Asymptotic of $_3F_2(1, \frac{1}{2}+d+n, 1+c+n; 1+2c, 2+2n;1)$ as $n\to \infty$ Let $c,d$ be in a small neighborhood of $0$; I think the limit
$$\begin{aligned}&\quad \lim_{n\to \infty} 4^{-n} n^{3c-d} {_3F_2}(1, \frac{1}{2}+d+n, 1+c+n; 1+2c, 2+2n;1) \\
&= \lim_{n\to \infty} 4^{-n} n^{3c-d} \sum_{k\geq 0} \frac{(1/2+d+n)_k (1+c+n)_k}{(1+2c)_k (2+2n)_k} := f(c,d)
\end{aligned}$$
exists and is nonzero. I would like to find it explicitly in terms of $c,d$.

Question: how to find $f(c,d)$?

Using integral representation of $_3F_2$, we have (here $c>0$):
$$\int_{[0,1]^2} (1-x)^{2c-1} y^c (1-y)^{-c} (1-xy)^{-1/2-d} \left(\frac{y(1-y)}{1-xy}\right)^n dxdy \sim \sqrt{\frac{\pi}{n}}\frac{1}{4c} n^{d-3c} f(c,d)$$
this seems amendable via Laplace's method: the maximum of $\frac{y(1-y)}{1-xy}$ in $[0,1]^2$ is $1$; however, this maximum occurs at boundary.
I'm not so familiar with such techniques, but perhaps an expert on asymptotic analysis here can say something. Any suggestion is appreciated.

When $c=0$, we have $f(0,d) = \frac{2 \Gamma \left(\frac{1}{2}-d\right)}{\sqrt{\pi }}$, because in this case, $_3F_2$ can be exactly evaluated in terms of gamma function.
 A: The main idea is to try to factorise the integral.
By means of several substitutions
$$I=\int_0^1\int_0^1ds\,dt'\,(1-t')^{2c-1}(1-t's)^{-1/2-d}\Big(\frac{s}{1-s}\Big)^c\Big(\frac{s(1-s)}{1-t's}\Big)^n\overset{s=\frac{1}{2}(x+1); \,t=1-t'}{=}$$
$$=\frac{2^{\frac{1}{2}+d}}{2^{n+1}}\int_{-1}^1\Big(\frac{1+x}{1-x}\Big)^c\frac{(1+x)^n}{(1-x)^{\frac{1}{2}+d}}dx\int_0^1t^{2c-1}dt\Big(1+t\frac{1+x}{1-x}\Big)^{-\frac{1}{2}-d}\frac{dt}{\big(1+t\frac{1+x}{1-x}\big)^n}$$
$$\overset{z=\frac{1+x}{1-x}}{=}\int_0^\infty z^c\Big(\frac{1}{1+z}\Big)^{\frac{3}{2}-d}\Big(\frac{z}{1+z}\Big)^ndz\int_0^1 t^{2c-1}(1+tz)^{-\frac{1}{2}-d}\frac{dt}{(1+tz)^n}\tag{1}$$
Heuristically (though, it can be also formulated mathematically) the integral over $z$ has the dominated contribution at $z\to\infty$. The integral over $t$, in turn, has the main contribution at $tz\to 0$. If desired, we can calculate several asymptotic terms; for the main term we use
$$\int_0^1 t^{2c-1}(1+tz)^{-\frac{1}{2}-d}\frac{dt}{(1+tz)^n}=\frac{1}{(nz)^{2c}}\int_0^{nz}x^{2c-1}\Big(1+\frac{x}{n}\Big)^{-\frac{1}{2}-d}e^{-n\ln\big(1+\frac{x}{n}\big)}$$
$$=\frac{\Gamma(2c)}{(nz)^{2c}}\Big(1+O\Big(\frac{1}{n}\Big)\Big)\tag{2}$$
We see that effectively we got the integral factorisation (for every asymptotic term). Putting (2) into (1) and making the substitution $z=\frac{1}{x}$
$$I=\frac{\Gamma(2c)}{n^{2c}}\Big(1+O\Big(\frac{1}{n}\Big)\Big)\int_0^\infty\frac{x^{c-\frac{1}{2}-d}}{(1+x)^{\frac{3}{2}-d}}e^{-n\ln(1+x)}dx$$
$$=\frac{1}{\sqrt n}\frac{\Gamma(2c)\,\Gamma\big(c-d+\frac{1}{2}\big)}{n^{3c-d}}\Big(1+O\Big(\frac{1}{n}\Big)\Big)$$
