Limits & Continuity
Recursive sequence; limit via trigonometric substitution
MMTS_Full_Test_16
Grade 12
Question:
If $x_1=\sqrt{3}$ and $x_{n+1}=\dfrac{x_n}{1+\sqrt{1+x_n^2}}$ for all $n\in\mathbb{N}$, then $\displaystyle\lim_{n\to\infty} 2^n x_n$ is equal to
(A) $\dfrac{3}{2\pi}$
(B) $\dfrac{2}{3\pi}$
(C) $\dfrac{2\pi}{3}$
(D) $\dfrac{3\pi}{2}$
Step-by-Step Solution
Key Concept: Substitute $x_n=\tan\theta_n$. The recursion becomes $\tan\theta_{n+1}=\frac{\tan\theta_n}{1+\sec\theta_n}=\tan(\theta_n/2)$. So $\theta_n=\theta_1/2^{n-1}$.
$x_n=\tan(\pi/(3\cdot2^{n-1}))$. $\lim 2^n x_n=2\pi/3$.
Correct Answer: (C) $\dfrac{2\pi}{3}$