设函数f(x)连续,g(x)=∫¹.f(xt)dt,且当x趋向于0时f(x)/x的极限为A,A为常数,求g'(x)并讨论g'(x

设函数f(x)连续,g(x)=∫¹.f(xt)dt,且当x趋向于0时f(x)/x的极限为A,A为常数,求g'(x)并讨论g'(x
设函数f(x)连续,g(x)=∫¹.f(xt)dt,且当x趋向于0时f(x)/x的极限为A,A为常数,求g'(x)并讨论g'(x

设函数f(x)连续,g(x)=∫¹.f(xt)dt,且当x趋向于0时f(x)/x的极限为A,A为常数,求g'(x)并讨论g'(x
显然f(0)=0.且f'(0)=lim (f(x)-f(0)/(x-0)=A.对积分做变量替换xt＝y,得g(x)=积分（从0到x）f(y)dy/x,x不等于0时.g(0)=0.因此
g'(0)=lim (g(x)-g(0)]/(x-0)=洛必达法则lim f(x)/(2x)=A/2.g'(x)=[xf(x)-积分（从0到x）f(y)dy]/x^2,于是lim g'(x)=洛必达法则lim xf'(x)/(2x)=lim f'(x)/2=A/2＝g'(0).综上知g'(x)连续.