解法一：∵lim(x->π/2)[(sinx-1)tanx]=lim(x->π/2){[(sinx-1)/cosx]sinx}=lim(x->π/2)[(sinx-1)/cosx]*lim(x->π/2)(sinx)=lim(x->π/2){[sin(x/2)-cos(x/2)]/[cos(x/2)+sin(x/2)]}*1=0*1=0lim(x->π/2){(sinx)^[...
