Behavior of the Gaussian Hypergeometric function when one of its arguments approaches $0$ or $1$ For two positive integers $a,b$, denote by $_2F_1(a,1-b;a+1;z)$ the Gaussian Hypergeometric function whose first three parameters are fixed at $a,1-b$ and $a+1$, respectively. such function is linked to the regularized incomplete Beta function (namely the CDF of a Beta distribution) by the relation:
$$
B(z;a,b)=\frac{z^a}{a} {}_2F_1(a,1-b;a+1;z).
$$
I was wondering whether there exist functions $f,\, g$, such that
$$
f(z)\sim {}_2F_1(a,1-b;a+1;z) \quad (z\downarrow 0),\\
g(z)\sim {}_2F_1(a,1-b;a+1;z) \quad (z\uparrow 1).
$$
In particular, I would be interested in the case $a,b\geq 1$.
 A: We have
$$
\eqalign{
  & F(z,a,b) = {}_2F_{\,1} \left( {\left. {\matrix{
   {a,\;1 - b}  \cr 
   {a + 1}  \cr 
 } \;} \right|\;z} \right) =   \cr 
  &  = \sum\limits_{0\, \le \,k} {{{a^{\,\overline {\,k\,} } \left( {1 - b} \right)^{\,\overline {\,k\,} } } \over {\left( {a + 1} \right)^{\,\overline {\,k\,} } }}
{{z^{\,k} } \over {k!}}}  =   \cr 
  &  = a\sum\limits_{0\, \le \,k} {{{\left( {1 - b} \right)^{\,\overline {\,k\,} } } \over {\left( {a + k} \right)}}{{z^{\,k} } \over {k!}}}  =   \cr 
  &  = a\sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,k} \left( {b - 1} \right)^{\,\underline {\,k\,} } } \over {\left( {a + k} \right)}}{{z^{\,k} } \over {k!}}}   \cr 
  &  = a\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{
  b - 1 \cr 
  k \cr}  \right){{z^{\,k} } \over {\left( {a + k} \right)}}}  \cr} 
$$
Then, for the first limit you have
$$
\mathop {\lim }\limits_{z\, \to \,0 + } F(z,a,b) = 1
$$
Continuing the development
$$
\eqalign{
  & F(z,a,b) = a\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{
  b - 1 \cr 
  k \cr}  \right){{z^{\,k} } \over {\left( {a + k} \right)}}}  =   \cr 
  &  = a\,z^{\, - \,a} \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{
  b - 1 \cr 
  k \cr}  \right){{z^{\,a + k} } \over {\left( {a + k} \right)}}}  =   \cr 
  &  = a\,z^{\, - \,a} \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{
  b - 1 \cr 
  k \cr}  \right)\int_{t = 0}^z {t^{\,a + k - 1} } dt}  =   \cr 
  &  = a\,z^{\, - \,a} \int_{t = 0}^z {t^{\,a - 1} \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{
  b - 1 \cr 
  k \cr}  \right)} \left( { - t} \right)^{\,k} } dt =   \cr 
  &  = a\,z^{\, - \,a} \int_{t = 0}^z {t^{\,a - 1} \left( {1 - t} \right)^{\,b - 1} } dt \cr} 
$$
which gives the relation with the Incomplete Beta function you already know.
$$
F(z,a,b) = a\,z^{\, - \,a} \,{\rm B}(z\,;a,b)\quad \left| {\;0 < a,b} \right.
$$
From here we get
$$
\eqalign{
  & \mathop {\lim }\limits_{z\, \to \,1 - } F(z,a,b) = a\int_{t = 0}^1 {t^{\,a - 1} \left( {1 - t} \right)^{\,b - 1} } dt =   \cr 
  &  = a\,{\rm B}(a,b) \cr} 
$$
and since (re. for instance to this)
$$
{\rm B}(z\,;a,b) = {{z^{\,\,a} } \over a}\left( {1 + O(z)} \right)
$$
also 
$$
\mathop {\lim }\limits_{z\, \to \,0 + } \left( {F(z,a,b) = a\,z^{\, - \,a} \,{\rm B}(z\,;a,b)} \right) = 1\quad \left| {\;0 < a,b} \right.
$$
and you can use the above expression for the whole domain of your interest.
