JEE Main 2019 — Mathematics Calculus

Given limit can be written as

$$L=\underset{x\rightarrow {1}^{-}}{\mathrm{lim}}\frac{\sqrt{\pi }-\sqrt{2{\mathrm{sin}}^{-1}x}}{\sqrt{1-x}}\times \frac{\sqrt{\pi }+\sqrt{2{\mathrm{sin}}^{-1}x}}{\sqrt{\pi }+\sqrt{2{\mathrm{sin}}^{-1}x}}$$L=x→1−lim​1−x​π​−2sin−1x​​×π​+2sin−1x​π​+2sin−1x​​

$$\Rightarrow L=\underset{x\rightarrow {1}^{-}}{\mathrm{lim}}\frac{\pi -2{\mathrm{sin}}^{-1}x}{\sqrt{1-x} (\sqrt{\pi }+\sqrt{2{\mathrm{sin}}^{-1}x})}=\underset{x\rightarrow {1}^{-}}{\mathrm{lim}}\frac{2{\mathrm{cos}}^{-1}x}{\sqrt{1-x} (\sqrt{\pi }+\sqrt{2{\mathrm{sin}}^{-1}x})}$$⇒L=x→1−lim​1−x​(π​+2sin−1x​)π−2sin−1x​=x→1−lim​1−x​(π​+2sin−1x​)2cos−1x​

Let $K=\underset{x\rightarrow {1}^{-}}{\mathrm{lim}}\frac{{\mathrm{cos}}^{-1}x}{\sqrt{1-x}}$K=x→1−lim​1−x​cos−1x​ and put $x=\mathrm{cos}\theta$x=cosθ, we get

$$K=\underset{\theta \rightarrow 0}{\mathrm{lim}}\frac{\frac{\theta }{2}.2}{\sqrt{2}.\mathrm{sin}\frac{\theta }{2}}=\underset{\theta \rightarrow 0}{\mathrm{lim}}\frac{\sqrt{2}}{\frac{\mathrm{sin}\frac{\theta }{2}}{\frac{\theta }{2}}}=\sqrt{2}[\because \underset{x\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}=1]$$K=θ→0lim​2​.sin2θ​2θ​.2​=θ→0lim​2θ​sin2θ​​2​​=2​[∵x→0lim​xsinx​=1]

$\therefore L=\frac{2\sqrt{2}}{2 \sqrt{\pi }}=\sqrt{\frac{2}{\pi }}$∴L=2π​22​​=π2​​.