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

limn→∞ (n+1)2πcos−1⁡1n−n=limn→∞ 2π(n+1)cos−1⁡1n−π2n=limn→∞ 2πncos−1⁡1n−π2+cos−1⁡1n=limn→∞ 2π−nsin−1⁡1n+cos−1⁡1n=limn→∞ 2π−sin−11n1n+cos−11n=2π−1+π2=π−2π

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

limn→∞ (n+1)2πcos−1⁡1n−n=limn→∞ 2π(n+1)cos−1⁡1n−π2n=limn→∞ 2πncos−1⁡1n−π2+cos−1⁡1n=limn→∞ 2π−nsin−1⁡1n+cos−1⁡1n=limn→∞ 2π−sin−11n1n+cos−11n=2π−1+π2=π−2π

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If $|\cos^{- 1} \frac{1}{n}| < \frac{π}{2},$ then $lim_{n \to \infty} \{(n + 1) \frac{2}{π} \cos^{- 1} \frac{1}{n} - n\}$ is equal to

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$\frac{2 - π}{π}$

$\frac{π - 2}{π}$

answer is B.

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Detailed Solution

$lim_{n \to \infty} \{(n + 1) \frac{2}{π} \cos^{- 1} \frac{1}{n} - n\}$

$\begin{array}{l} = lim_{n \to \infty} \frac{2}{π} \{(n + 1) \cos^{- 1} \frac{1}{n} - \frac{π}{2} n\} \\ = lim_{n \to \infty} \frac{2}{π} \{n (\cos^{- 1} \frac{1}{n} - \frac{π}{2}) + \cos^{- 1} \frac{1}{n}\} \\ = lim_{n \to \infty} \frac{2}{π} \{- n \sin^{- 1} \frac{1}{n} + \cos^{- 1} \frac{1}{n}\} \end{array}$

$= lim_{n \to \infty} \frac{2}{π} \{- \frac{\sin^{- 1} \frac{1}{n}}{\frac{1}{n}} + \cos^{- 1} \frac{1}{n}\} = \frac{2}{π} (- 1 + \frac{π}{2}) = \frac{π - 2}{π}$