不是f(0)=0嘛 为什么原函数也等于0?
lim(x-0) f(x)/x =1
g(x) =∫(0->1) f(xt) dt
To find: g'(x)
To prove: x=0, g'(x) 连续
solution:
f(0) =0
g(x) =∫(0->1) f(xt) dt
g(0)
=∫(0->1) f(0) dt
=∫(0->1) 0 dt
=0
//
let
u=xt
du= xdt
t=0, u=0
t=1, u=x
g(x)
=∫(0->1) f(xt) dt
=∫(0->x) f(u) (du/x)
=(1/x)∫(0->x) f(u) du
=(1/x)∫(0->x) f(t) dt
x≠0
g'(x) =(1/x)f(x) -(1/x^2)∫(0->x) f(t) dt
lim(x->0) g'(x)
=lim(x->0) [(1/x)f(x) -(1/x^2)∫(0->x) f(t) dt ]
=lim(x->0) [xf(x) -∫(0->x) f(t) dt ]/ x^2
洛必达
=lim(x->0) [xf'(x)+f(x) -f(x) ]/ (2x)
=lim(x->0) xf'(x)/ (2x)
=(1/2)f'(0)
x=0
g'(0)
=lim(h->0) [g(h) -g(0) ]/h
=lim(h->0) [(1/h)∫(0->h) f(t) dt -g(0) ]/h
=lim(h->0) ∫(0->h) f(t) dt /h^2
洛必达
=lim(h->0) f(h) /(2h)
洛必达
=lim(h->0) f'(h)/2
=(1/2)f'(0)
=lim(x->0) g'(x)
x=0, g'(x) 连续
=∫(0到1) 0 dt
=0 ∫(0到1)dt
=0