Đáp án:$\lim \dfrac{n^3-n^2.\sin n-1}{2n^4-n^2+7}=0$
Giải thích các bước giải:
$\lim \dfrac{n^3-n^2.\sin n-1}{2n^4-n^2+7}$
$=\lim \dfrac{\dfrac{1}{n}-\dfrac{\sin n}{n^2}-\dfrac{1}{n^4}}{2-\dfrac{1}{n^2}+\dfrac{7}{n^2}}$
$=\lim \dfrac{0-0-0}{2-0+0}$
$=0$
Vì $0\le \lim |\dfrac{\sin n}{n^2}|=\lim\dfrac{|\sin n|}{n^2}\le \lim \dfrac{1}{n^2}=0$
$\rightarrow \lim \dfrac{\sin n}{n^2}=0$
