$\lim_{} \sqrt[3]{n^{3}-3n^{2}+1}$- $\sqrt[]{n^{2}+4n }$ 30/10/2021 Bởi Ariana $\lim_{} \sqrt[3]{n^{3}-3n^{2}+1}$- $\sqrt[]{n^{2}+4n }$
$\lim (\sqrt[3]{n^3 – 3n^2 + 1} – \sqrt{n^2 + 4n})$ $= lim (\sqrt[3]{n^3 – 3n^2 + 1} – n) + lim (n – \sqrt{n^2 + 4n})$ $= lim \dfrac{n^3 – 3n^2 + 1 – n^3}{(\sqrt[3]{n^3 – 3n^2 + 1})^2 + n\sqrt[3]{n^3 – 3n^2 + 1} + n^2} + lim \dfrac{n^2 – n^2 – 4n}{n + \sqrt{n^2 + 4n}}$ $= lim \dfrac{\dfrac{1}{n^2} – 3}{(\sqrt[3]{1 – \dfrac{3}{n} + \dfrac{1}{n^3}})^2 + \sqrt[3]{1 – \dfrac{3}{n} + \dfrac{1}{n^3}} + 1} + lim \dfrac{-4}{1 + \sqrt{1 + \dfrac{4}{n}}}$ `= (0 – 3)/(1 + 1 + 1) + (-4)/(1 + 1)` `= -1 – 2` `= -3` Bình luận
$\lim (\sqrt[3]{n^3 – 3n^2 + 1} – \sqrt{n^2 + 4n})$
$= lim (\sqrt[3]{n^3 – 3n^2 + 1} – n) + lim (n – \sqrt{n^2 + 4n})$
$= lim \dfrac{n^3 – 3n^2 + 1 – n^3}{(\sqrt[3]{n^3 – 3n^2 + 1})^2 + n\sqrt[3]{n^3 – 3n^2 + 1} + n^2} + lim \dfrac{n^2 – n^2 – 4n}{n + \sqrt{n^2 + 4n}}$
$= lim \dfrac{\dfrac{1}{n^2} – 3}{(\sqrt[3]{1 – \dfrac{3}{n} + \dfrac{1}{n^3}})^2 + \sqrt[3]{1 – \dfrac{3}{n} + \dfrac{1}{n^3}} + 1} + lim \dfrac{-4}{1 + \sqrt{1 + \dfrac{4}{n}}}$
`= (0 – 3)/(1 + 1 + 1) + (-4)/(1 + 1)`
`= -1 – 2`
`= -3`
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