$L=\lim \frac{n+2\sin(n+1)}{(n+2\sqrt[3]{n})}=\lim \frac{1+2\frac{\sin(n+1)}n}{1+\frac{2}{\sqrt[3]{n^2}}}$.
Ta có
$-\frac1n \le \frac{\sin(n+1)}n \le \frac1n \Rightarrow \lim \frac{\sin(n+1)}n =0$ và $\lim \frac{2}{\sqrt[3]{n^2}} =0$. Suy ra
$L=\frac{1+0}{1+0}=1.$