因为f(x)为(-1,1)上的奇函数
故,f(0)=0
f'+(0)
=lim(x→0+) [f(x)-f(0)] / (x-0)
=lim f(x)/x
=A
f'-(0)
=lim(x→0-) [f(x)-f(0)] / (x-0)
换元t=-x
=lim(t→0+) [f(-t)-f(0)] / (-t-0)
=lim(t→0+) [-f(t)-f(0)] / (-t-0)
=lim f(t)/t
=A
因为左右导数都存在且相等
因此,f'(0)存在,且f'(0)=A
有不懂欢迎追问