Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
\lim {u_n} = L \Rightarrow \lim \frac{1}{{\sqrt[3]{{{u_n} + 8}}}} = \frac{1}{{\sqrt[3]{{L + 8}}}}\\
5,\\
\lim \frac{{2 - {5^{n + 2}}}}{{{3^n} + {{2.5}^n}}} = \lim \frac{{\frac{2}{{{5^n}}} - {5^2}}}{{{{\left( {\frac{3}{5}} \right)}^n} + 2}} = \frac{{ - {5^2}}}{2} = - \frac{{25}}{2}\\
6,\\
{u_n} = \left( {n + 1} \right)\sqrt {\frac{{2n + 2}}{{{n^4} + {n^2} - 1}}} = \sqrt {\frac{{\left( {2n + 2} \right){{\left( {n + 1} \right)}^2}}}{{{n^4} + {n^2} - 1}}} \\
= \sqrt {\frac{{\frac{{2n + 2}}{n}.\frac{{{{\left( {n + 1} \right)}^2}}}{{{n^2}}}}}{{\frac{{{n^4} + {n^2} - 1}}{{{n^3}}}}}} \\
= \sqrt {\frac{{\left( {2 + \frac{1}{n}} \right){{\left( {1 + \frac{1}{n}} \right)}^2}}}{{n + \frac{1}{n} - \frac{1}{{{n^3}}}}}} \\
\Rightarrow \lim {u_n} = \lim \sqrt {\frac{{\left( {2 + \frac{1}{n}} \right){{\left( {1 + \frac{1}{n}} \right)}^2}}}{{n + \frac{1}{n} - \frac{1}{{{n^3}}}}}} = \lim \sqrt {\frac{{2.1}}{n}} = 0
\end{array}\)
