Giải thích các bước giải:
\(\begin{array}{l}
56;\\
\lim \left( {\sqrt[3]{{{n^3} + 3{n^2}}} - \sqrt {{n^2} + 2n} } \right)\\
= \lim \left[ {\left( {\sqrt[3]{{{n^3} + 3{n^2}}} - n} \right) + \left( {n - \sqrt {{n^2} + 2n} } \right)} \right]\\
= \lim \left[ {\frac{{{n^3} + 3{n^2} - {n^3}}}{{\sqrt[3]{{{{\left( {{n^3} + 3{n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} + 3{n^2}}} + {n^2}}} + \frac{{{n^2} - {n^2} - 2n}}{{n + \sqrt {{n^2} + 2n} }}} \right]\\
= \lim \left[ {\frac{3}{{\sqrt[3]{{{{\left( {1 + \frac{3}{n}} \right)}^2}}} + \sqrt[3]{{1 + \frac{3}{n}}} + 1}} + \frac{{ - 2}}{{1 + \sqrt[3]{{1 + \frac{2}{n}}}}}} \right]\\
= \lim \left( {\frac{3}{3} + \frac{{ - 2}}{2}} \right) = 0\\
58,\\
\lim \frac{{1 + 2 + {2^2} + .... + {2^n}}}{{3 + {3^2} + {3^3} + .... + {3^{n + 2}}}}\\
= \lim \frac{{\frac{{{2^{n + 1}} - 1}}{{2 - 1}}}}{{\frac{{{3^{n + 3}} - 3}}{{3 - 1}}}} = \lim \frac{{{2^{n + 1}} - 1}}{{\frac{{{3^{n + 3}} - 3}}{2}}} = \lim \frac{{{2^{n + 2}} - 2}}{{{3^{n + 3}} - 3}}\\
= \lim \frac{{{{\left( {\frac{2}{3}} \right)}^{n + 2}} - \frac{2}{{{3^{n + 2}}}}}}{{3 - \frac{1}{{{3^{n + 1}}}}}} = 0
\end{array}\)
