7.f(0)=1,设u=1/n→0，对y-x=e^[x(1-y)]求导数得y'-1=e^[x(1-y)]*(1-y-xy'),∴{1+xe^[x(1-y)]}y'=1+(1-y)e^[x(1-y)],∴y'(0)=1,∴原式→[f(u)-1]/u→f'(0)=1.
