Asymptotic approximation of $_{2}F_{1}(\{1/2- n/2, -n/2\},\{3/2 - n\};z\}$ for $-1/z\rightarrow0$ I'm looking at hypergeometric functions at the moment in relation to the $n$'th term of the Taylor series of $\sqrt{1-a x^2}$. From this consideration a $_{2}F_{1}$ arises that I'd like to approximate using for example techniques outlined in this answer (or similar).
The specific case of interest is $_{2}F_{1}\left(\frac12 - \frac n2, -\frac n2 ,\frac32 - n,z\right)$ where $-\frac1z\rightarrow0$ and $n\in\mathbb{Z}^+$.
Finding a useful transformation to something of an integral representation does not seem straightforward: reduction and integral formulas always seem to divide by $\Gamma(b)$, $\Gamma(c)$, or spawn Pochhammer symbols that I can't justify due to negative integer values appearing in the Gamma functions through which I will divide, leading to divergences.
Something close to a solution that I could see would be the Gegenbauer polynomial double $\nu$ relations inspired by this answer:
\begin{align}
C_{2\nu}^{(\lambda)}(x) &= \binom{2(\nu+\lambda)-1}{2\nu} {}_2F_{1}\left(-\nu,\nu+\lambda,\lambda+\frac12,1-x^2\right)
\end{align}
with the obvious caveat that $c=1+a+b$ such that the exact parameters of the ${}_2F_1$ are not satisfied. Parameter $\lambda$ may become negative with the extension mentioned in this paper so this should not be a problem.
Something else that came close was the limit representation from corollary 6.1  in this paper:
\begin{align}
\lim_{q\rightarrow-\gamma} \frac{{}_2F_1\left(\alpha,\beta,q,x\right)}{\Gamma(q)} &= x^{\gamma+1} \frac{(\alpha)_{\gamma+1}(\beta)_{\gamma+1}}{(\gamma+1)!} {}_2F_1\left(\alpha+\gamma+1,\beta+\gamma+1,\gamma+2,x\right)
\end{align}
So just to be clear: the goal is to find an asymptotic approximation to the aforementioned ${}_2F_1$ with the lowest error possible. I myself thought the integral representation would lend itself to that the best, but I'm very much open to other suggestions. (Expanding the hypergeometric function to first order in $1/z$ is too imprecise to me and I feel like I'm leaving something on the table that way.)
 A: Partial answer but too long for comment.
Using the following relations from Prudnikov the hypergeometric equation can be rewritten into Gamma functions and associated Legendre polynomials. Some of the identities I used include the following:
\begin{align}
\begin{split}
    {}_2F_{1}\left(a,b,c,z\right) &= \frac{\Gamma\left(c\right)\Gamma\left(b-a\right)}{\Gamma\left(b\right)\Gamma\left(c-a\right)}(-z)^{-a} {}_2F_{1}\left(a,1+a-c,1+a-b,\frac1z\right) +\cdots\\
    &\phantom=\cdots+\frac{\Gamma\left(c\right)\Gamma\left(a-b\right)}{\Gamma\left(a\right)\Gamma\left(c-b\right)}(-z)^{-b} {}_2F_{1}\left(b,1+b-c,1+b-a,\frac1z\right)
\end{split}\\
\begin{split}
    {}_2F_{1}\left(a,b,\frac12,z\right) &= \frac{2^{a-b-1}}{\sqrt{\pi}}\Gamma\left(a+\frac12\right) \Gamma\left(1-b\right) (1-z)^{-(a+b)/2}\times\cdots\\
    &\phantom=\cdots\times\left[P^{b-a}_{a+b-1}\left(-\sqrt{\frac{z}{z-1}}\right) + P^{b-a}_{a+b-1}\left(\sqrt{\frac{z}{z-1}}\right)\right]
\end{split}\\
\begin{split}
    {}_2F_{1}\left(a,b,\frac32,z\right) &= \frac{2^{a-b-5/2}}{\sqrt{\pi z}}\Gamma\left(a-\frac12\right) \Gamma\left(b-\frac12\right) (1-z)^{3/4-(a+b)/2}\times\cdots\\
    &\phantom=\cdots\times\left[P^{3/2-(a+b)}_{a-b-1/2}\left(-\sqrt{z}\right) - P^{3/2-(a+b)}_{a-b-1/2}\left(\sqrt{z}\right)\right]
\end{split}
\end{align}
which can be found in Prudnikov et al. Integrals and Series Vol.3, page 454 equation 6 and page 458 equations 74 through 78. Note that the above relations are subject to conditions mentioned there.
I may update this answer, going in a different direction without the associated Legendre polynomials if the asymptotics turn out a little nicer.
