Toán $\lim_{n \to \infty} \frac{n^6+4n}{3n^6-3n^4+2n^2-1}$ 09/09/2021 By Cora $\lim_{n \to \infty} \frac{n^6+4n}{3n^6-3n^4+2n^2-1}$
Đáp án: $\lim\dfrac{n^6 + 4n}{3n^6 – 3n^4 + 2n^2 – 1}=\dfrac13$ Giải thích các bước giải: $\quad \lim\dfrac{n^6 + 4n}{3n^6 – 3n^4 + 2n^2 – 1}$ $=\lim\dfrac{1 + \dfrac{4}{n^5}}{3- \dfrac{3}{n^2} + \dfrac{2}{n^4} -\dfrac{1}{n^6}}$ $=\dfrac{1 + 0}{3- 0 + 0 – 0}$ $=\dfrac13$ Trả lời
Đáp án:
$\lim\dfrac{n^6 + 4n}{3n^6 – 3n^4 + 2n^2 – 1}=\dfrac13$
Giải thích các bước giải:
$\quad \lim\dfrac{n^6 + 4n}{3n^6 – 3n^4 + 2n^2 – 1}$
$=\lim\dfrac{1 + \dfrac{4}{n^5}}{3- \dfrac{3}{n^2} + \dfrac{2}{n^4} -\dfrac{1}{n^6}}$
$=\dfrac{1 + 0}{3- 0 + 0 – 0}$
$=\dfrac13$