(1+sinx-cosx)/(1+sinβx-cosβx)=（sinx+2(sin(x/2))^2)/（sinbx+2(sin(bx/2))^2)=[(sin(x/2)+cos(x/2))/(sin(bx/2)+cos(bx/2))]*(sin(x/2)/sin(bx/2))极限与下式的极限相等(cos(x/2)/cos(bx/2))*(sin(x/2)/sin(bx/...