如图如图求详解 10
f(x)=lim(n->∞) { x^2.e^[n(x-1)] +ax +b }/ { e^[n(x-1)] +1 }
case 1: x<1
f(x)
=lim(n->∞) { x^2.e^[n(x-1)] +ax +b }/ { e^[n(x-1)] +1 }
= ax+b
case 2: x=1
f(x)
=lim(n->∞) { x^2.e^[n(x-1)] +ax +b }/ { e^[n(x-1)] +1 }
=( 1^2 +a(1) +b )/ ( 1 +1 )
= (1/2)(a+b+1)
case 3: x>1
f(x)
=lim(n->∞) { x^2.e^[n(x-1)] +ax +b }/ { e^[n(x-1)] +1 }
分子，分母同时除以 e^[n(x-1)]
=lim(n->∞) { x^2 + (ax+b)/e^[n(x-1)] } / { 1+ 1/e^[n(x-1)] }
= x^2
f(1-) = lim(x->1) ax+b = a+b
f(1) = (1/2)(a+b+1)
f(1-) = lim(x->1) x^2 = 1
x=1 , f(x) 连续
f(1)=f(1+)=f(1-)
a+b=1
ie
a+b =1 , x=1, f(x) 连续
f'(1-)
=lim(h->0) [a(1+h) +b -f(1) ]/h
=lim(h->0) [a(1+h) +b -1 ]/h
=lim(h->0) [a+b-1 +ah ]/h
=lim(h->0) ah/h
=a
f'(1+)
=lim(h->0) [(1+h)^2 -f(1) ]/h
=lim(h->0) [(1+h)^2 -1 ]/h
=lim(h->0) (2h+h^2)/h
=2
=>
x=1, f(x) 可导
f'(1+) = f'(1-)
=> a=2
a+b=1
b= -1
ie
(a,b) = ( 2, -1)
