_ x → 1 ^ - π - 2 ^ - 1 x 1 - x is equal to

Given limit can be written as

$L = \lim_{x \to {1}^{-}}\dfrac{\sqrt{\pi} - \sqrt{2\sin^{- 1}x}}{\sqrt{1 - x}} \times \dfrac{\sqrt{\pi} + \sqrt{2\sin^{- 1}x}}{\sqrt{\pi} + \sqrt{2\sin^{- 1}x}}$

$⇒ L = \lim_{x \to {1}^{-}}\dfrac{\pi - 2\sin^{- 1}x}{\sqrt{1 - x}\left(\sqrt{\pi} + \sqrt{2\sin^{- 1}x}\right)} = \lim_{x \to {1}^{-}}\dfrac{2\cos^{- 1}x}{\sqrt{1 - x}\left(\sqrt{\pi} + \sqrt{2\sin^{- 1}x}\right)}$

Let $K = \lim_{x \to {1}^{-}}\dfrac{\cos^{- 1}x}{\sqrt{1 - x}}$ and put $x = cos\theta$, we get

$K = \lim_{\theta \to 0}\dfrac{\dfrac{\theta}{2} . 2}{\sqrt{2} . sin\dfrac{\theta}{2}} = \lim_{\theta \to 0}\dfrac{\sqrt{2}}{\dfrac{sin\dfrac{\theta}{2}}{\dfrac{\theta}{2}}} = \sqrt{2}\left[∵ \lim_{x \to 0}\dfrac{\sin x}{x} = 1\right]$

$∴ L = \dfrac{2\sqrt{2}}{2\sqrt{\pi}} = \sqrt{\dfrac{2}{\pi}}$.