<s>[INST] Prove $\frac{2n+\sin(n)}{n+2}$ converges to 2. Here's my attempt.
Let $\epsilon >0.$ Then $N \geq \frac{5}{\epsilon}$, with $n \geq N \implies$
$| \frac{2n+\sin(n)}{n+2} -2|=| \frac{2n+\sin(n)-2n-4}{n+2}|=| \frac{\sin(n)-4}{n+2}|= \frac{|\sin(n)-4|}{n+2} \leq \frac{5}{n+2} \leq \frac{5}{n} \leq \epsilon$
 [/INST] Alternatively,
$$
\frac{2n-1}{n+2}
\le
\frac{2n+\sin(n)}{n+2}
\le
\frac{2n+1}{n+2}
$$
and both bounds go to $2$ as $n \to \infty$.
</s>