Lời giải:
b)
$lim\frac{(-1)^n.sin(3n+n^2)}{3n-1}$
Ta có:
$\frac{sin(3n+n^2)}{3n-1}\leq \frac{1}{3n-1}$(Vì $sinx\leq1$)
Mà:$lim\frac{1}{3n-1}=0$
=>$lim\frac{(-1)^n.sin(3n+n^2)}{3n-1}=0$
c)
$lim\frac{3sin^6n+5cos^2(n+2)}{n^2+1}$
Ta có:
$\frac{3sin^6n+5cos^2(n+2)}{n^2+1}\leq\frac{8}{n^2+1}$(Vì $sinx \leq 1;cosx \leq 1$)
Mà:$lim\frac{8}{n^2+1}=0$
=>$lim\frac{3sin^6n+5cos^2(n+2)}{n^2+1}=0$
