Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\\
\lim \left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 - \frac{1}{{{3^2}}}} \right)\left( {1 - \frac{1}{{{4^2}}}} \right)....\left( {1 - \frac{1}{{{n^2}}}} \right)\\
= \lim \left( {\frac{{{2^2} - 1}}{{{2^2}}}.\frac{{{3^2} - 1}}{{{3^2}}}.\frac{{{4^2} - 1}}{{{4^2}}}.....\frac{{{n^2} - 1}}{{{n^2}}}} \right)\\
= \lim \left( {\frac{{1.3}}{{{2^2}}}.\frac{{2.4}}{{{3^2}}}.\frac{{3.5}}{{{4^2}}}.....\frac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}}} \right)\\
= \lim \left( {\frac{{\left( {1.2.3.....\left( {n - 1} \right)} \right).\left( {3.4.5....\left( {n + 1} \right)} \right)}}{{\left( {2.3.4.....n} \right)\left( {2.3.4.....n} \right)}}} \right)\\
= \lim \frac{{1.\left( {n + 1} \right)}}{{n.2}} = \lim \frac{{n + 1}}{{2n}} = \frac{1}{2}\\
e,\\
\lim \frac{{1 + 2 + 3 + ... + n}}{{{n^2} + 3n}}\\
= \lim \frac{{\frac{{n\left( {n + 1} \right)}}{2}}}{{{n^2} + 3n}}\\
= \lim \frac{{{n^2} + n}}{{2{n^2} + 6n}}\\
= \lim \frac{{1 + \frac{1}{n}}}{{2 + \frac{6}{n}}} = \frac{1}{2}\\
f,\\
\lim \frac{{1 + 2 + {2^2} + .... + {2^n}}}{{1 + 3 + {3^2} + ..... + {3^n}}}\\
= \lim \frac{{\frac{{{2^{n + 1}} - 1}}{{2 - 1}}}}{{\frac{{{3^{n + 1}} - 1}}{{3 - 1}}}} = \lim \frac{{{2^{n + 1}} - 1}}{{\frac{{{3^{n + 1}} - 1}}{2}}} = \lim \frac{{{2^{n + 2}} - 2}}{{{3^{n + 1}} - 1}} = \lim \frac{{2.{{\left( {\frac{2}{3}} \right)}^n} - \frac{2}{{{3^{n + 1}}}}}}{{1 - \frac{1}{{{3^{n + 1}}}}}} = 0
\end{array}\)
