Help with a limit involving incomplete beta integral In trying to prove that the limit of a certain function approaches 1 as the  positive integer parameter $n$ approaches infinity, I have ended up with the following intermediate expressions:
$$f(n)=2^{1+2n}B_{1/2}(n,n+2)$$
$$g(n)=4^nB_{1/2}(n+1,n)$$
$$ h(n)=n(n-1)/2 \left( \frac{f(n)}{n+1}-\frac{g(n)}{n-1}\right)$$
Can somebody kindly help me with the evaluation of $ \lim_{n \to \infty} h(n)$? If somebody could also plug it in Mathematica, I would be highly obliged. Thanks for any help in advance.
P.S.: In the above the notation $B_z(a,b)$ stands for the incomplete beta function defined by:
$$B_z(a,b)=\int\limits_0^z u^{a-1}(1-u)^{b-1} \mathrm{d}u.$$
 A: Using the substitution $u=(1-s)/2$ in the relevant integrals $\int_0^{1/2} u^{a-1}(1-u)^{b-1} \,du$, then using the substitution $s^2=t$, and finally noting that $\Gamma(x+1/2)/\Gamma(x)\sim\sqrt x$ as $x\to\infty$, we have
$$h(n)=\frac n{2(n+1)}\,\int_0^1 ds\,(1 - s^2)^{n - 1} ((n - 1) s^2 + (3 n - 1) s - 2) \\ 
=\frac n{8(n+1)}\,{\left(6-\frac2n-\frac{3 \sqrt{\pi }\, \Gamma (n+2)}{n\Gamma
   \left(n+3/2\right)}\right)}\to\frac68$$
as $n\to\infty$.
A: The blue line is the numerical evaluation of $h(n)$, the gold line is
$$H(n)=\frac{3}{4}-\frac{3 \sqrt{\pi } n \Gamma (n)}{8 \Gamma \left(n+\frac{3}{2}\right)}\rightarrow \frac{3}{4}$$

A: Python gives values upto $n=1000$. Values of $h(n)$ for $997 \leq n \leq 1000$ are $f(997)= 0.728174227325497 ,f(998)=0.7281857363386501 ,f(999)= 0.7281972275318511 ,f(1000)=0.7282087009506559$. Also, $f(900)=0.7269653136612485 $

The Code:
import matplotlib.pyplot as plt 
import numpy as np 
import math as m 

def f(x):
    f=0
    g=0
    h=0.001
    for i in range(0,2000):
        u=i*h
        f=f+2*(u**(x-1))*((1-u/4)**(x+1))*h*(x-1)
        g=g+0.25*(u**x)*((1-u/4)**(x-1))*h*(x+1)  

    v=x/(x+1)*(f-g)/2
    return v

x=np.linspace(2, 1000, 1000)
F=np.vectorize(f)
plt.plot(x,F(x),'r')
plt.xlabel('n')
plt.ylabel('h(n)')
plt.show()

