就是用最直接最简单的方法做,下面提供2种不同的方法:参考定理lim[x→0]
sinx/x=1
lim[x→0]
(tanx-sinx)/x³
=lim[x→0]
(sinx/cosx-sinx)/x³
=lim[x→0]
(sinx-sinxcosx)/(x³cosx)
=lim[x→0]
sinx(1-cosx)/(x³cosx)
=lim[x→0]
sin³x(1-cosx)/(x³sin²xcosx)
=lim[x→0]
(sinx/x)³·(1-cosx)/(sin²xcosx)
=lim[x→0]
(sinx/x)³·(1-cosx)/[(1-cos²x)cosx]
=lim[x→0]
(sinx/x)³·(1-cosx)/[(1+cosx)(1-cosx)cosx]
=lim[x→0]
(sinx/x)³·1/[(1+cosx)cosx]
=1·1/(1+1)
=1/2
lim[x→0]
(tanx-sinx)/x³
=lim[x→0]
(sinx/cosx-sinx)/x³
=lim[x→0]
(sinx/x)·(1-cosx)/(x²cosx)
=lim[x→0]
(sinx/x)·[1-(1-2sin²x/2)]/(x²cosx)
=lim[x→0]
(sinx/x)·2sin²(x/2)/(x²cosx)
=lim[x→0]
(sinx/x)·2sin²(x/2)/[4(x/2)²cosx]
=lim[x→0]
(1/2)·(sinx/x)·[sin(x/2)/(x/2)]²·(1/cosx)
=1/2·1·1²·1
=1/2