I need to prove that: $\lim \limits_{n \to \infty} \frac{n \sin(n)}{2+n^2}=0$

so here is what I did:

For each $\epsilon>0$ ,there is an $N \in$ $\mathbb N$, such that for each $n>N$:

$$ \left\lvert \frac{n\sin n}{2+n^2} \right\rvert = \frac{\lvert n\sin n\rvert}{\left\lvert n^2+2\right\rvert} \le \frac{n\lvert\sin n\rvert}{n^2+2} \le \frac{n}{n^2+2} $$

but I got stick here, can someone give me just a HINT for how to continue?

for some reason the part $\frac{n}{n^2+2}$ is hard to make it less than $\epsilon$ and find the $N$ im searching for.


Just note that for all $n \geq 1$ we have $$ \bigg| \frac{n\sin n}{2 + n^{2}} \bigg| \leq \frac{n}{n^{2}} = \frac{1}{n}. $$


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