Question

设f(x)为连续函数，g(x)=∫(0,1)f(xt)dt,且当x趋于0时，limf(x)/x=A，求g'(x)并讨论g'(x)在x=0的连续性

Answer 1

由题设，知f(0)=0,g(0)=0,令u=xt,得g(x)=∫(0,x)f(u)du/x,(x≠0)，从而g'(x)=[xf(x)-∫(0,x)f(u)du]/x^2,(x≠0)，由导数定义有，g'(0)=
Lim(x->0) ∫(0,x)f(u)du/x^2=
Lim(x->0)f(x)/2x=A/2,
由于
Lim(x->0)g'(0)=Lim(x->0)[xf(x)-∫(0,x)f(u)du]/x^2=Lim(x->0)f(x)/x-Lim(x->0)∫x0 f(u)du/x^2=A-A/2=A/2= g'(0)
从而知g'(x)在x=0连续