Question

Let \[ a_n=\frac{\sin n}{n} \] be a sequence of real numbers. Then, \[ \lim_{n\to\infty} a_n \] is

• A. 0
• B. 1
• C. Does not exist
• D. \(\frac{1}{2}\)

Hint:

Do not confuse \( \lim_{n \to \infty} \frac{\sin n}{n} \) with the famous limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).
When the variable approaches infinity, any bounded term divided by infinity always goes to 0:
\[ \frac{\text{Bounded}}{\infty} = 0 \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question requires finding the limit of a sequence containing a trigonometric function as \( n \) approaches infinity.
This is a classical problem that is easily solved using the Squeeze Theorem (or Sandwich Theorem) in real analysis.

Step 2: Detailed Explanation:
Let us analyze the behavior of the terms in the sequence:
\[ a_n = \frac{\sin n}{n} \] - The sine function is a bounded function. Regardless of how large \( n \) becomes, the value of \( \sin n \) always remains restricted within the closed interval \( [-1, 1] \):
\[ -1 \le \sin n \le 1 \quad \text{for all } n \in \mathbb{R} \] - Since the sequence index \( n \) is a positive integer (\( n > 0 \)), we can divide the entire inequality by \( n \) without reversing the direction of the inequality signs:
\[ -\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n} \] - Now, let us evaluate the limits of the lower and upper bounding sequences as \( n \to \infty \):
- Lower bound limit:
\[ \lim_{n \to \infty} \left(-\frac{1}{n}\right) = 0 \] - Upper bound limit:
\[ \lim_{n \to \infty} \left(\frac{1}{n}\right) = 0 \] - According to the Squeeze Theorem, if a sequence is bounded between two other sequences that both converge to the same limit \( L \), then the squeezed sequence must also converge to \( L \).
Since both bounding limits are 0, we conclude:
\[ \lim_{n \to \infty} \frac{\sin n}{n} = 0 \]

Step 3: Final Answer:
The limit of the sequence as \( n \to \infty \) is 0.
Therefore, the correct choice is Option (A).