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

问题补充：

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

答案：

(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.