Let $F(a,n,x)$ be a function defined as $$ F(a,n,x):={_2}F_2\begin{pmatrix}{\begin{matrix} 1, & an+2+ix \\ an+2, & an+3+ix \\ \end{matrix}} & ; -n \end{pmatrix}\\ $$ where $i=\sqrt{-1};a>1 \text{ and is fixed}$;$n\in\mathbb{N}$ and $x\in\mathbb{R}$.
Question (1): If we want an asymptotic expansion of $F(a,n,x)$ (as $n\to\infty$) valid for all $x\in \mathbb{R}$, how many sub cases do we need to consider? Why?
sub case (A) $0\leqslant|x|<n$ as $n\to\infty$;
sub case (B) $|x|\geqslant n$ as $n\to\infty$.
Question (2): Is there a specific name, like "uniform asymptotic expansion", for the asymptotic expansion that we are looking for?
Initially I thought that we only need to consider the sub case where |x|=O(1) as $n\to\infty$. But I am little bit confused now.
Update Here is one such example.
Let $a,b$ be fixed constants and $a>|b|>0;b\in \mathbb{R}$; let $c\in\mathbb{C}$, then for all $c$, we have
$$ \begin{equation} {_2}F_2\begin{pmatrix}{\begin{matrix} 1, & ay+2+c \\ ay+2, & ay+3+c \\ \end{matrix}} & ; b y \end{pmatrix}\\ =\frac{a}{a-b}-\frac{(2a-b)b}{(a-b)^3 y}-\frac{ab}{(a-b)^2 (ay+c) }+O(y^{-2})\quad \text{ as } y\to +\infty \end{equation} $$
Proof: $$ \begin{equation} \frac{(b y)^k}{(ay+2)_k}= \frac{b y}{(ay+2)}\frac{b y}{(ay+2+1)}\cdots\frac{b y}{(ay+2+k-1)}\\ =(b/a)^{k}\left(1-\sum_{j=0]}^{k-1}\frac{2+j}{ay}\right)+O(y^{-2})\\ =(b/a)^{k}\left(1-\frac{k(k+3)}{2ay}\right)+O(y^{-2})\\ \end{equation} $$
sub case (B): If $|c|\geqslant y$, then we set $c=y/\mu,0<|\mu| \leqslant 1$. Thus
$$ \begin{equation} \frac{(ay+2+y/\mu+j)}{(ay+3+y/\mu+j)} =1-\frac{\mu}{(1+a\mu)y}+O(y^{-2})\quad j\in[0,k-1], \end{equation} $$
$$ \begin{equation} \frac{(ay+2+y/\mu)_k}{(ay+3+y/\mu)_k} =\frac{(ay+2+y/\mu)}{(ay+3+y/\mu)}\frac{(ay+2+y/\mu+1)}{(ay+3+y/\mu+1)}\cdots\frac{(ay+2+y/\mu+k-1)}{(ay+3+y/\mu+k-1)}\\ =1-\sum_{j=0}^{k-1}\frac{\mu}{(1+a\mu)y}+O(y^{-2})\\ =1-\frac{k\mu}{(1+a\mu)y}+O(y^{-2})\\ =1-\frac{k}{(a+\mu^{-1})y}+O(y^{-2})\\ \end{equation} $$
$$ \begin{equation} {_2}F_2\begin{pmatrix}{\begin{matrix} 1, & ay+2+c \\ ay+2, & ay+3+c \\ \end{matrix}} & ; b y \end{pmatrix}\\ =\sum_{k=0}^{\infty}\frac{(1)_k}{k!}\frac{(ax+2+c)_k}{(ay+3+c)_k}\frac{(b y)^k}{(ay+2)_k}\\ =\sum_{k=0}^{\infty}(b/a)^k\left(1-\frac{k(k+3)}{2ay}-\frac{k}{(a+\mu^{-1})y}\right)+O(y^{-2})\\ =\frac{a}{a-b}-\frac{(2a-b)b}{(a-b)^3 y}-\frac{1}{(a+\mu^{-1} )}\frac{ab}{(a-b)^2 y}+O(y^{-2})\\ =\frac{a}{a-b}-\frac{(2a-b)b}{(a-b)^3 y}-\frac{ab}{(a-b)^2 (ay+c) }+O(y^{-2}) \end{equation}\tag{1} $$
sub case (A): If $0\leqslant |c|\leqslant y$, then we set $c=y \nu,0\leqslant|\nu| \leqslant 1$. Thus
$$ \begin{equation} \frac{(ay+2+y\nu+j)}{(ay+3+y\nu+j)} =1-\frac{1}{(\nu+a)y}+O(y^{-2})\quad j\in[0,k-1], \end{equation} $$
$$ \begin{equation} \frac{(ay+2+y\nu)_k}{(ay+3+y\nu)_k}\\ =\frac{(ay+2+y\nu)}{(ay+3+y\nu)}\frac{(ay+2+y\nu+1)}{(ay+3+y\nu+1)}\cdots\frac{(ay+2+y\nu+k-1)}{(ay+3+y\nu+k-1)}\\ =1-\sum_{j=0}^{k-1}\frac{1}{(a+\nu)y}+O(y^{-2})\\ =1-\frac{k}{(a+\nu)y}+O(y^{-2})\\ \end{equation} $$
$$ \begin{equation} {_2}F_2\begin{pmatrix}{\begin{matrix} 1, & ay+2+c \\ ay+2, & ay+3+c \\ \end{matrix}} & ; b y \end{pmatrix}\\ =\sum_{k=0}^{\infty}\frac{(1)_k}{k!}\frac{(ax+2+c)_k}{(ay+3+c)_k}\frac{(b y)^k}{(ay+2)_k}\\ =\sum_{k=0}^{\infty}(b/a)^k\left(1-\frac{k(k+3)}{2ay}-\frac{k}{(a+\nu)y}\right)+O(y^{-2})\\ =\frac{a}{a-b}-\frac{(2a-b)b}{(a-b)^3 y}-\frac{1}{(a+\nu )}\frac{ab}{(a-b)^2 y}+O(y^{-2})\\ =\frac{a}{a-b}-\frac{(2a-b)b}{(a-b)^3 y}-\frac{ab}{(a-b)^2 (ay+c) }+O(y^{-2})\\ \end{equation}\tag{2} $$
We remark that the change order of summation with expansion above is justified because ${_2}F_2(a_1,a_2;b_1,b_2;z)$ are absolutely convergent series.
Notice that the last line of (1) is identical to the last line of (2).