大一一道关于夹逼定理的题目

2024-11-25 15:53:58
推荐回答(1个)
回答1:

这像定积分定义的题。
lim(π/n)((cosπ/n)^2+(cos2π/n)^2+...+(cos(n-1)π/n)^2)
=lim(π/n)((cosπ/n)^2+(cos2π/n)^2+...+(cos(n-1)π/n)^2+1)
考虑函数(cosx)^2,该函数在区间[0,π]连续,分区间为n等分,取右端点
lim(π/n)((cosπ/n)^2+(cos2π/n)^2+...+(cos(n-1)π/n)^2)
=lim(π/n)((cosπ/n)^2+(cos2π/n)^2+...+(cos(n-1)π/n)^2+1)
=∫(0,π)(cosx)^2dx
=∫(0,π)(1+cos2x)/2 dx
=π/2