gấp lắm r.mn giúp em với

e,

$\lim0=0$

$\lim\dfrac{3}{2-3n^2}=\lim\dfrac{\dfrac{3}{n^2}}{\dfrac{2}{n^2}-3}=0$

Mà $0\le \sin^2(n^3+2)\le 1$

$\Leftrightarrow 0\le \dfrac{3\sin^2(n^3+1)}{2-3n^2}\le \dfrac{3}{2-3n^2}$

$\Rightarrow \lim\dfrac{3\sin^2(n^3+2)}{2-3n^2}=0$

$\to \lim\dfrac{3\sin^2(n^3+2)+n^2}{2-3n^2}=\lim\dfrac{n^2}{2-3n^2}=\lim\dfrac{1}{\dfrac{2}{n^2}-3}=\dfrac{-1}{3}$

f,

$\lim\dfrac{3n^2-2n+2}{3n\cos n+2n}$

$=\lim\dfrac{3-\dfrac{2}{n}+\dfrac{2}{n^2}}{\dfrac{3\cos n}{n}+\dfrac{2}{n}}$

Ta có:

$\Big|\dfrac{3\cos n}{n}\Big|\le \dfrac{3}{n}$

$\Rightarrow lim\dfrac{3\cos n}{n}=\lim\dfrac{3}{n}=0$

$-1\le \cos n\le 1\Leftrightarrow -n\le 3n\cos n+2n\le 5n$

$n>0\Rightarrow 3n\cos n+2n<0$ hoặc $>0$

$\to \lim\dfrac{3n^2-2n+3}{3n\cos n+2n}=\pm\infty$