Lời giải của Tự Học 365

Ta có: $\lim n\left( {\sqrt {4{n^2} + 3}  - \sqrt[3]{{8{n^3} + n}}} \right)$$ = \lim n\left[ {\left( {\sqrt {4{n^2} + 3}  - 2n} \right) + \left( {2n - \sqrt[3]{{8{n^3} + n}}} \right)} \right]$

$ = \lim \left[ {n\left( {\sqrt {4{n^2} + 3}  - 2n} \right) + n\left( {2n - \sqrt[3]{{8{n^3} + n}}} \right)} \right]$.

Ta có: $\lim n\left( {\sqrt {4{n^2} + 3}  - 2n} \right)$$ = \lim \dfrac{{3n}}{{\left( {\sqrt {4{n^2} + 3}  + 2n} \right)}}$$ = \lim \dfrac{3}{{\left( {\sqrt {4 + \dfrac{3}{{{n^2}}}}  + 2} \right)}} = \dfrac{3}{4}$.

Ta có: $\lim n\left( {2n - \sqrt[3]{{8{n^3} + n}}} \right)$$ = \lim \dfrac{{ - {n^2}}}{{\left( {4{n^2} + 2n\sqrt[3]{{8{n^3} + n}} + \sqrt[3]{{{{\left( {8{n^3} + n} \right)}^2}}}} \right)}}$

$ = \lim \dfrac{{ - 1}}{{\left( {4 + 2\sqrt[3]{{8 + \dfrac{1}{{{n^2}}}}} + \sqrt[3]{{{{\left( {8 + \dfrac{1}{{{n^2}}}} \right)}^2}}}} \right)}} =  - \dfrac{1}{{12}}$.

Vậy $\lim n\left( {\sqrt {4{n^2} + 3}  - \sqrt[3]{{8{n^3} + n}}} \right) = \dfrac{3}{4} - \dfrac{1}{{12}}$$ = \dfrac{2}{3}$.

Đáp án cần chọn là: d