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

 

Đáp án

\(\frac{2}{3}\).

Giải thích

Cách 1.

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

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

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

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

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

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

Cách 2. Sử dụng chức năng CALC của máy tính Casio.