Giá trị của \(K = {\rm{lim}}\left( {\sqrt[3]{{{n^3} + {n^2} - 1}} - 3\sqrt {4{n^2} + n + 1} + 5n} \right)\) bằng

Đáp án

\( - \frac{5}{{12}}\).

Giải thích

Ta có: \(K = {\rm{lim}}\left( {\sqrt[3]{{{n^3} + {n^2} - 1}} - n} \right) - 3{\rm{lim}}\left( {\sqrt {4{n^2} + n + 1} - 2n} \right)\)

Mà

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

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

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

\({\rm{lim}}\left( {\sqrt {4{n^2} + n + 1} - 2n} \right) = {\rm{lim}}\frac{{{{\left( {\sqrt {4{n^2} + n + 1} } \right)}^2} - {{(2n)}^2}}}{{\sqrt {4{n^2} + n + 1} + 2n}} = \frac{{n + 1}}{{\sqrt {4{n^2} + n + 1} + 2n}} = \frac{{1 + \frac{1}{n}}}{{\sqrt {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} + 2}} = \frac{1}{4}\)

Do đó, \(K = \frac{1}{3} - \frac{3}{4} = - \frac{5}{{12}}\).