0^0 = 1

1) 0^0 = 1
2) Suppose that the sequence f(1) = x, f(2)=x^x, ..., f(n+1)=f(n)^x, then

as x->+0, f(2) -> 1. Suppose f(n)^x -> y, as n->00, x->0, then

y= f(2)^y, and f(2) ->1 so y->1. Done.

所有跟帖： 

• 回复：0^0 = 1 -yma16- ♂ (257 bytes) () 04/29/2007 postreply 15:53:32

• solution -emlx- ♂ (353 bytes) () 04/29/2007 postreply 16:51:38

• yma16 does have a good question here.. -乱弹- ♂ (215 bytes) () 04/30/2007 postreply 11:27:57

• 回复：yma16 does have a good question here.. -yma16- ♂ (363 bytes) () 04/30/2007 postreply 11:42:41

• 回复：0^0 = 1 -yma16- ♂ (31 bytes) () 04/30/2007 postreply 11:47:36

• 看错题了, 以为 f(n+1)=f(n)^x. 应是f(n+1)=x^f(n). -emlx- ♂ (0 bytes) () 04/30/2007 postreply 12:01:46

• 数学功底太差, 不好意思. -emlx- ♂ (0 bytes) () 04/30/2007 postreply 13:51:57

• 你很强了。 我也常看错题目 -乱弹- ♂ (0 bytes) () 04/30/2007 postreply 14:36:20

• 不对，不对。。你不是功底问题，是态度问题！这个你要 -idiot94- ♂ (122 bytes) () 04/30/2007 postreply 18:28:13