How can I find the limit $\lim_{x\to \frac12}\frac{4x^2 - 1}{\arcsin(1 - 2x)} $

Question

How can I find the limit

$$\lim_{\left(x\to 1/2\right)}\ \frac{4x^2 - 1}{\arcsin(1 - 2x)}\quad ? $$

Answer 1

HINT

We have

$$\frac{4x^2 - 1}{\arcsin(1 - 2x)}=\frac{(2x-1)(2x+1)}{1-2x}\frac{1-2x}{\arcsin(1 - 2x)}$$

then refer to the standard limit as $t\to 0\quad \frac{\arcsin t}{t}\to 1$.

Answer 2

Hint:

Obvious substitution

$\arcsin(1-2x)=y,2x=1-\sin y$

$y\to0,$ as $x\to1/2$