How to find $\lim_{x\to0^+}\frac{1 - \frac{2}{\pi} \arcsin(_{2}F_{1}(1/2, 1/2, 3/2, x^4))}{x^2}$? Does the limit exist? 
$$\lim_{x\to0^+}\frac{1 - \displaystyle \frac{2}{\pi} \arcsin(_{2}F_{1}(1/2, 1/2, 3/2, x^4))}{x^2}$$
 A: We may use a classical hypergeometric identity (equation $(14)$ from Cook's 'Notes on hypergeometric functions') :
$$_{2}F_{1}\left(\frac 12, \frac 12; \frac 32; z^2\right)=\frac {\arcsin(z)}{z}$$
to rewrite your limit as :
$$\lim_{x\to0^+}\frac{1 - \displaystyle \frac{2}{\pi} \arcsin\left(\frac {\arcsin\left(x^2\right)}{x^2}\right)}{x^2}$$
and use the Taylor expansion : $\,\displaystyle\frac {\arcsin\left(x^2\right)}{x^2}=1+\frac{x^4}6+O\bigl(x^8\bigr)$
Note that you could obtain this expansion directly from the hypergeometric expansion :
$$_2F_1\left(\begin{array}{c} \frac 12, & \frac 12\ \\\qquad \frac 32  \end{array};\ x^4\right)=\sum_{j=0}^{\infty}\frac{\bigl(\frac 12\bigr)_j\bigl(\frac 12\bigr)_j}{\bigl(\frac 32\bigr)_j}\frac {x^{4j}}{j!}$$
with $(a)_0:=1,\ (a)_j:=a(a+1)(a+2)\cdots(a+j-1)\,$ for $j>0$ .
Anyway you need only to compute the limit :
$$\lim_{x\to0^+}\frac{1 - \displaystyle \frac{2}{\pi} \arcsin\left(1+\frac{x^4}6+O\bigl(x^8\bigr)\right)}{x^2}$$
The numerator becomes $0$ as $x\to 0^+$ and you may apply for example l'Hôpital's rule to get :
$$L=\lim_{x\to0^+}\frac {-\frac 2{\pi} \frac{4x^3+O\bigl(x^7\bigr)}{6\sqrt{1-\left(1+\frac{x^4}6+O\bigl(x^8\bigr)\right)^2}}}{2x}=\frac {-2}{\sqrt{-3}\pi}$$
The choice of $\sqrt{3}\;i$ or $-\sqrt{3}\;i$ for the root of the negative value $-3$ remains to you (this imaginary value comes from evaluating $\arcsin$ for a value greater than $1$, this could be too a sign missing in your problem...)
Hoping this helped,
