Đáp án:
\[\lim \left[ {\left( {1 - \dfrac{1}{{{2^2}}}} \right)\left( {1 - \dfrac{1}{{{3^2}}}} \right).....\left( {1 - \dfrac{1}{{{n^2}}}} \right)} \right] = \dfrac{1}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1 - \dfrac{1}{{{n^2}}} = \dfrac{{{n^2} - 1}}{{{n^2}}} = \dfrac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}}\\
\lim \left[ {\left( {1 - \dfrac{1}{{{2^2}}}} \right)\left( {1 - \dfrac{1}{{{3^2}}}} \right).....\left( {1 - \dfrac{1}{{{n^2}}}} \right)} \right]\\
= \lim \left[ {\dfrac{{{2^2} - 1}}{{{2^2}}}.\dfrac{{{3^2} - 1}}{{{3^2}}}.....\dfrac{{{n^2} - 1}}{{{n^2}}}} \right]\\
= \lim \left[ {\dfrac{{\left( {2 - 1} \right).\left( {2 + 1} \right)}}{{{2^2}}}.\dfrac{{\left( {3 - 1} \right)\left( {3 + 1} \right)}}{{{3^2}}}.....\dfrac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}}} \right]\\
= \lim \left[ {\dfrac{{1.3}}{{{2^2}}}.\dfrac{{2.4}}{{{3^2}}}......\dfrac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}}} \right]\\
= \lim \left[ {\dfrac{{1.2.3.....\left( {n - 1} \right)}}{{2.3.4.....n}}.\dfrac{{3.4.5....\left( {n + 1} \right)}}{{2.3.4.....n}}} \right]\\
= \lim \left[ {\dfrac{1}{n}.\dfrac{{n + 1}}{2}} \right]\\
= \lim \dfrac{{n + 1}}{{2n}} = \lim \dfrac{{1 + \dfrac{1}{n}}}{2} = \dfrac{{1 + 0}}{2} = \dfrac{1}{2}
\end{array}\)
