$\limsup$ and $\liminf $ of $\sin ({1\over {n^2}}) $

Question

The sequence given is $a_n=\sin ({1\over {n^2}}) $. Now question what are it's limit supremum and infimum?

I know that for $x\ge 0$, $|\sin x|\le x $.Using that we can write that $$\left|\sin \left({1\over n^2}\right)\right|\le {1\over n^2}\implies -{1\over n^2}\le \sin\left({1\over n^2}\right) \le {1\over n^2}$$ So we have the possible infimum and supremum both of which go to $0$ if the limit is taken to $\infty$.

From this can I say $\liminf$ and $\limsup$ are equal and the sequence converges to $0$?

Seems really easy. Is it correct? If not please explain to me what I need to do.

Thank You.

Answer 1

If $(x_n)$ is a convergent sequence, then

$$ \lim \sup x_n= \lim \inf x_n= \lim x_n.$$

Your sequence $(a_n)$ is covergent, $ \lim a_n=0$