solution

It's enough to show x^x->1 as x->+0, since then the sequence become 0,1,1,1,.... all 1's.

Here log = ln

let y=x^x, then log(y) = x log(x) = log(x)/(1/x) (*)
log(x) ->-infy and 1/x->infy as x->+0, by L'Hospital's Rule,
lim log(y) = lim (1/x)/(-1/x^2) = lim(-x) = 0.
lim(y) = lim(e^log(y)) = 1, as x->0. Done.

所有跟帖： 

• 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