原式=lim(x->0)[(sinx/cosx-sinx)/sin³x]
=lim(x->0)[(1-cosx)/(sin²xcosx)]
=lim(x->0)[2sin²(x/2)/(sin²xcosx)]
=lim(x->0)[(x/sinx)²*(sin(x/2)/(x/2))²*(1/(2cosx))]
=1*1*(1/2) (利用重要极限lim(x->0)(sinx/x)=1)
=1/2