Limit involving a hypergeometric function I am new to hypergeometric function and am interested in evaluating the following limit:
$$L(m,n,r)=\lim_{x\rightarrow 0^+} x^m\times {}_2F_1\left(-m,-n,-(m+n);1-\frac{r}{x}\right)$$
where $n$ and $m$ are non-negative integers, and $r$ is a positive real constant.
However, I don't know where to start.  I did have Wolfram Mathematica symbolically evaluate this limit for various values of $m$, and the patters seems to suggest the following expression for $L(m,n,r)$:
$$L(m,n,r)=r^m\prod_{i=1}^m\frac{n-i+1}{n+i}$$
which one can re-write using the Pochhammer symbol notation as follows:
$$L(m,n,r)=r^mn\frac{(n)_m}{n^{(m)}}$$
If the above is in fact correct, I am interested in learning how to derive it using "first principles" as opposed to the black box that is Wolfram Mathematica.  I am really confused by the definition of hypergeometric function ${}_2F_1(a,b,c;z)$, as the defition that uses the Pochhammer symbol in the wikipedia page excludes the case that I have where $c$ is a non-positive integer.  Any help would be appreciated.
 A: OK, let's start with an integral representation of that hypergeometric:
$$_2F_1\left(-m,-n,-(m+n);1-\frac{r}{z}\right) \\= \frac{1}{B(-n,-m)} \int_0^1 dx \: x^{-(n+1)} (1-x)^{-(m+1)} \left[1-\left(1-\frac{r}{z}\right)x\right]^m$$
where $B$ is the beta function.  Please do not concern yourself with poles involved in gamma functions of negative numbers for now: I will address this below.  
As $z \rightarrow 0^+$, we find that
$$\begin{align}\lim_{z \rightarrow 0^+} z^m\ _2F_1\left(-m,-n,-(m+n);1-\frac{r}{z}\right) &= \frac{r^m}{B(-n,-m)} \int_0^1 dx \: x^{-(n-m+1)} (1-x)^{-(m+1)}\\ &= r^m \frac{B(-(n-m),-m)}{B(-n,-m)}\\ &= r^m \lim_{\epsilon \rightarrow 0^+} \frac{\Gamma(-n+m+\epsilon) \Gamma(-n-m+\epsilon)}{\Gamma(-n+\epsilon)^2} \end{align} $$
Note that, in that last line, I used the definition of the Beta function, along with a cautionary treatment of the Gamma function near negative integers, which are poles.  (I am assuming that $n>m$.)  The nice thing is that we have ratios of these Gamma function values, so the singularities will cancel and leave us with something useful.
I use the following property of the Gamma function (see Equation (41) of this reference):
$$\Gamma(x) \Gamma(-x) = -\frac{\pi}{x \sin{\pi x}}$$
Also note that, for small $\epsilon$
$$\sin{\pi (n-\epsilon)} \approx (-1)^{n+1} \pi \epsilon$$
Putting this all together (I leave the algebra as an exercise for the reader), I get that
$$\lim_{z \rightarrow 0^+} z^m \ _2F_1\left(-m,-n,-(m+n);1-\frac{r}{z}\right) = r^m \frac{n!^2}{(n+m)! (n-m)!}$$
which I believe is equivalent to the stated result.  
That last statement is readily seen from writing out the product above:
$$\prod_{i=1}^m \frac{n-(i-1)}{n+1} = \frac{n}{n+1} \frac{n-1}{n+2}\frac{n-2}{n+3}\ldots\frac{n-(m-1)}{n+m}$$
The numerator of the above product is 
$$\frac{n!}{(n-m)!}$$
and the denominator is
$$\frac{(n+m)!}{n!}$$
The result follows.
