Evaluate $\lim_{n\to-\infty}a_{n}$ and determine whether the sequence converges or diverges.
$a_{n}=300\sin\left(\frac{n^2-120}{n}\right)$

• A. $a_{n}=300$, converges.
• B. $a_{n}=300$, diverges.
• C. $a_{n}=DNE$, converges.
• D. $a_{n}=DNE$, diverges.

Answer: (D)

Verified step by step guidance

1. First, identify the sequence given: a_n = 300\sin\left(\frac{n^2 - 120}{n}\right).
2. To evaluate the limit as n approaches negative infinity, consider the expression inside the sine function: \frac{n^2 - 120}{n}.
3. Simplify \frac{n^2 - 120}{n} to n - \frac{120}{n}. As n approaches negative infinity, \frac{120}{n} approaches 0, so the expression simplifies to n.
4. The sine function, \sin(n), oscillates between -1 and 1 for all real numbers n. Therefore, \sin(n) does not settle to a single value as n approaches negative infinity.
5. Since \sin(n) oscillates indefinitely, the sequence a_n = 300\sin(n) does not converge to a single value, and thus the limit does not exist (DNE), indicating divergence.