$Suppose x_{1} = \tan^{- 1} 2 > x_{2} > x_{3} > \dots \dots \dots . are positive real numbers satisfying$ $\sin (x_{n + 1} - x_{n}) + 2^{- (n + 1)} \sin x_{n} \sin x_{n + 1} = 0 for n > 1 then \lim_{n \to \infty} x_{n} is equal to$

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$\frac{π}{2}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{6}$

answer is B.

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Detailed Solution

$\begin{array}{l} \cot x_{n + 1} - \cot x_{n} = 2^{- (n + 1)} \\ \cot x_{n} - \cot x_{n - 1} = 2^{- n} \\ \cot x_{2} - \cot x_{1} = 2^{- 2} \\ \cot x_{n + 1} - \cot x_{1} = \frac{1}{2^{2}} + \frac{1}{2^{3}} + ....... + \frac{1}{2^{n + 1}} \\ \cot x_{n + 1} = \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ....... + \frac{1}{2^{n + 1}} \\ \lim_{n \to \infty} \cot x_{n + 1} = 1 \\ \lim_{n \to \infty} x_{n + 1} = \lim_{n \to \infty} x_{n} = \frac{π}{4} \end{array}$