f(x)=f(x+0)=f(x*0)=f(0),f(x)在R上处处可导,且f'(x)=[f(0)]' = 0.
1+xg(x)=f(x)=f(0).
lim_{x->-1}[1+xg(x)]=1-1=0=lim_{x->-1}[f(x)]=f(0)
所以,
f'(x)=0=f(0)=f(x)