The function stirling automates the expansion suggested by Stavros.
(%i1) load(stirling)$
(%i2) gamma ((x+1)/2)/(sqrt(x)*gamma (x/2));
(%o2) gamma((x+1)/2)/(sqrt(x)*gamma(x/2))
(%i3) stirling(%);
(%o3) ((x+1)^((x+1)/2-1/2)*2^(x/2-(x+1)/2)*%e^(x/2-(x+1)/2))/x^(x/2)
(%i4) limit(%,x,inf);
(%o4) 1/sqrt(2)
And for fun:
(%i9) gamma ((x+1)/2)/(sqrt(x)*gamma (x/2));
(%o9) gamma((x+1)/2)/(sqrt(x)*gamma(x/2))
(%i10) taylor(stirling(%,4),x,inf,3);
(%o10) sqrt(2)/2-sqrt(2)/(8*x)+sqrt(2)/(64*x^2)+(5*sqrt(2))/(256*x^3)+...
(%i7) taylor(%,x,inf,3);
(%o7) sqrt(2)/2-sqrt(2)/(8*x)+sqrt(2)/(64*x^2)+(5*sqrt(2))/(256*x^3)+...
BW
-----maxima-bounces at math.utexas.edu wrote: -----
>To: "David Joyner" <wdjoyner at gmail.com>
>From: "Stavros Macrakis" <macrakis at alum.mit.edu>
>Sent by: maxima-bounces at math.utexas.edu
>Date: 11/24/2007 03:20PM
>cc: maxima <maxima at math.utexas.edu>
>Subject: Re: [Maxima] a limit in Maxima
>
>I'd think you could use the truncated asymptotic expansion here. Define
>
> gg(x):=x^x*%e^-x*sqrt(2*%pi*x);
>
>then calculate
>
> limit (gg((x+1)/2)/(sqrt(x)*gg(x/2)),x,inf)
>
>
>getting 1/sqrt(2) as expected.
>
> -s
>
>
>On Nov 24, 2007 3:03 PM, David Joyner <wdjoyner at gmail.com> wrote:
>
>A very low-priority question: Out of curiousity,
>a friend asked if Maxima could compute the limit
>limit (gamma ((x+1)/2)/(sqrt(x)*gamma (x/2)),x,inf)
>(I believe that the answer should be 1/sqrt(2)).
>
>
>
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