①充分性:f(0)=0,则:
F'(0)
=lim(x->0)[f(x)(1+|sinx|)-f(0)(1+|sin0|]/x
=lim(x->0)f(x)(1+|sinx|)/x
=lim(x->0)[f(x)-f(0)]/x*(1+|sinx|)
=f'(0)*1
=f'(0)
②必要性:F(x)在x=0处可导,则:
F'(0+0)=F'(0-0)
由导数极限定理【此处也可改为极限式计算】:
F'(0+0)
=lim(x->0+)[f(x)(1+sinx)]'
=lim(x->0+)[f'(x)(1+sinx)+f(x)*cosx]
=f'(0)+f(0)
F'(0-0)
=lim(x->0+)[f(x)(1-sinx)]'
=lim(x->0+)[f'(x)(1-sinx)-f(x)*cosx]
=f'(0)-f(0)
∵F'(0+0)=F'(0-0)
∴f'(0)+f(0)=f'(0)-f(0)
∴f(0)=0