a. $|2\sin n + 3\cos n| \le 2|\sin n|+3|\cos n|=5$. Suy ra
$-\frac{5}{n^2+1} \le \frac{2\sin n + 3\cos n}{n^2+1} \le \frac{5}{n^2+1}$.
Mặt khác
$\lim \frac{5}{n^2+1} = \lim\left ( - \frac{5}{n^2+1} \right )=0\Rightarrow \lim \frac{2\sin n + 3\cos n}{n^2+1}=0.$