∵lim(x->π/3)[(1-2cosx)/sin(x-π/3)]
=lim(x->π/3)[2sinx/cos(x-π/3)] (0/0型极限,应用罗比达法则)
=2sin(π/3)/cos(π/3-π/3)
=2*(√3/2)
=√3
∴lim(x->π/3){[(1-2cosx)ln(1+x)]/sin(x-π/3)]
=lim(x->π/3)[(1-2cosx)/sin(x-π/3)]*lim(x->π/3)ln(1+x)
=√3*ln(1+π/3).