrayEisenberg, this can be used to explain why the limit is what it is. Without going through all details we'll just investigate the series for Gamma[1+z] at z=0. This amounts to looking at Gamma[z] and first derivative at z=1. Recall gamma definition as

Gamma[z] == Integrate[x^(z-1) Exp[-x], {x,0,Infinity}]

So Gamma[1] is given as

In[68]:= Integrate[x^(z - 1)*Exp[-x] /. z -> 1, {x, 0, Infinity}]

Out[68]= 1

For the first derivative, we differentiate under the integral sign (omitting details as to why this is valid).

In[69]:= Integrate[D[x^(z - 1)*Exp[-x], z] /. z -> 1, {x, 0, Infinity}]

Out[69]= -EulerGamma

The rest involving the log and dividing by z is straightforward using the series of the logarithm centered at 1.

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