$\lim \frac{sin(n!)}{n^2+1}$

$\sin(n!)\in [-1;1]$

$\to \Big|\sin (n!)\Big|\le 1$

$\to \Big| \dfrac{\sin (n!)}{n^2+1}\Big|\le \dfrac{1}{|n^2+1|}=\dfrac{1}{n^2+1}$

$\to \lim\dfrac{\sin (n!)}{n^2+1}=\lim\dfrac{1}{n^2+1}=\lim\dfrac{\dfrac{1}{n^2}}{1+\dfrac{1}{n^2}}=0$

Đáp án:

$\lim\dfrac{\sin(n!)}{n^2+1}= 0$

Giải thích các bước giải:

$\quad \lim\dfrac{\sin(n!)}{n^2+1}$

Ta có:

$\quad - 1 \leqslant \sin(n!)\leqslant 1 \forall n$

$\Leftrightarrow -\dfrac{1}{n^2+1} \leqslant \dfrac{\sin(n!)}{n^2+1} \leqslant \dfrac{1}{n^2+1}$

$\Leftrightarrow \lim\left(-\dfrac{1}{n^2+1}\right) \leqslant \lim\dfrac{\sin(n!)}{n^2+1} \leqslant \lim\dfrac{1}{n^2+1}$

Ta lại có:

$\lim\left(-\dfrac{1}{n^2+1}\right)= \lim\dfrac{1}{n^2+1}= 0$

Do đó:

$\lim\dfrac{\sin(n!)}{n^2+1}= 0\quad$ (định lý giới hạn kẹp)

Vậy $\lim\dfrac{\sin(n!)}{n^2+1}= 0$