Cho dãy số \[({u_n})\]với \[{u_n} = \frac{{\left( {2n + 1} \right)\left( {1 - 3n} \right)}}{{\sqrt[3]{{{n^3} + 5n - 1}}}}\] Khi đó \[lim\,{u_n}\] bằng?

\[\lim {u_n} = \lim \frac{{\left( {2n + 1} \right)\left( {1 - 3n} \right)}}{{\sqrt[3]{{{n^3} + 5n - 1}}}} = \lim \frac{{ - 6{n^2} - n + 1}}{{\sqrt[3]{{{n^3} + 5n - 1}}}}\]

\[ = \lim \frac{{\frac{{ - 6{n^2} - n + 1}}{{{n^2}}}}}{{\sqrt[3]{{\frac{{{n^3} + 5n - 1}}{{{n^6}}}}}}} = \lim \frac{{ - 6 - \frac{1}{n} + \frac{1}{{{n^2}}}}}{{\sqrt[3]{{\frac{1}{{{n^3}}} + \frac{5}{{{n^5}}} - \frac{1}{{{n^6}}}}}}} = - \infty .\]

Đáp án cần chọn là: A