Đáp án:
$1/ - \infty$
$2/ \dfrac{1}{8}$
$3/ \dfrac{11}{18}$
Giải thích các bước giải:
1/ $\lim(-n^3-3n^2+n+2021)$
$= \lim\Big[n^3.\Big(-1-\dfrac{3}{n} +\dfrac{1}{n^2}+\dfrac{2021}{n^3}\Big)\Big]$
$= - \infty$. Vì : $\begin{cases} \lim n^3 = +\infty \\\lim(-1-\dfrac{3}{n}+\dfrac{1}{n^2}+\dfrac{2021}{n^3})=-1<0\end{cases}$
2/ $\lim\Big(\sqrt{16^{n+1}+4^n}-\sqrt{16^{n+1}+3^n}\Big)$
$= \lim\dfrac{16^{n+1}+4^n-16^{n+1}-3^n}{\sqrt{16^{n+1}+4^n}+\sqrt{16^{n+1}+3^n}}$
$= \lim\dfrac{4^n-3^n}{\sqrt{16+\Big(\dfrac{1}{4}\Big)^n}+\sqrt{16+\Big(\dfrac{3}{16}\Big)^n}}$
$= \lim\dfrac{1}{4+4}=\dfrac{1}{8}$
3/ $\lim\Big[\dfrac{1}{1.4}+\dfrac{1}{2.5}+... +\dfrac{1}{n.(n+3)}\Big]$ $(*)$
Ta có : $\dfrac{1}{1.4}+\dfrac{1}{2.5}+... +\dfrac{1}{n.(n+3)}$
$= \dfrac{1}{3}.\Big[1-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{6}+.....+\dfrac{1}{n}-\dfrac{1}{n+3}\Big]$
$= \dfrac{1}{3}.\Big[\Big(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\Big)-\Big(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+....+\dfrac{1}{n+3}\Big)\Big]$
$= \dfrac{1}{3}.\Big(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\Big)$
$= \dfrac{1}{3}.\Big(\dfrac{11}{6}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\Big)$ $(**)$
Thay $(**)$ vào $(*)$ ta có :
$(*) \Leftrightarrow \lim\dfrac{1}{3}.\Big(\dfrac{11}{6}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\Big)$
$= \lim\dfrac{11}{18} = \dfrac{11}{18}$
