Question

微积分，求解

微积分，求解极限和导数的知识

Answer 1

当x<=1时，n(x-1)<=0
n->∞时，e^[n(x-1)]->0
f(x)=(x^2*0+ax+b)/(0+1)
=ax+b
当x>1时，n(1-x)<0
n->∞时，e^[n(1-x)]->0
f(x)=lim(n->∞) {x^2+(ax+b)e^[n(1-x)]}/{1+e^[n(1-x)]}
=(x^2+0)/(1+0)
=x^2
因为f(x)在x=1处可导，所以
①f(1)=f(1-)=f(1+)
(ax+b)|(x=1)=(x^2)|(x=1)
a+b=1
②f'(1-)=f'(1+)
(ax+b)'|(x=1)=(x^2)'|(x=1)
a=2
综上所述，a=2，b=-1