Đáp án: $\lim \dfrac{3\sin n+4\cos n}{n+1}=0$
Giải thích các bước giải:
Ta có :
$\lim \dfrac{3\sin n+4\cos n}{n+1}$
$=\lim 5.\dfrac{\dfrac{3}{5}.\sin n+\dfrac{4}{5}.\cos n}{n+1}$
$=\lim 5.\dfrac{\cos \alpha.\sin n+\sin \alpha.\cos n}{n+1}$
(Đặt $\alpha=\arccos \dfrac{3}{5},0\le \alpha\le \dfrac{\pi}{2})$
$=\lim 5.\dfrac{\sin( n+\alpha)}{n+1}$
$\rightarrow A=\lim 5.\dfrac{\sin( n+\alpha)}{n+1}$
$\rightarrow |A|=\lim 5.\dfrac{|\sin( n+\alpha)|}{n+1}$
$\rightarrow 0\le |A|=\lim 5.\dfrac{1}{n+1}=0$
$\rightarrow \lim A=0$
$\rightarrow \lim \dfrac{3\sin n+4\cos n}{n+1}=0$
