求极限lim(x→0)(tanx-sinx)/sin³x

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求极限lim(x→0)(tanx-sinx)/sin³x

tanx-sinx=sinx/cosx-sinx
=[sinx(1-cosx)]/cosx
(tanx-sinx)/sin3x=(1-cosx/cosx)sin2x
=(1-cosx/cosx)/1-cos2x
=(1-cosx/cosx)/[(1-cosx)(1+cosx)]
=1/[cosx(1+cosx)]
lim趋向于0时应该=1/2

原式=(sinx/cosx-sinx)/sin3x
=(1/cosx-1)/(1-cos2x)
另t=cosx:
原式=(1/t-1)/(1-t^2)化简=1/(t+t^2)
x---->0,t--->1
所以原式极限:1/2