Limit involving a hypergeometric function
OK, let's start with an integral representation of that hypergeometric:
$$_2F_1\left(m,n,(m+n);1\frac{r}{z}\right) \\= \frac{1}{B(n,m)} \int_0^1 dx \: x^{(n+1)} (1x)^{(m+1)} \left[1\left(1\frac{r}{z}\right)x\right]^m$$
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
$$\begin{align}\lim_{z \rightarrow 0^+} z^m\ _2F_1\left(m,n,(m+n);1\frac{r}{z}\right) &= \frac{r^m}{B(n,m)} \int_0^1 dx \: x^{(nm+1)} (1x)^{(m+1)}\\ &= r^m \frac{B((nm),m)}{B(n,m)}\\ &= r^m \lim_{\epsilon \rightarrow 0^+} \frac{\Gamma(n+m+\epsilon) \Gamma(nm+\epsilon)}{\Gamma(n+\epsilon)^2} \end{align} $$
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
$$\lim_{z \rightarrow 0^+} z^m \ _2F_1\left(m,n,(m+n);1\frac{r}{z}\right) = r^m \frac{n!^2}{(n+m)! (nm)!}$$
which I believe is equivalent to the stated result.
That last statement is readily seen from writing out the product above:
$$\prod_{i=1}^m \frac{n(i1)}{n+1} = \frac{n}{n+1} \frac{n1}{n+2}\frac{n2}{n+3}\ldots\frac{n(m1)}{n+m}$$
The numerator of the above product is
$$\frac{n!}{(nm)!}$$
and the denominator is
$$\frac{(n+m)!}{n!}$$
The result follows.
M.B.M.
Updated on August 01, 2022Comments

M.B.M. over 1 year
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 nonnegative 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 $L(m,n,r)$:
$$L(m,n,r)=r^m\prod_{i=1}^m\frac{ni+1}{n+i}$$
which one can rewrite 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 nonpositive integer. Any help would be appreciated.

M.B.M. over 10 yearsThanks for the derivation! I have a question: in your first step you bring $z^m$ into the integral and multiply $(1rx/z)^m$ by it. This results in $(zrx)^m$, which goes to $(rx)^m$ as $z\rightarrow 0$ and $r^m$ comes out of the integral. What happens to the negative sign in $(x)^m$ in that step then? Perhaps I am misunderstanding the Beta function...

Ron Gordon over 10 years@M.B.M.: thanks for the catch! I made a mistake in writing down the integral representation, so it all works out.

M.B.M. over 10 years@Ror Gordon: no problem! Another question: what happens when $n=m$? I think your approach still works, unless one needs to treat the pole of the gamma function at zero differently from its poles at negative numbers.

Ron Gordon over 10 years@M.B.M.: I think it works out OK; the pole at $0$ in this case is the same as the others. Just take the limit as $\epsilon \rightarrow 0$ through positive values.

M.B.M. over 10 yearsIt turns out that I used the Pochhammer notation incorrectly in the last line of my question, so I fixed that. Thanks for your help!

M.B.M. about 10 yearsjust want to draw your attention to a related question I posted not long ago. Any help would be greatly appreciated... Thanks!