lim(x趋向于0）(1+sinx-cosx)/(1+sinβx-cosβx)

网友回答

(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/2))
=sinx/sinbx极限为1/b.