I apoligize if this is a stupid/obvious question, but last night I was wondering how we can compute limits for factorials of negative integers, for instance, how do we evaluate:
$\lim_{x\to-3}\frac{x!}{(2x)!}=-120$
Neither $x!$, nor $(2x)!$ are defined for $x\in\mathbb{Z}^{-}$, and indeed, both are singularities according to the graph of $\Gamma(x+1)$.
The book I am reading calculates this using a previously shown identity that:
$F\left(\left.{1-c-2n,-2n \atop c}\right|-1\right)=(-1)^{n}\frac{(2n)!}{n!}\frac{(c-1)!}{(c+n-1)!},\space\forall n\in\mathbb{Z}^{*}$
And then, the more general Kummer's Formula:
$F\left(\left.{a,b \atop 1+b-a}\right|-1\right)=\frac{(b/2)!}{b!}(b-a)^{\underline{b/2}}$
It then shows that they would only produce consistent results if:
$(-1)^{n}\frac{(2n)!}{n!}=\lim_{b\to-2n}{\frac{(b/2)!}{b!}}=\lim_{x\to-n}{\frac{x!}{(2x)!}},\space n\in\mathbb{Z}^{*}$
It then gives the example of $n=3$, proving that:
$\lim_{x\to-3}{\frac{x!}{(2x)!}}=-\frac{6!}{3!}=-120$
However, using Wolfram|Alpha, I can see that there are other such limits defined (such as $\lim_{x\to-3}{\frac{x!}{(8x)!}}=-103408066955539906560000$.
Without using the hypergeometric series, how could we evaluate limits such as these?
Again, sorry if this is a stupid question, thanks in advance!