1个回答
展开全部
你好!
∵x→0时,1-cosx~ x²/2
∴1-cos√x ~ x/2
lim<x→0+> [1- √(cosx) ] / [x(1- cos√x )]
= lim<x→0+> [1- √(cosx)] / (x²/2)
= lim<x→0+> [1-√(cosx)][1+√cosx] / (x²/2)[1+√cosx]
= lim<x→0+> (1- cosx) / (x²/2)(1+√cosx) ...展开全文你好!
∵x→0时,1-cosx~ x²/2
∴1-cos√x ~ x/2
lim<x→0+> [1- √(cosx) ] / [x(1- cos√x )]
= lim<x→0+> [1- √(cosx)] / (x²/2)
= lim<x→0+> [1-√(cosx)][1+√cosx] / (x²/2)[1+√cosx]
= lim<x→0+> (1- cosx) / (x²/2)(1+√cosx)
= lim<x→0+> (x²/2) / (x²/2)(1+√cosx)
= 1/2收起
∵x→0时,1-cosx~ x²/2
∴1-cos√x ~ x/2
lim<x→0+> [1- √(cosx) ] / [x(1- cos√x )]
= lim<x→0+> [1- √(cosx)] / (x²/2)
= lim<x→0+> [1-√(cosx)][1+√cosx] / (x²/2)[1+√cosx]
= lim<x→0+> (1- cosx) / (x²/2)(1+√cosx) ...展开全文你好!
∵x→0时,1-cosx~ x²/2
∴1-cos√x ~ x/2
lim<x→0+> [1- √(cosx) ] / [x(1- cos√x )]
= lim<x→0+> [1- √(cosx)] / (x²/2)
= lim<x→0+> [1-√(cosx)][1+√cosx] / (x²/2)[1+√cosx]
= lim<x→0+> (1- cosx) / (x²/2)(1+√cosx)
= lim<x→0+> (x²/2) / (x²/2)(1+√cosx)
= 1/2收起
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询