Đáp án:
\[\frac{1}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 - \frac{1}{{{3^2}}}} \right)\left( {1 - \frac{1}{{{4^2}}}} \right).....\left( {1 - \frac{1}{{{n^2}}}} \right)\\
= \lim \left( {\frac{{{2^2} - 1}}{{{2^2}}}} \right).\left( {\frac{{{3^2} - 1}}{{{3^2}}}} \right).\left( {\frac{{{4^2} - 1}}{{{4^2}}}} \right)....\left( {\frac{{{n^2} - 1}}{{{n^2}}}} \right)\\
= \lim \left( {\frac{{1.3}}{{{2^2}}}.\frac{{2.4}}{{{3^2}}}.\frac{{3.5}}{{{4^2}}}......\frac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}}} \right)\\
= \lim \frac{{\left( {1.2.3....\left( {n - 1} \right)} \right)\left( {3.4.5....\left( {n + 1} \right)} \right)}}{{\left( {2.3.4....n} \right)\left( {2.3.4.....n} \right)}}\\
= \lim \frac{{1.\left( {n + 1} \right)}}{{n.2}} = \lim \frac{{n + 1}}{{2n}} = \frac{1}{2}
\end{array}\)
