(2sinx+cosx-1)/x=[4sin(x/2)cos(x/2)+1-2sin(x/2)sin(x/2)-1]/x=sin(x/2)[sin(x/2)+2cos(x/2)]/(x/2)
显然,当x→0,sin(x/2)[sin(x/2)+2cos(x/2)]/(x/2)=sin(x/2)+2cos(x/2)=2.
或者
因为x→0,(2sinx+cosx-1)/x为0/0型,可以上下同时求导,既有2cosx-sinx=2