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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
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