Consider a function sequence $\{f_n(t)\}$ ($n\in\mathbb{N}^+$) defined on the interval $(\frac{1}{2},1)$, where \begin{eqnarray}\label{eqn:constraint1} f_n(t)=\frac{\exp\left(n\left(\log R(h_t) - th_t\right)\right)}{h_t\sigma(h_t)\sqrt{2\pi n}}\left(1+o\left(1\right)\right) \end{eqnarray} as $n\to \infty$ uniformly in $t$ in any closed subinterval of the interval $(\frac{1}{2},1)$, \begin{eqnarray} R(h)=\frac{e^h-1}{h}, \end{eqnarray} \begin{eqnarray} m(h)=\frac{R'(h)}{R(h)}=\frac{e^h}{e^h-1}-\frac{1}{h}, \end{eqnarray} \begin{eqnarray} \sigma(h)=m'(h), \end{eqnarray} and $h_t$ is the positive real root of the equation $m(h)=t$.

Then we study the following equation of $g_n$ \begin{eqnarray} c=g_n+\frac{\int_{g_n}^1f_n(t)~dt}{f_n(g_n)},~\frac{1}{2}< g_n<1, \end{eqnarray} where the constant number $\frac{1}{2}< c<1$.

Try to prove:

- the convergence of the sequence $\{g_n\}$ and compute $A=\lim_{n\to \infty}g_n$;
- $\lim_{n\to \infty}n(g_n-A)=0$.