Question

f(x)在x=0处可微,且f(0)=0,证明存在在x=0处连续的g(x)使得f(x)=xg(x)成立.

Answer 1

f(x)在x=0处可微,则

f'(0)=lim[（f(x)-f(0)]/x x趋向0

所以可知有

f(x)=f'(0)x+f(0).x趋向0

因为f(0)=0.所以

f(x)=f'(0)x

令f'(x)=g(x)

所以存在在x=0处连续的g(x)

使得f(x)==f'(0)x=xg(0)成立.