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limx趋近于π/2tAnx/tAn3x

tanx的导数是(secx)^2,tan3x的导数是3(sec3x)^2洛比达法则要用两次原式=(1/3)*lim[(cos3x)/(cosx)]^2=(1/3)*lim[(-3sin3x)/(-sinx)]^2=3*lim{[sin(3π/2)/sin(π/2)]^2}=3

求极限 当x趋向于π/2时 limtanx/tan3x 解:lim(x→π/2)tanx/tan3x =lim(x→π/2)(sinx/cosx)/(sin3x/cos3x) =lim(x→π/2)(1/cosx)/((-1)/cos3x) =-lim(x→π/2)(cos3x/cosx) =-lim(x→π/2)(-3sin3x)/(-sinx) =3

:lim(x->π/2) (tanx/tan3x) (∞/∞) =lim(x->π/2) (secx)^2/[ 3(sec3x)^2] =lim(x->π/2) (cos3x)^2/[ 3(cosx)^2 ] (0/0) =lim(x->π/2) -3sin6x/( -3sin2x) =lim(x->π/2) sin6x/sin2x (0/0) =lim(x->π/2) 6cos6x/(2cos2x) =-6/(-2) =3

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lim(x->π/2) (tanx/tan3x) (∞/∞) =lim(x->π/2) (secx)^2/[ 3(sec3x)^2] =lim(x->π/2) (cos3x)^2/[ 3(cosx)^2 ] (0/0) =lim(x->π/2) -3sin6x/( -3sin2x) =lim(x->π/2) sin6x/sin2x (0/0) =lim(x->π/2) 6cos6x/(2cos2x) =-6/(-2) =3

等价无穷小,从名称上看,都应该知道,是无穷小才有可能使用的方法埃 而无穷小,是指函数的极限为0的情况。 现在当x→π/2的时候,无论是tanx,还是tan3x,极限都是无穷大,不是无穷校当然不能使用等价无穷小啦。又不存在等价无穷大的玩意,数学中...

答案

令t=π/2-x,则t→0 原式=lim(t→0)tan(π/2-t)/tan(3π/2-3t) =lim(t→0)cott/cot3t =lim(t→0)tan3t/tant =lim(t→0)3t/t =3

lim tanx/tan3x=lim ( 1/cosx^2) / (3/cos3x^2)= lim (cos3x^2)/3cosx^2= lim 6cos3x(-sin3x)/6cosx(-sinx)= lim sin6x/sin2x= lim 6cos6x/2cos2x= -6 / -2 = 3

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