Question

变上限积分函数求极限

第二题怎么求极限

Answer 1

lim(x->0) x^2.∫(0->x) f(t) dt / ∫(0->x^2) f(t) dt (0/0) :分子，分母分别求导
=lim(x->0) [x^2.f(x) +2x.∫(0->x) f(t) dt] / [2x.f(x^2) ]
=lim(x->0) [xf(x) +2∫(0->x) f(t) dt] / [2f(x^2) ] (0/0) :分子，分母分别求导

=lim(x->0) [xf'(x)+ f(x) +2f(x)] / [4x.f'(x^2) ]
=lim(x->0) [xf'(x)+ 3f(x) ] / [4x.f'(x^2) ]
=(1/4) lim(x->0) f'(x)/ f'(x^2) + [3/(4f'(0) )]. lim(x->0) [ f(x)/x]
=1/4 +[3/(4f'(0)) ].f'(0)
=1/4 + 3/4
=1