$\lim_{n \to 1} (3n^4 -n -2)/(3n^2 – 4n +1)

Lời giải:

$lim_{n->1}\frac{3n^4-n-2}{3n^2-4n+1}$

$=lim_{n->1}\frac{(3n^3+3n^2+3n+2)(n-1)}{(3n-1)(n-1)}$

 $=lim_{n->1}\frac{3n^3+3n^2+3n+2}{3n-1}$

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

$=\frac{3n^3}{3n}$

$=n^2$

$=1^2$

$=1$