Question

求极限:lim(x→0)(tanx-sinx)/x^3

Answer 1

简单计算一下即可，答案如图所示

Answer 2

那我就不用洛必达法则了呵呵~,用定理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