lim(x→0)(x-sinx)/[x(1-cosx)]
=lim(x→0)(1-cosx)/[(1-cosx)+xsinx] 罗必塔法则
=lim(x→0)sinx/[sinx+sinx+xcosx]
=lim(x→0)sinx/[2sinx+xcosx]
=lim(x→0)1/[2+xcosx/sinx]
=lim(x→0)1/lim(x→0)[2+xcosx/sinx]
=1/[2+1]
=1/3
附加说明：
lim(x→0)xcosx/sinx
=lim(x→0)[cosx-xsinx]/cosx
=[1-0]/1
=1