Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
8,\\
\lim \dfrac{{3 - 4n}}{{5n}} = \lim \dfrac{{\dfrac{3}{n} - 4}}{5} = \dfrac{{0 - 4}}{5} = - \dfrac{4}{5}\\
9,\\
\lim \dfrac{{3{n^4} - 2n + 3}}{{4{n^4} + 2n + 1}} = \lim \dfrac{{3 - \dfrac{2}{{{n^3}}} + \dfrac{3}{{{n^4}}}}}{{4 + \dfrac{2}{{{n^3}}} + \dfrac{1}{{{n^4}}}}} = \dfrac{{3 - 0 + 0}}{{4 + 0 + 0}} = \dfrac{3}{4}\\
10,\\
\lim \left( { - 3{n^3} + 2{n^2} - 5} \right) = \lim \left[ {{n^3}\left( { - 3 + \dfrac{2}{n} - \dfrac{5}{{{n^3}}}} \right)} \right] = - \infty \\
\left( \begin{array}{l}
\lim {n^3} = + \infty \\
\lim \left( { - 3 + \dfrac{2}{n} - \dfrac{5}{{{n^3}}}} \right) = - 3 + 0 - 0 = - 3
\end{array} \right)\\
11,\\
\lim \left( {2{n^4} + {n^2} - 5n} \right) = \lim \left[ {{n^4}\left( {2 + \dfrac{1}{{{n^2}}} - \dfrac{5}{{{n^3}}}} \right)} \right] = + \infty \\
\left( \begin{array}{l}
\lim {n^4} = + \infty \\
\lim \left( {2 + \dfrac{1}{{{n^2}}} - \dfrac{5}{{{n^3}}}} \right) = 2 + 0 - 0 = 2
\end{array} \right)
\end{array}\)
