Đáp án:
\[\lim \frac{{\sqrt {4{n^6} + n + 1} }}{{1 - 2n - 3{n^3}}} = \frac{{ - 2}}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \frac{{\sqrt {4{n^6} + n + 1} }}{{1 - 2n - 3{n^3}}}\\
= \lim \frac{{\sqrt {{n^6}.\left( {4 + \frac{1}{{{n^5}}} + \frac{1}{{{n^6}}}} \right)} }}{{{n^3}\left( {\frac{1}{{{n^3}}} - \frac{2}{{{n^2}}} - 3} \right)}}\\
= \lim \frac{{{n^3}\sqrt {4 + \frac{1}{{{n^5}}} + \frac{1}{{{n^6}}}} }}{{{n^3}\left( {\frac{1}{{{n^3}}} - \frac{2}{{{n^2}}} - 3} \right)}}\\
= \lim \frac{{\sqrt {4 + \frac{1}{{{n^5}}} + \frac{1}{{{n^6}}}} }}{{\left( {\frac{1}{{{n^3}}} - \frac{2}{{{n^2}}} - 3} \right)}}\\
= \frac{{\sqrt 4 }}{{ - 3}} = \frac{{ - 2}}{3}
\end{array}\)
