b,
Ta có: $-1\le \sin n\le 1$
$\to -2\le -2\sin n\le 2$
$\to \sqrt{n}-2\le \sqrt{n}-2\sin n\le \sqrt{n}+2$
Theo định lí kẹp:
$\lim(\sqrt{n}-2\sin n)=\lim(\sqrt{n}-2)=\lim(\sqrt{n}+2)=+\infty$
d,
$L=\lim \sqrt[3]{n\cos n-8n^3}$
$=n.\lim\sqrt[3]{\dfrac{\cos n}{n^2}-8}$
Ta có:
$\left| \dfrac{\cos n}{n^2}\right|\le \dfrac{1}{n^2}$
Theo định lí kẹp:
$\lim\dfrac{\cos n}{n^2}=\lim\dfrac{1}{n^2}=0$
$\to \lim\sqrt[3]{\dfrac{\cos n}{n^2}-8}=\sqrt[3]{0-8}=-2<0$
Mà $\lim n=+\infty$
Vậy $L=-\infty$
b,
$\lim\dfrac{\sqrt{n^3-n^2+1}}{2n+3}$
$=\lim\dfrac{ n\sqrt{n}.\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^3}}}{2n+3}$
$=\lim\sqrt{n}.\left( \dfrac{\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^3}}}{2+\dfrac{3}{n}}\right)$
$=+\infty$
