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limx趋向于0tAnx/tAn3x

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

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

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

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(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

如图中::

答: x→π,tanx→0 属于0-0型,可以应用洛必达法则lim(tanx/tan3x) =lim {(1/cosx)^2/[3/(cos3x)^2]} =1/3

因为x趋近于π/2所以cos3x和cosx都是趋近于0的但是sin3x和sinx不是趋近于0,此时可以将x=π/2直接代入所以sin3x/sinx=-1 在后面一步使用了求导,上下都求导数cos3x导数是-3sin3xcosx导数是-sinx希望对您有所帮助

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