If cos−1⁡1n<π2, then limn→∞ (n+1)2πcos−1⁡1n−n is equal to

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

1. A

$\frac{2-\pi }{\pi }$

2. B

$\frac{\pi -2}{\pi }$

3. C

1

4. D

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### Solution:

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

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

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

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