Solve this limit $\lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}$ I am trying to figure out how to make this limit, even with the hopital. I've tried using hopital two times, but the situation 0/0 is still there. I've tried to solve it using wolfram, but I don't the solution. Even with rationalization + hopital nothing comes out. $$\lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}$$
I wonder if there is some way to solve it, and would really appreciate any suggestion.

\begin{align}
\lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}&=\lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1-4x^2}}\frac{2\sqrt{x-2x^2}}{1-4x}, \text{  L'hopital}\\
&= \lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1-2x}\sqrt{1+2x}}\frac{2\sqrt{x}\sqrt{1-2x}}{1-4x}\\
&=\lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1+2x}}\frac{2\sqrt{x}}{1-4x}\\
&=\frac{2}{\sqrt{2}}\frac{2\sqrt{\frac{1}{2}}}{1-2}
\\&=-2\end{align}
