Đáp án:
\[\frac{3}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \frac{{\left( {{3^{n + 2}} + 1} \right)\left( {{2^{3n}} - 2} \right)}}{{\left( {{2^{2n}} - 1} \right)\left( {{6^{n + 1}} + 1} \right)}}\\
= \lim \left( {\frac{{\left( {{3^{n + 2}} + 1} \right)\left( {{2^{3n}} - 2} \right)}}{{{3^{n + 1}}{{.2}^{3n + 1}}}}.\frac{{{3^{n + 1}}{{.2}^{3n + 1}}}}{{\left( {{2^{2n}} - 1} \right)\left( {{6^{n + 1}} + 1} \right)}}} \right)\\
= \lim \left( {\frac{{{3^{n + 2}} + 1}}{{{3^{n + 1}}}}.\frac{{{2^{3n}} - 2}}{{{2^{3n + 1}}}}.\frac{{{2^{2n}}}}{{{2^{2n}} - 1}}.\frac{{{6^{n + 1}}}}{{{6^{n + 1}} + 1}}} \right)\\
= \lim \left[ {\left( {3 + \frac{1}{{{3^{n + 1}}}}} \right)\left( {\frac{1}{2} - \frac{1}{{{2^{3n}}}}} \right).\left( {1 + \frac{1}{{{2^{2n}} - 1}}} \right)\left( {1 - \frac{1}{{{6^{n + 1}} + 1}}} \right)} \right]\\
= 3.\frac{1}{2}.1.1 = \frac{3}{2}
\end{array}\)
