Question

Let $x_0=2\cos \frac{\pi }{6}$ and $x_n=\sqrt{2+x_{n-1}}$, $n=1,2,3,...............,$ find $\underset{n\to \infty }{Lim}{2}^{(n+1)}.\sqrt{2-x_n}$

• A. $\frac{\pi }{3}$
• B. $\frac{-\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{-\pi }{6}$

Answer 1

The correct option is A $\frac{\pi }{3}$

Let $x_0=2cos\theta$
$x_1=\sqrt{2+2\cos \theta }=2\cos \theta /2$
$x_2=\sqrt{2+2\cos \theta /2}=2\cos \theta /4$
$x_n=2\cos \frac{\theta }{2^n}$
Now $\underset{n\to \infty }{lim}{2}^{n+1}\sqrt{2-2\cos \frac{\theta }{2^n}}=\underset{n\to \infty }{lim}\frac{2\sin \frac{\theta }{2^{n+1}}}{\frac{1}{2^{n+1}}}$
$=\theta =2.\frac{\pi }{6}=\pi /3$