Question

Let the sequence $\{x_n\}_{n \ge 1}$ be given by $x_n = \sin \dfrac{n\pi}{6}$, $n = 1, 2, \ldots$. Then which of the following statements is/are TRUE?
 

• A. The sequence $\{x_n\}$ has a subsequence that converges to $\dfrac{1}{2}$
• B. $\limsup_{n \to \infty} x_n = 1$
• C. $\liminf_{n \to \infty} x_n = -1$
• D. The sequence $\{x_n\}$ has a subsequence that converges to $\dfrac{1}{\sqrt{2}}$

Hint:

For trigonometric sequences like $\sin(n\pi/k)$, note their periodicity and find limits from distinct repeating values.

Answer 1

The Correct Option is A, B, C

Step 1: Determine periodicity.
Since $x_n = \sin\left(\dfrac{n\pi}{6}\right)$, the sequence is periodic with period 12 because $\sin(\theta + 2\pi) = \sin(\theta)$.

Step 2: List the values for one full cycle.
\[ x_1 = \tfrac{1}{2}, x_2 = \tfrac{\sqrt{3}}{2}, x_3 = 1, x_4 = \tfrac{\sqrt{3}}{2}, x_5 = \tfrac{1}{2}, x_6 = 0, \] \[ x_7 = -\tfrac{1}{2}, x_8 = -\tfrac{\sqrt{3}}{2}, x_9 = -1, x_{10} = -\tfrac{\sqrt{3}}{2}, x_{11} = -\tfrac{1}{2}, x_{12} = 0. \]

Step 3: Identify subsequences.
- Subsequence with $x_3, x_{15}, \ldots$ gives limit $1$. - Subsequence with $x_9, x_{21}, \ldots$ gives limit $-1$. - Subsequence with $x_1, x_5, \ldots$ gives limit $\frac{1}{2}$.

Step 4: Compute $\limsup$ and $\liminf$.
\[ \limsup_{n \to \infty} x_n = 1, \liminf_{n \to \infty} x_n = -1. \]

Step 5: Conclusion.
\[ \boxed{\text{(A), (B), and (C) are correct.}} \]