Question

g（x）在x=0二阶可导，g（0）=0求a 使 当x不等于0，f（x）=g（x）/x当x=0，f（x）=a；在x=0处可导

再求f在x=0处的导数

Answer 1

由导数的定义
f'(0) = lim(x->0) [f(x) -f(0)] / (x -0) = lim [g(x)/x -a] /x 记为1式
又g'(0) =lim(x->0) g(x) -g(0)/x-0 = lim(x->0) g(x)/x
因此当a = g'(0)时，1式的极限存在

又f'(0)=lim [g(x)/x -a] /x =lim [g(x) - ax]/x^2，运用洛必达法则，
f'(0) = lim [g'(x) - a] /2x (再次洛必达） = lim(x->0) g''(x) /2 = g''(0) /2

另外，若g(x) = e^x -1,满足题目可导条件且g(0) = 0 ，但g'(0) = 1不为0，楼上有问题

Answer 2

相当于求 x→0 lim g（x）/x
x→0 lim g（x）/x=x→0 lim g'(x)=g'(0)=
g（0）=0 g'(0)=O a=0
f'（x）=(g'(x)*x-g（x）)/x^2
f在x=0处的导数= x→0 lim (g'(x)*x-g（x）)/x^2
=x→0 lim (g''(x)*x+g'(x)-g''(x))/2x
=x→0 lim (g'''(x)*x+g''(x)+g''(x)-g'''(x))/2
=g''(0)+g''(0)-g'''(0))/2
g（0）=g'(0)=g''(0)=g'''(0)=0
f在x=0处的导数=0