Answer:
limit n→[infinity], sin(3n) neither converges nor diverges.
Detailed explanation below
Step-by-step explanation:
The trigonometric function/sequence sin(3n) continues to assume a periodic values as the values of n continue to increase. That is, n→[infinity]. sin(3n) neither diverges nor converges to a limit, uniformly. Rather, the sequence continue to diverge and converge repeatedly as n→[infinity].
This is because sin(3n) will form a periodic and oscillatory sequence as n tends to infinity.
