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

Posted 5 months ago
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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!! Answer
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Posted 5 months ago
 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 *) Answer
Posted 5 months ago
 In:= Limit[Log[x!]/x, x -> 0] Out= -EulerGamma In:= N[%] Out= -0.577216 Answer
Posted 5 months ago
 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]. Answer
Posted 5 months ago
 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 is given as In:= Integrate[x^(z - 1)*Exp[-x] /. z -> 1, {x, 0, Infinity}] Out= 1 For the first derivative, we differentiate under the integral sign (omitting details as to why this is valid). In:= Integrate[D[x^(z - 1)*Exp[-x], z] /. z -> 1, {x, 0, Infinity}] Out= -EulerGamma The rest involving the log and dividing by z is straightforward using the series of the logarithm centered at 1. Answer
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