Giải thích các bước giải:
Câu 1:
Ta có: $\lim_{u_n}=\lim\dfrac{n}{4^n}=0$
$\to B$
Câu 2:
Ta có:
$A=\lim\dfrac{n\cos2n}{n^2+1}$
$\to\lim\dfrac{-n}{n^2+1}\le A\le \lim\dfrac{n}{n^2+1}$ vì $-1\le \cos2n\le 1$
$\to 0\le A\le 0$
$\to A=0$
$\to \lim\dfrac{n\cos2n}{n^2+1}=0$
$\to \lim(5-\dfrac{n\cos2n}{n^2+1})=5$
$\to B$
Câu 3:
Ta có:
$A=\lim\dfrac{2n+1}{1-3n}=\lim\dfrac{2+\dfrac1n}{\dfrac1n-3}=-\dfrac23$
$\to C$
Câu 4:
Ta có:
$\lim\dfrac{4n^2+3n+1}{(3n-1)^2}$
$=\lim\dfrac{4+\dfrac3n+\dfrac1{n^2}}{(3-\dfrac1n)^2}$
$=\dfrac{4+0+0}{(3-0)^2}$
$=\dfrac49$
$\to C$
Câu 5:
Ta có:
$\lim\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}$
$=\lim\dfrac{-1+\dfrac2n+\dfrac1{n^2}}{\sqrt{3+\dfrac2{n^2}}}$
$=\dfrac{-1+0+0}{\sqrt{3+0}}$
$=-\dfrac{1}{\sqrt{3}}$
$\to A$
Câu 6:
Ta có:
$\lim u_n=\lim\dfrac{2n-n^4}{4n-5}$
$\to \lim u_n=\lim\dfrac{\dfrac2{n^3}-1}{\dfrac4{n^3}-\dfrac5{n^4}}$
$\to \lim u_n=\dfrac{0-1}{0-0}$
$\to \lim u_n=-\infty$
$\to A$