How to calculate this limit (if exist)? I have just asked the calculation of the following summation see here $$S(a,b,m,n_1,n_2)=\sum_{k=0}^m a^k b^{m-k} {n_1\choose k} {n_2\choose m-k}, $$
which is motivated by the calculation of the following limit (if exist)
$$L_0=\lim_{m \to \infty}\frac{\sqrt{ab}S(a,b,m-1,n_1-1,n_2-1)}{S(a,b,m,n_1,n_2)}$$
where $a>0, b>0, \frac{m}{n_1},\frac{m}{n_2}$ are kept fixed.
My question is: Whether the above limit exists ($a\neq b$)? If it exists, then how to calculate it? When $a=b$, the result is simple $\lim_{m \to \infty}\frac{a\cdot a^{m-1} {n_1+n_2-2\choose m-1}}{a^{m} {n_1+n_2\choose m}}=\lim_{m \to \infty}\frac{m(n_1+n_2-m)}{(n_1+n_2)(n_1+n_2-1)}$ and this limit obviously exists, where we have used $S(a,b,m,n_1,n_2)=a^m {n_1+n_2\choose m}$ for $a=b$.
Despite the nice comments and answers in the previous question, I still don't know how to calculate the limit.
 A: $\newcommand\ka\kappa\newcommand{\de}{\delta}\newcommand\ep\varepsilon$The problem obviously reduces to this one: find
\begin{equation}
    L:=\lim_{m\to\infty}\frac{S(c,m,n_1,n_2)}{S(c,m-1,n_1-1,n_2-1)},
\end{equation}
where $c:=a/b\ne1$ and
\begin{equation}
    S(c,m,n_1,n_2):=\sum_{k=0}^m c^k A_k(m,n_1,n_2), 
\end{equation}
where
\begin{equation}
    A_k(m,n_1,n_2):=\binom{n_1}k \binom{n_2}{m-k}. 
\end{equation}
In accordance with the linked post, assume that  $m<n_1$ and $m<n_2$, so that $m\to\infty$,
\begin{equation}
    n_1/m\to\nu_1\in[1,\infty),\quad n_2/m\to\nu_2\in[1,\infty). 
\end{equation}
By considering the ratios $r_k:=c^{k+1}A_{k+1}(m,n_1,n_2)/(c^k A_k(m,n_1,n_2))$, one sees that
\begin{equation}
    S(c,m,n_1,n_2)\sim\sum_{k\sim \ka m} c^k A_k(m,n_1,n_2), 
\end{equation}
where
\begin{equation}
    \ka:=\frac{c \nu _1-\sqrt{\left(c \nu _1+c+\nu _2-1\right){}^2-4 (c-1) c \nu _1}+c+\nu _2-1}{2 (c-1)}\in(0,1). 
\end{equation}
(More specifically, note that $r_k\ge1$ for $k\le k_*$ and $r_k\le1$ for $k\ge k_*$, for a certain integer $k_*\sim\ka m$. Next, note that for each real $\ep>0$ there is some $\de>0$ such that for all large enough $m$ and all $k=0,\dots,m-1$ we have the implication
if $k/m-\ka\ge\de\implies r_k<1-\ep$. So, for integers $k\ge(\ka+3\de)m$ we will have $c^k A_k(m,n_1,n_2)\le(1-\ep)^{\de m}c^{k_*+1} A_{k_*+1}(m,n_1,n_2)$ and hence
$\sum_{k\ge (\ka+3\de)m} c^k A_k(m,n_1,n_2)\le m (1-\ep)^{\de m} c^{k_*+1} A_{k_*+1}(m,n_1,n_2)
=o(\sum_{|k-\ka m|\le3\de m} c^k A_k(m,n_1,n_2))$.
Similarly, $\sum_{k\le (\ka-3\de)m} c^k A_k(m,n_1,n_2)
=o(\sum_{|k-\ka m|\le3\de m} c^k A_k(m,n_1,n_2))$.
Finally here, note that we can choose $\de>0$ to be however small.)
Also, for $k\sim \ka m$,
\begin{equation}
    \frac{A_k(m-1,n_1-1,n_2-1)}{A_k(m,n_1,n_2)}
    =\frac{(m-k)(n_1-k)}{n_1 n_2}\sim\frac{(1-\ka )(\nu_1-\ka )}{\nu_1\nu_2}. 
\end{equation}
So,
\begin{equation}
    L=\frac{(1-\ka )(\nu_1-\ka )}{\nu_1\nu_2}.
\end{equation}
