If $\left|{\cos }^{-1}\frac{1}{n}\right|<\frac{\pi }{2},$ then $\underset{n\to \mathrm{\infty }}{lim} \left\{(n+1)\frac{2}{\pi }{\cos }^{-1}\frac{1}{n}-n\right\}$ is equal to

Solution:

   $\underset{n\to \mathrm{\infty }}{lim} \left\{(n+1)\frac{2}{\pi }{\cos }^{-1}\frac{1}{n}-n\right\}$

$\begin{array}{l}=\underset{n\to \mathrm{\infty }}{lim} \frac{2}{\pi }\left\{(n+1){\cos }^{-1}\frac{1}{n}-\frac{\pi }{2}n\right\}\\ =\underset{n\to \mathrm{\infty }}{lim} \frac{2}{\pi }\left\{n\left({\cos }^{-1}\frac{1}{n}-\frac{\pi }{2}\right)+{\cos }^{-1}\frac{1}{n}\right\}\\ =\underset{n\to \mathrm{\infty }}{lim} \frac{2}{\pi }\left\{-n{\sin }^{-1}\frac{1}{n}+{\cos }^{-1}\frac{1}{n}\right\}\end{array}$

$=\underset{n\to \mathrm{\infty }}{lim} \frac{2}{\pi }\left\{-\frac{{\sin }^{-1}{\displaystyle \frac{1}{n}}}{\frac{1}{n}}+{\cos }^{-1}\frac{1}{n}\right\}=\frac{2}{\pi }\left(-1+\frac{\pi }{2}\right)=\frac{\pi -2}{\pi }$