Question

求照片上这个题解答过程

Answer 1

g'(0)
=lim(h->0) [ f(h)/h -f'(0) ]/h (0/0)
=lim(h->0) [ hf'(h)- f(h) ]/h^2 (0/0)
=lim(h->0) [ hf''(h) +f'(h)- f'(h) ]/(2h)
=lim(h->0) f''(h)/2
=f''(0)/2

追问

可以给个完整全面的答案不，我有点看不懂

追答

只要证明g'(0)存在 =>g'(x) 连续
g'(0)
=lim(h->0) [g(h)-g(0)]/h
=lim(h->0) [ f(h)/h -f'(0) ]/h (0/0)
分子，分母分别取导
=lim(h->0) [ hf'(h)- f(h) ]/h^2 (0/0)
分子，分母分别取导
=lim(h->0) [ hf''(h) +f'(h)- f'(h) ]/(2h)
=lim(h->0) f''(h)/2
=f''(0)/2
ie
g'(0) = f''(0)/2 存在