$\underset{x→-1}{\lim}\frac{\sqrt{π}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}}$

$\frac{1}{\sqrt{2π}}$

$\underset{x→-1}{\lim}\frac{\sqrt{π}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}}=\underset{x→-1}{\lim}\frac{π-\cos^{-1}x}{\sqrt{x+1}[\sqrt{π}+\sqrt{\cos^{-1}x}]}=\frac{1}{2\sqrt{π}}\underset{x→-1}{\lim}\frac{π-\cos^{-1}x}{\sqrt{x+1}}$

$=\frac{1}{2\sqrt{π}}\underset{x→-1}{\lim}\frac{(\frac{1}{\sqrt{1-x^2}})}{\frac{1}{2\sqrt{1+x}}}$ [Apply LH rule]

$=\frac{1}{2\sqrt{π}}\underset{x→-1}{\lim}\frac{1}{\sqrt{1-x^2}}×2\sqrt{1+x}=\frac{1}{\sqrt{π}}\underset{x→-1}{\lim}\frac{1}{\sqrt{1-x}}=\frac{1}{\sqrt{2π}}$