Đáp án:
\[ - \infty \]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \left( {\frac{{ - {n^6} - 2{n^2} - \sqrt 3 n}}{{4{n^4} + 4{n^3} + 9{n^2}}}} \right)\\
= \lim \left( {\frac{{ - {n^2} - \frac{2}{{{n^2}}} - \frac{{\sqrt 3 }}{{{n^3}}}}}{{4 + \frac{4}{n} + \frac{9}{{{n^2}}}}}} \right)\\
n \to + \infty \Rightarrow \lim \left( { - {n^2}} \right) = - \infty ,\,\,\,\lim \frac{{ - 2}}{{{n^2}}} = \lim \frac{{ - \sqrt 3 }}{{{n^3}}} = \lim \frac{4}{n} = \lim \frac{9}{{{n^2}}} = 0\\
\Rightarrow \lim \left( {\frac{{ - {n^6} - 2{n^2} - \sqrt 3 n}}{{4{n^4} + 4{n^3} + 9{n^2}}}} \right) = - \infty
\end{array}\)
