作业帮求极限：lim(x→0）(1/x^2)[1-cosx(cos2x)^(1/2)(cos3x)^(1/3)...(cosnx)^(1/n)]
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 04:36:48
![求极限：lim(x→0）(1/x^2)[1-cosx(cos2x)^(1/2)(cos3x)^(1/3)...(cosnx)^(1/n)]](/uploads/image/z/2985646-22-6.jpg?t=%E6%B1%82%E6%9E%81%E9%99%90%EF%BC%9Alim%28x%E2%86%920%EF%BC%89%281%2Fx%5E2%29%5B1-cosx%28cos2x%29%5E%281%2F2%29%28cos3x%29%5E%281%2F3%29...%28cosnx%29%5E%281%2Fn%29%5D)
求极限：lim(x→0）(1/x^2)[1-cosx(cos2x)^(1/2)(cos3x)^(1/3)...(cosnx)^(1/n)]
求极限：lim(x→0）(1/x^2)[1-cosx(cos2x)^(1/2)(cos3x)^(1/3)...(cosnx)^(1/n)]
求极限：lim(x→0）(1/x^2)[1-cosx(cos2x)^(1/2)(cos3x)^(1/3)...(cosnx)^(1/n)]
需要用到的知识点,等价无穷小+重要极限+洛必达法则
首先证明：当x→0,(cosnx)^(1/n) 1-(n/2)*x^2 (等价无穷小）
这是因为,lim(x→0) cosnx/[1-(n/2)*x^2]^n,应用洛必达法则,上下同时求导,得
上式 = lim(x→0)(-nsinnx)/[n*[(1-(n/2)*x^2)^(n-1)]*(-n*x) = lim(x→0)sinnx/nx = 1
于是
所求极限的分子可等价于
1-cosx(cos2x)^(1/2)(cos3x)^(1/3)...(cosnx)^(1/n)
1-(1-(1/2)*x^2)*(1-(2/2)*x^2)*...*(1-(n/2)*x^n) 展开
1-(1-(1/2+2/2+3/2+...+n/2)*x^2 + o(x^4)) (1/2+2/2+3/2+...+n/2)*x^2 - o(x^4)
因此所求极限=(1/2+2/2+3/2+...+n/2)=(n+1)*n/2/2 = n*(n+1)/4
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