lim (3n + sin2n)/(3n + 5) bằng: 3/8; 1/5; 0; 1

* Hướng dẫn giải

Chọn D

$\lim \frac{3n+\sin 2n}{3n+5}=\lim \frac{3n}{3n+5}+\lim \frac{\sin 2n}{3n+5}$

Ta có: $\lim \frac{3n}{3n+5}=\lim \frac{3}{3+\frac{5}{n}}=1$

Vì $0<\left|\frac{\sin 2n}{3n+5}\right|\le \frac{1}{3n+5}$  mà $\lim \frac{1}{3n+5}=0$

Do đó $\lim \frac{\sin 2n}{3n+5}=0$ .

Vậy  $\lim \frac{3n+\sin 2n}{3n+5}=1$.