Đáp án:
\[\frac{1}{2}\]
Giải thích các bước giải:
Ta có CTTQ:
\(1 - \frac{1}{{{n^2}}} = \frac{{{n^2} - 1}}{{{n^2}}} = \frac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}} = \frac{{n - 1}}{n}.\frac{{n + 1}}{n}\)
Do đó, ta có:
\(\begin{array}{l}
\lim \left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 - \frac{1}{{{3^2}}}} \right)\left( {1 - \frac{1}{{{4^2}}}} \right)....\left( {1 - \frac{1}{{{n^2}}}} \right)\\
= \lim \left( {\frac{1}{2}.\frac{3}{2}} \right).\left( {\frac{2}{3}.\frac{4}{3}} \right).\left( {\frac{3}{4}.\frac{5}{4}} \right).....\left( {\frac{{n - 1}}{n}.\frac{{n + 1}}{n}} \right)\\
= \lim \left[ {\left( {\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{{n - 1}}{n}} \right)\left( {\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{{n + 1}}{n}} \right)} \right]\\
= \lim \left( {\frac{1}{n}.\frac{{n + 1}}{2}} \right)\\
= \lim \frac{{n + 1}}{{2n}}\\
= \lim \frac{{1 + \frac{1}{n}}}{2} = \frac{1}{2}
\end{array}\)
