Is $\lim\limits_{n\to\infty}\Bigl(\frac{2+\sin n}{3}\Bigr)^n =0 $; $n \in \mathbb{N}$?

$$\lim_{n\to\infty}\Bigl(\frac{2+\sin n}{3}\Bigr)^n $$ where $n$ is a natural number.

The problem is if the numerator is less than 3. I think it is because $\sin n$ is never $1$ otherwise $\pi$ would be rational. But i am not sure if the limit exists because $\sin$ oscillates. How to solve this ?

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