\(\lim \frac{{{{\left( {2 - 3n} \right)}^2}\left( {n + 1} \right)}}{{1 - 4{n^3}}}\) bằng:
A. \(\frac{3}{4}\)
B. 0
C. \(\frac{9}{4}\)
D. \(-\frac{9}{4}\)
Ta có:
\(\begin{array}{l}
\lim \frac{{{{\left( {2 - 3n} \right)}^2}\left( {n + 1} \right)}}{{1 - 4{n^3}}} = \lim \frac{{{n^2}{{\left( {\frac{2}{n} - 3} \right)}^2}.n\left( {1 + \frac{1}{n}} \right)}}{{{n^3}\left( {\frac{1}{{{n^3}}} - 4} \right)}}\\
= \lim \frac{{{{\left( {2n - 3} \right)}^2}\left( {1 + \frac{1}{n}} \right)}}{{\frac{1}{{{n^3}}} - 4}} = \frac{{{{\left( { - 3} \right)}^2}}}{{ - 4}} = \frac{{ - 9}}{4}
\end{array}\)
Đáp án: D
-- Mod Toán 11