Question

If $x_1=\sqrt{3}$ and $x_{n+1}=\frac{x_n}{1+\sqrt{1+x_n^2}}$, $\forall nϵN$, then $\underset{n\to \infty }{lim}{2}^n{x}_n$ equal to

• A. $\frac{3}{2\pi }$
• B. $\frac{2}{3\pi }$
• C. $\frac{2\pi }{3}$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is C $\frac{2\pi }{3}$

$x_{n+1}=\frac{x_n}{1+\sqrt{1+x_n^2}}$
Put $x_1=\tan \theta ;\theta =\frac{\pi }{3}$
$x_2=\frac{\tan \theta }{1+\sec \theta }=\tan \frac{\theta }{2}$
$x_n=\frac{\tan \theta }{1+\sec \theta }=\tan \frac{\theta }{2^{n-1}}$
$\underset{n\to \infty }{lim}{2}^n{x}_n=\underset{n\to \infty }{lim}{2}^n\tan \frac{\theta }{2^{n-1}}={lim}_{n\to \infty }2\theta \frac{\tan \frac{\theta }{2^{n-1}}}{\frac{\theta }{2^{n-1}}}=2\theta =\frac{2\pi }{3}$