But the limit of a^x as x-> infinity depends on what a is.

If a >1 then the limit doesn't exist.

If 0<a<1, then the limit is 0.

If a=0, then the limit is 0.

If a =1, then the limit is 1.

If a<0 , then all hell breaks loose and so no limit exists.

Jhevon, you 've said that "1^infinity is indeterminate because the base of 1 should give you 1, but thinking of exponential functions, they go off to infinity as the power goes to infinity." But, as you later said, that only applies if the base is greater than 1. So there's no contradiction there as there is for 0^0.0

I don't agree that the limit of 1^x as x-> infinity is indeterminate. Even though that's what wikipedia says - who believes them anyway!

1^infinity = 1x1x1x1x...... How can it not be 1? Where does it change from being 1 to being indeterminate?

Topsquark, I find it interesting that you learnt that it is indeterminate and not 1.

I'm open to anyone convincing me otherwise. Interesting thread!!