JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Limits

As $x\to 1^{-}$ both numerator and denominator vanish, so rationalise the numerator by multiplying above and below by $\sqrt{\pi }+\sqrt{2{\sin }^{-1}x}$. The numerator becomes $\pi -2{\sin }^{-1}x=2\left(\frac{\pi }{2}-{\sin }^{-1}x\right)=2{\cos }^{-1}x$, giving ${\lim }_{x\to 1^{-}}\frac{2{\cos }^{-1}x}{\sqrt{1-x}\left(\sqrt{\pi }+\sqrt{2{\sin }^{-1}x}\right)}$. Substitute $x=\cos \theta$ with $\theta \to 0^{+}$, so ${\cos }^{-1}x=\theta$ and $\sqrt{1-x}=\sqrt{1-\cos \theta }=\sqrt{2}\sin \frac{\theta }{2}$. As $\theta \to 0^{+}$, $\frac{\theta }{\sin (\theta /2)}\to 2$ and $\sqrt{\pi }+\sqrt{2{\sin }^{-1}x}\to 2\sqrt{\pi }$. Hence the limit is $\frac{2\theta }{\sqrt{2}\sin (\theta /2)\cdot 2\sqrt{\pi }}\to \frac{2\cdot 2}{2\sqrt{2}\sqrt{\pi }}=\sqrt{\frac{2}{\pi }}$.