1. Tìm lim (n^3 – 2n) / (3n^2 + n – 2) 2. Tìm lim (-2 + 3n – 2n^3) / (3n – 2) 05/11/2021 Bởi Liliana 1. Tìm lim (n^3 – 2n) / (3n^2 + n – 2) 2. Tìm lim (-2 + 3n – 2n^3) / (3n – 2)
1. $\lim\dfrac{n^3-2n}{3n^2+n-2}$ $=\lim\dfrac{n^3\Big(1-\dfrac{2}{n^2}\Big)}{n^2\Big(3+\dfrac{1}{n}-\dfrac{2}{n^2}\Big)}$ $=\lim n.\dfrac{1-\dfrac{2}{n^2}}{3+\dfrac{1}{n}-\dfrac{2}{n^2}}$ $=+\infty$ 2. $\lim\dfrac{-2+3n-2n^3}{3n-2}$ $=\lim \dfrac{n^3\Big(\dfrac{-2}{n^3}+\dfrac{3}{n^2}-2\Big)}{n\Big(3-\dfrac{2}{n}\Big)}$ $=\lim n^2.\dfrac{\dfrac{-2}{n^3}+\dfrac{3}{n^2}-2}{3-\dfrac{2}{n}}$ $=-\infty$ Bình luận
Đáp án:lim (n^3 – 2n) / (3n^2 + n – 2)=n^3(1-2/n^2)/n^2(3+1/n-2/n^2)=limn.lim(1-2/n^2)/n^2(3+1/n-2/n^2)=+∞.(1-0)/(3+0-0)=+∞.1/3=+∞ 2. lim (-2 + 3n – 2n^3) / (3n – 2)=lim n^3(-2/n^3+3/n^2-2)/n(3-2/n)=lim n^2.lim(-2/n^3+3/n^2-2)/(3-2/n)=+∞.(0+0-2)/(3-0)=+∞.-2/3=-∞ Giải thích các bước giải: Bình luận
1.
$\lim\dfrac{n^3-2n}{3n^2+n-2}$
$=\lim\dfrac{n^3\Big(1-\dfrac{2}{n^2}\Big)}{n^2\Big(3+\dfrac{1}{n}-\dfrac{2}{n^2}\Big)}$
$=\lim n.\dfrac{1-\dfrac{2}{n^2}}{3+\dfrac{1}{n}-\dfrac{2}{n^2}}$
$=+\infty$
2.
$\lim\dfrac{-2+3n-2n^3}{3n-2}$
$=\lim \dfrac{n^3\Big(\dfrac{-2}{n^3}+\dfrac{3}{n^2}-2\Big)}{n\Big(3-\dfrac{2}{n}\Big)}$
$=\lim n^2.\dfrac{\dfrac{-2}{n^3}+\dfrac{3}{n^2}-2}{3-\dfrac{2}{n}}$
$=-\infty$
Đáp án:lim (n^3 – 2n) / (3n^2 + n – 2)=n^3(1-2/n^2)/n^2(3+1/n-2/n^2)=limn.lim(1-2/n^2)/n^2(3+1/n-2/n^2)=+∞.(1-0)/(3+0-0)=+∞.1/3=+∞
2. lim (-2 + 3n – 2n^3) / (3n – 2)=lim n^3(-2/n^3+3/n^2-2)/n(3-2/n)=lim n^2.lim(-2/n^3+3/n^2-2)/(3-2/n)=+∞.(0+0-2)/(3-0)=+∞.-2/3=-∞
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