[交流] 如下

2楼: Originally posted by 花痕逸韵 at 2015-07-26 22:46:18
f'(ξ)=f(x+1)-f(x)，两边取极限....这里假定了极限的存在，没有考虑极限不存在的情况

3楼: Originally posted by dfdx at 2015-07-26 22:50:48
你说错了，右边能取极限，当然左边能取极限啊！
...

4楼: Originally posted by 花痕逸韵 at 2015-07-26 22:55:09
右边不一定能取极限...

5楼: Originally posted by dfdx at 2015-07-26 22:57:20
右边能取极限，f(+∞)=0
...

7楼: Originally posted by liuqh at 2015-07-27 00:21:12
一开始就错了，
存在ξ∈(x，x+1)，使得f'(ξ)=f(x+1)-f(x)，两边取极限有f'(ξ=+∞)=0，
即f'(x)在无穷远的极限存在且为0
这只能说明f'(x)中存在一个收敛于0的序列，并不能保证f'(x)在无穷远的极限存在且为0
...

7楼: Originally posted by liuqh at 2015-07-27 00:21:12
一开始就错了，
存在ξ∈(x，x+1)，使得f'(ξ)=f(x+1)-f(x)，两边取极限有f'(ξ=+∞)=0，
即f'(x)在无穷远的极限存在且为0
这只能说明f'(x)中存在一个收敛于0的序列，并不能保证f'(x)在无穷远的极限存在且为0
...