Let $\sin^{-1}x = \pi/2-\epsilon$ where $\epsilon\to0^+$ as $x\to1^-$. Then $x=\sin(\pi/2-\epsilon)=\cos\epsilon\approx1-\epsilon^2/2$, so $1-x\approx\epsilon^2/2$.

$\sqrt{\pi}-\sqrt{2(\pi/2-\epsilon)}=\sqrt{\pi}-\sqrt{\pi-2\epsilon}=\sqrt{\pi}\left(1-\sqrt{1-\frac{2\epsilon}{\pi}}\right)\approx\sqrt{\pi}\cdot\dfrac{\epsilon}{\pi}=\dfrac{\epsilon}{\sqrt{\pi}}$

$\sqrt{1-x}\approx\sqrt{\epsilon^2/2}=\epsilon/\sqrt{2}$

Limit $=\dfrac{\epsilon/\sqrt{\pi}}{\epsilon/\sqrt{2}}=\sqrt{\dfrac{2}{\pi}}$