lim π/n[sinπ/n+sin2π/n+...+sin(n-1)π/n] 当n趋近于无穷
来源:百度知道 编辑:UC知道 时间:2024/05/30 00:26:37
谢谢!
lim π/n[sinπ/n+sin2π/n+...+sin(n-1)π/n]
=π*lim[sinπ/n+sin2π/n+...+sin(n-1)π/n]/n
=π*∫sinxdx(对0到π求定积分)
=π*2
=2π
显然是0呀,令 π/n=x,当n趋近于无穷,x趋近于0,
lim xsinkx当x趋近于0时,结果均为0,所以原极限结果为0
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