Đáp án:
$\lim_{n \to \infty} \dfrac{n^6-1}{n^2+4n+4}$
$=\lim_{n \to \infty} \dfrac{n^6(\dfrac{n^6}{n^6}-\dfrac{1}{n^6})}{n^6(\dfrac{n^2}{n^6}+\dfrac{4n}{n^6}+\dfrac{4}{n^6})}$
$=\lim_{n \to \infty} \dfrac{1-\dfrac{1}{n^6}}{\dfrac{n^2}{n^6}+\dfrac{4n}{n^6}+\dfrac{4}{n^6}}=+∞$
$Vì:$
$\lim_{n \to \infty}(1-\dfrac{1}{n^6})>0$
$\lim_{n \to \infty}(n^2+4n+4)>0$
BẠN THAM KHẢO.
