Question
Find the limit of the following sequence or determine that the sequence diverges. \( \left.\qquad \frac{\sin 9 n}{4 n}\right\} \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The limit of the sequence is (Type an exact answer.) B. The sequence diverges.
Ask by Romero Gibson. in the United States
Mar 15,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The limit of the sequence is 0.
Solution
Let \( a_n = \frac{\sin 9n}{4n} \).
1. Since the sine function is bounded, we have:
\[
-1 \leq \sin 9n \leq 1.
\]
2. Dividing by \( 4n \) (for \( n > 0 \)), we obtain:
\[
-\frac{1}{4n} \leq \frac{\sin 9n}{4n} \leq \frac{1}{4n}.
\]
3. The limits of the bounding sequences as \( n \to \infty \) are:
\[
\lim_{n \to \infty} \left(-\frac{1}{4n}\right) = 0 \quad \text{and} \quad \lim_{n \to \infty} \frac{1}{4n} = 0.
\]
4. By the Squeeze Theorem, we conclude:
\[
\lim_{n \to \infty} \frac{\sin 9n}{4n} = 0.
\]
Thus, the limit of the sequence is \( 0 \).
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Simplify this solution The Deep Dive
To find the limit of the sequence \( \frac{\sin(9n)}{4n} \), we can analyze its behavior as \( n \) approaches infinity. We know that the sine function oscillates between -1 and 1. Therefore, we can establish the following bounds: \[ -\frac{1}{4n} \leq \frac{\sin(9n)}{4n} \leq \frac{1}{4n}. \] As \( n \) approaches infinity, both \( -\frac{1}{4n} \) and \( \frac{1}{4n} \) converge to 0. By the Squeeze Theorem, since \( \frac{\sin(9n)}{4n} \) is squeezed between two bounds that both converge to 0, we conclude that: \[ \lim_{n \to \infty} \frac{\sin(9n)}{4n} = 0. \] Thus, the correct choice is: A. The limit of the sequence is 0.
