Đáp án:
$\lim\limits_{n\to \infty}\dfrac{2n^4 - 4n^2 +1}{(n-1)^2(3n^2+1)} =\dfrac23$
Giải thích các bước giải:
$\lim\limits_{n\to \infty}\dfrac{2n^4 - 4n^2 +1}{(n-1)^2(3n^2+1)}$
$=\lim\limits_{n\to \infty}\dfrac{2 - \dfrac{4}{n^2}+\dfrac{1}{n^4}}{\left(1-\dfrac1n\right)^2\left(3 +\dfrac{1}{n^2}\right)}$
$=\dfrac{2 - 4.0 + 0}{(1-0)^2(3+0)}$
$= \dfrac23$