Limit of ratio of two Gamma functions with negative integer arguments When using the hypergeometric representation for a Legendre polynomial, I encounter, for integer n and l, the following ratio: $$\frac{\Gamma(n-l)}{\Gamma(-l)}$$
Where $n \leq l$ (the quantity is definitely zero for $n > l$, as it should be in the definition of a Legendre polynomial). I am unsure as to how to evaluate this ratio; as it stands, it is indeterminate. My original idea was to use: $$\Gamma(k+1) = (k+1)\Gamma(k)$$
Multiple times to reduce $\Gamma(n-l)$ to: $$(n-l)(n-l-1)(...)(-l+1) \ \Gamma(-l)$$
Then the $\Gamma(-l)$ terms would cancel and I'd be left only with some sensible terms. Unfortunately, I do not think that this approach is valid, as it does not yield the correct representation for the Legendre polynomial. Secondly, the above can be written as: $$(-1)^n\frac{(l-1)!}{(l-n-1)!} \ \Gamma(-l)$$
Which now no longer permits us to set $n=l$ as is necessary to obtain a polynomial of order $l$. I'm trying to remain brief on the references to Legendre polynomials as it is specifically the ratio of the Gamma functions listed at the start of this post that I am interested in evaluating.

We have for $k\in\mathbf{N}$ and $x\rightarrow 0$
$$\Gamma(-k+x) \sim \frac{1}{k!x} + O(1)$$
and therefore 
$$\lim_{x\rightarrow 0}\frac{\Gamma(n-l+x)}{\Gamma(-l+x)} = \frac{1}{(l-n)!}/\frac{1}{l!} = \frac{l!}{(l-n)!}$$
Of course this is only a very special limit. It is finite for $n=l$ but I do not know what you expect for the value. If the above does reproduce it, you will at least have a simple arugument.
