Kết quả của giới hạn \[{\mathop{\rm li}\nolimits} {\rm{m}}\frac{{{{\rm{3}}^{\rm{n}}} - {\rm{2}}{\rm{.}}{{\rm{5}}^{{\rm{n + 1}}}}}}{{{{\rm{2}}^{{\rm{n + 1}}}}{\rm{ + }}{{\rm{5}}^{\rm{n}}}}}\] bằng:

\[\lim {{\rm{u}}_{\rm{n}}} = \lim \left( {\sqrt {{{\rm{n}}^{\rm{2}}}{\rm{ + an + 5}}} - \sqrt {{{\rm{n}}^{\rm{2}}}{\rm{ + 1}}} } \right)\]

\[{\rm{ = lim}}\frac{{{\rm{an + 4}}}}{{\sqrt {{{\rm{n}}^{\rm{2}}}{\rm{ + an + 5}}} {\rm{ + }}\sqrt {{{\rm{n}}^{\rm{2}}}{\rm{ + 1}}} }}{\rm{ = lim}}\frac{{{\rm{a + }}\frac{{\rm{4}}}{{\rm{n}}}}}{{\sqrt {{\rm{1 + }}\frac{{\rm{a}}}{{\rm{n}}}{\rm{ + }}\frac{{\rm{5}}}{{{{\rm{n}}^{\rm{2}}}}}} {\rm{ + }}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{{{\rm{n}}^{\rm{2}}}}}} }}{\rm{ = }}\frac{{\rm{a}}}{{\rm{2}}}\]

Để \[\lim {{\rm{u}}_{\rm{n}}} = - 1 \Leftrightarrow \frac{{\rm{a}}}{{\rm{2}}} = - 1 \Rightarrow {\rm{a}} = - 2\]

Đáp án cần chọn là: C

\[{\rm{L = lim}}\left( {{\rm{5n}} - 3\left( {{{\rm{a}}^2} - 2} \right){{\rm{n}}^3}} \right) = {\rm{lim}}{{\rm{n}}^{\rm{3}}}\left( {\frac{5}{{{{\rm{n}}^2}}} - 3\left( {{{\rm{a}}^2} - 2} \right)} \right) = - \infty \]

\[{\rm{lim}}\left( {\frac{5}{{{{\rm{n}}^2}}} - 3\left( {{{\rm{a}}^2} - 2} \right)} \right) = - 3\left( {{{\rm{a}}^2} - 2} \right) < 0\]

\( \Leftrightarrow \left( {{{\rm{a}}^2} - 2} \right) > 0\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{{\rm{a}} > \sqrt 2 }\\{{\rm{a}} < \sqrt 2 }\end{array}} \right. \to {\rm{a}} = - 9; - 2;...;2;...9\)

Đáp án cần chọn là: B