Giá trị giới hạn của dãy số D = lim căn {{n^2} + n + 1}  - 2 căn [3]{{{n^3} + {n^2} - 1}} + n

Đáp án

\( - \frac{1}{6}\).

Giải thích

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

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

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

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

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

Vậy \(D = \frac{1}{2} - \frac{2}{3} = - \frac{1}{6}\).