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IP属地:重庆来自Android客户端1楼2024-12-16 14:03回复
    顶🔝


    IP属地:重庆来自Android客户端2楼2024-12-16 14:15
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      2026-01-25 06:25:15
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      IP属地:上海来自Android客户端3楼2024-12-16 14:23
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        x->0
        分子
        ln(2-cosx)
        ~ln(1+(1/2)x^2-(1/24)x^4)
        ~[(1/2)x^2-(1/24)x^4] -(1/2)[(1/2)x^2-(1/24)x^4]^2
        =[(1/2)x^2-(1/24)x^4] -(1/2)[(1/4)x^4+...]
        =(1/2)x^2- (1/6)x^4
        sinx ~ x -(1/6)x^3
        [sinx]^2 ~ [x -(1/6)x^3]^2 ~ x^2 -(1/3)x^4
        [1+ (sinx)^2]^(1/3)
        ~[ 1+x^2 -(1/3)x^4 ]^(1/3)
        ~ 1 +(1/3)[x^2 -(1/3)x^4] -(1/9)[x^2 -(1/3)x^4]^2
        = 1 +(1/3)[x^2 -(1/3)x^4] -(1/9)[x^4+...]
        =1 +(1/3)x^2 -(2/9)x^4
        2ln(2-cosx) -3[[1+ (sinx)^2]^(1/3) -1]
        ~2[(1/2)x^2- (1/6)x^4 ] -3[(1/3)x^2 -(2/9)x^4]
        =(1/3)x^4
        分母
        [xln(1+x)]^2~ (x^2)^2 =x^4
        //
        lim(x->0) { 2ln(2-cosx) -3[[1+ (sinx)^2]^(1/3) -1] }/[xln(1+x)]^2
        =lim(x->0) (1/3)x^4/x^4
        =1/3


        IP属地:中国香港4楼2024-12-16 15:24
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          硬头皮做两次洛必塔


          IP属地:北京5楼2024-12-16 19:00
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