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

Chọn D

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

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

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

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