Đáp án:
$\begin{array}{l}
b)\lim \left( {\sqrt {{n^2} + 4n + 2} - n} \right)\\
= \lim \dfrac{{{n^2} + 4n + 2 - {n^2}}}{{\sqrt {{n^2} + 4n + 2} + n}}\\
= \lim \dfrac{{4n + 2}}{{\sqrt {{n^2} + 4n + 2} + n}}\\
= \lim \dfrac{{4 + \dfrac{2}{n}}}{{\sqrt {1 + \dfrac{4}{n} + \dfrac{2}{{{n^2}}}} + 1}}\\
= \dfrac{4}{{1 + 1}} = 2\\
c)\lim \dfrac{{\left( {2n - {n^3}} \right)\left( {3{n^2} + 1} \right)}}{{\left( {2n - 1} \right)\left( {{n^4} - 7} \right)}}\\
= \lim \dfrac{{\dfrac{{2n - {n^3}}}{{{n^3}}}.\dfrac{{3{n^2} + 1}}{{{n^2}}}}}{{\dfrac{{2n - 1}}{n}.\dfrac{{{n^4} - 7}}{{{n^4}}}}}\\
= \lim \dfrac{{\left( {\dfrac{2}{{{n^2}}} - 1} \right).\left( {3 + \dfrac{1}{{{n^2}}}} \right)}}{{\left( {2 - \dfrac{1}{n}} \right).\left( {1 - \dfrac{7}{{{n^4}}}} \right)}}\\
= \dfrac{{ - 1.3}}{{2.1}} = \dfrac{{ - 3}}{2}\\
d)\lim \dfrac{{{n^3} - 2n + 1}}{{3n - 4}}\\
= \lim \dfrac{{{n^2} - 2 + \dfrac{1}{n}}}{{3 - \dfrac{4}{n}}}\\
= + \infty
\end{array}$
