I have a function
$$f(x)=x^{2m}\text{ }_2F_1\left(\frac{1}{2},-m;\frac{3}{2};-\frac{1}{x^2}\right)$$
where $x>0$. I am interested in asymptotics in the two extreme limits: $$\lim_{x\rightarrow 0} f(x);\qquad \lim_{x\rightarrow \infty}f(x).$$
Any idea on how to proceed? My idea was to use the series representation of hypergeometric function, but that did not help much.
As $x\to\infty$, the series representation of the hypergeometric function is the right approach (since then the argument of $_2F_1$ goes to $0$). We have $$f(x\to\infty)=x^{2m}\sum_{k=0}^{\infty}\frac{\left(-m\right)_k (-1)^kx^{-2k}}{k!(2k+1)}.$$ In particular, the leading asymptotic term is just $x^{2m}$.
As $x\to 0$, use this formula to transform $f(x)$ into a combination of two hypergeometric functions with argument $-x^2$ and then apply previous procedure to get full series expansion in powers of $x$. If you are interested only in the leading asymptotics, note that $a,b$ in $_2F_1(a,b,c;z)$ are critical exponents at $\infty$; therefore the asymptotic result will depend on whether $\Re m>-\frac12$ or $\Re m<-\frac12$. Namely: $$f(x\to0)\sim \begin{cases} \displaystyle \frac{\sqrt\pi\,\Gamma\left(-m-\frac12\right)}{2\Gamma(-m)}x^{2m+1},\qquad& \Re m<-\frac12,\\ \displaystyle \frac1{2m+1},\qquad & \Re m>-\frac12. \end{cases}$$ For $\Re m=-\frac12$, $\Im m\ne 0$, the asymptotics is given by the sum of two terms above. Finally, for $m=-\frac12$, the hypergeometric function in question becomes elementary, so that $$f(x)\Bigl|_{m=-\frac12}=\operatorname{arcsinh} x^{-1}\sim \ln\frac2x\quad \text{as }x\to 0.$$
