How to prove that $(1-x)^b$ $_2F_1(a,b;c;x)$ can be approximated to $1-\alpha x$ (with $\alpha \approx 1$) for $x\ll 1$ in this specific case After multiple plots I noticed that function $h(x)= (1-x)^b$ $_2F_1(a,b;c;x)$ can be approximated to  $1-\alpha x$ (with $\alpha \approx 1$), for $x\ll 1$ (specifically $0<x<0.1$) and $a=(K-1)d$, $b=K$, c=$Kd$; with $K \gg 1$, $K \gg d$ and $d>1$. 
In other words, I need to prove that the first derivative of $h(x)$ is constant ($\approx -1$) for $0<x<0.1$, Or equivalently that :$-b(1-x)^{b-1} \, _2F_1(a,b;c;x) + \frac{ab}{c} (1-x)^b  \, _2F_1(a+1,b+1;c+1;x) \approx \text{constant} (\approx -1)$.
 A: The first few terms of the Maclaurin series can be obtained explicitly:
$$1-x-{\frac { \left( d-1 \right)  \left( K-1 \right) }{2\,Kd+2}}{x}^{2
}-{\frac { \left( d-1 \right)  \left( d-2 \right)  \left( K-1 \right) 
 \left( K-2 \right) }{ \left( 6\,Kd+6 \right)  \left( Kd+2 \right) }}{
x}^{3}+O \left( {x}^{4} \right) 
$$
The coefficient of $x^2$ is not especially small, but of course on a small
interval around $0$ the $1-x$ dominates.  I guess the question is really about the behaviour as $K \to \infty$ for fixed $x$.
The Maclaurin coefficient of $x^n$ in $h(x)$ is
$$ A_n(K,d) = \sum_{ k=0}^n (-1)^k {K \choose k}  \dfrac{\Gamma(K+n-k) \Gamma(Kd-d+n-k) \Gamma(Kd)}{\Gamma(K) \Gamma(Kd-d) \Gamma(Kd+n-k) (n-k)!}$$
$A_n(K,d)$ is a rational function of $K$ and $d$ where it appears that numerator and denominator both have degree $n-1$ in $K$ for $n \ge 1$.
Thus as $K \to +\infty$, $h(x)$ may be approaching a smooth limit.  In 
fact, it looks to me like
$$ \lim_{K \to \infty} h(x) = \left( \dfrac{x-1}{(1-1/d) x - 1}\right)^d $$
For example, with $d=3$ and $K=20$ the maximum difference between $h(x)$ and $((x-1)/(2x/3-1))^3$ on the interval $(-1,1)$ is about $0.01388$. 
