41: Intuitively, the square in the denominator prevails, one power is then lost by cancelling with the numerator, so the whole fraction essentially behaves like 1/n and therefore tends to 0. Proof is simple, this is a textbook example of a ratio of polynomials:

So the sequence is something that goes to zero times sine. Since the oscillation of sine is bounded, the whole sequence tends to zero. After all, there is a theorem about "bounded times zero gives zero", see the box "comparison and oscillation".

Referring to the theorem actually completes the solution. If this did not convince you, or you want to try another argument, you can use another approach from the box on oscillating expressions. Since sine has a bounded oscillation, with a little bit of luck it can be proved by the Squeeze theorem that the limit is zero. In fact, for zero limits we have an easier version, the absolute value squeeze. Try to set it up.

Next hint
Answer