Asymptotic behavior of a hypergeometric function Can anybody see how to deduce an asymptotic formula for the hypergeometric function
$$ _3F_2\left(\frac{1}{2},x,x;x+\frac{1}{2},x+\frac{1}{2} \hskip2pt\bigg|\hskip2pt 1\right), \quad\mbox{ as } x\to\infty?$$
For the standard definition of the hypergeometric series, see here.
Remark: I've tried to combine the numerous transformations and integral representations which $_3F_2$-functions fulfill. I've also tried to get inspired by existing works on asymptotic expansions of similar (but simpler) $_2F_1$-functions but never ended up with the desired result.
 A: $\newcommand\Ga\Gamma$Let $f(x)$ denote your hypergeometric expression. Then
$$f(x)\sim\sqrt{\pi x}$$
(as $x\to\infty$).
Indeed,
$$f(x)=\sum_{k\ge0}\frac{(1/2)_k}{k!}r_k(x)^2,$$
where $(a)_k:=a(a+1)\cdots(a+k-1)=\Ga(a+k)/\Ga(a)$ and
$$r_k(x):=\frac{(x)_k}{(x+1/2)_k}=\frac{\Ga(x+k)}{\Ga(x)}\Big/\frac{\Ga(x+1/2+k)}{\Ga(x+1/2)}
\sim\sqrt{\frac x{x+k}}$$
uniformly in $k\ge0$ (as $x\to\infty$).
So,
\begin{equation*}
    f(x)\sim x\,\sum_{k\ge0}\frac{(1/2)_k}{k!} \frac1{x+k}
=\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)}
\sim\sqrt{\pi x} \tag{1}\label{1}
\end{equation*}
(as $x\to\infty$).

Details on the equality in \eqref{1}:
\begin{equation*}
\begin{aligned}
    \sum_{k\ge0}\frac{(1/2)_k}{k!} \frac1{x+k}
&   =\sum_{k\ge0}\frac{(1/2)_k}{k!} \int_0^1 dt\,t^{x+k-1} \\ 
&   =\int_0^1 dt\,t^{x-1}\sum_{k\ge0}\frac{(1/2)_k}{k!} t^k  \\ 
&   =\int_0^1 dt\,t^{x-1}(1-t)^{-1/2}, 
\end{aligned}
\end{equation*}
in view of the Maclaurin series for $(1-t)^{-1/2}$.
Therefore,
\begin{equation*}
\begin{aligned}
    x\,\sum_{k\ge0}\frac{(1/2)_k}{k!} \frac1{x+k}
=x\,B(x,1/2)= x\,\frac{\Ga(1/2)\, \Ga(x)}{\Ga(x+1/2)}
=\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)}, 
\end{aligned}
\end{equation*}
as claimed.
A: Use the identity
$$_3F_2\left(\frac{1}{2},x,x;x+\frac{1}{2},x+\frac{1}{2} \bigg|1\right)=\frac{\sqrt{\pi }\, \Gamma \left(x+\frac{1}{2}\right)}{\Gamma (x)}\,_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,x+\frac{1}{2};1\right),$$
and the large-$x$ limits
$$_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,x+\frac{1}{2};1\right)\rightarrow 1,$$
$$\frac{\sqrt{\pi }\, \Gamma \left(x+\frac{1}{2}\right)}{\Gamma (x)}\rightarrow\sqrt{\pi x},$$
to conclude that
$$_3F_2\left(\frac{1}{2},x,x;x+\frac{1}{2},x+\frac{1}{2} \bigg|1\right)\rightarrow \sqrt{\pi x}.$$
A numerical test shows that the $\sqrt{\pi x}$ asymptote (gold) is nearly indistinguishable from the exact function (blue) for $x$ larger than 4.

