Lời giải:
$lim_{n->1}\frac{3n^4-n-2}{3n^2-4n+1}$
$=lim_{n->1}\frac{(3n^3+3n^2+3n+2)(n-1)}{(3n-1)(n-1)}$
$=lim_{n->1}\frac{3n^3+3n^2+3n+2}{3n-1}$
$=lim_{n->1}\frac{3n^3.(1+\frac{1}{n}+\frac{1}{n^2}+\frac{2}{3n^3})}{3n.(1-\frac{1}{3n})}$
$=\frac{3n^3}{3n}$
$=n^2$
$=1^2$
$=1$