## c) lim ( căn bậc ba (n^3 – 2n^2) – n) d) lim căn bậc ba (2n^3 + n)/ (n + 2) g) lim ( căn bậc ba (n – n^3) + n + 2) Giúp mình câu c,d,g với ạ.

Question

c) lim ( căn bậc ba (n^3 – 2n^2) – n)
d) lim căn bậc ba (2n^3 + n)/ (n + 2)
g) lim ( căn bậc ba (n – n^3) + n + 2)
Giúp mình câu c,d,g với ạ.

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11 months 2021-02-20T07:13:19+00:00 2 Answers 144 views 0

$$\begin{array}{l} c,\\ \lim \left( {\sqrt[3]{{{n^3} – 2{n^2}}} – n} \right)\\ = \lim \left( {\frac{{{n^3} – 2{n^2} – {n^3}}}{{\sqrt[3]{{{{\left( {{n^3} – 2{n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} – 3{n^2}}} + {n^2}}}} \right)\\ = \lim \left( {\frac{{ – 2{n^2}}}{{\sqrt[3]{{{{\left( {{n^3} – 2{n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} – 3{n^2}}} + {n^2}}}} \right)\\ = \lim \left( {\frac{{ – 2}}{{\sqrt[3]{{{{\left( {1 – \frac{2}{n}} \right)}^2}}} + .\sqrt[3]{{1 – \frac{3}{n}}} + 1}}} \right) = – \frac{2}{3}\\ d,\\ \lim \left( {\frac{{\sqrt[3]{{2{n^3} + n}}}}{{n + 2}}} \right) = \lim \left( {\frac{{\sqrt[3]{{2 + \frac{1}{{{n^2}}}}}}}{{1 + \frac{2}{n}}}} \right) = \frac{{\sqrt[3]{2}}}{1} = \sqrt[3]{2}\\ g,\\ lim\left( {\sqrt[3]{{n – {n^3}}} + n + 2} \right)\\ = \lim \left( {\frac{{n – {n^3} + {n^3} + 6{n^2} + 12n + 8}}{{\sqrt[3]{{{{\left( {n – {n^3}} \right)}^2}}} – \left( {n + 2} \right).\sqrt[3]{{n – {n^3}}} + {{\left( {n + 2} \right)}^2}}}} \right)\\ = \lim \left( {\frac{{6 + \frac{{13}}{n} + \frac{8}{{{n^2}}}}}{{\sqrt[3]{{{{\left( {\frac{1}{{{n^2}}} – 1} \right)}^2}}} – \left( {1 + \frac{2}{n}} \right).\sqrt[3]{{\frac{1}{{{n^2}}} – 1}} + {{\left( {1 + \frac{2}{n}} \right)}^2}}}} \right)\\ = \frac{6}{{1 + 1 + 1}} = 2 \end{array}$$