technically the limit I found seems to suggest 1^∞ = ∞ but this is obviously not true. This is just the limit. 1 to any power = 1 no matter what. The nth root of n as n approaches infinity, tends to be 1, but 1 is the limit, it can never equal 1.

]]>It was shown somewhere else that this equation:

Had the solution of x = the nth root of n.

Surely, what you've just said means that:

]]>Anyways, the title of this thread is "who woulda thunk it?" clearly the answer is "everyone woulda thunk it.... except mikau..."

I feel stupid...

]]>1.01^ big number, goes up.]]>

At any rate, the reason I didn't conclude that from the start, I could have looked at the exponant 1/n and figured this comes to zero so its one no matter what, but singling out and isolating a piece of a limit and evaluating can be dangerous, ESPECIALLY when they are directly connected to infinitly large numbers. I repeat: In the definition of e: Limit of (1 + 1/n)^n as n approaches infinity. Your inclined to notice the 1/n term and rewrite it as zero, but this infinitly small fraction is raised to an infinitly large power! So it does make a difference! Here we have the reverse, an infinitly large number being raised to an infinitly small power. For all I knew it might have ended up being 1/e or something like that. Thats why I made no assumptions.

Like I said, when infinitly small and infinitly large come in contact, sometimes they "cancel eachother out" so isolating each individual limit can be dangerous. Don't you agree?

]]>#*#=2

#*#*#=3

#*#*#*#=4

#*#*#*#*#=5

#*#*#*#*#*#=6

The #'s go up slowly, but multiplying is so powerful, why wouldn't you expect to

approach one from above? I would.

'Course I can't remember L'Hopital's Rule by its name. I have a lot to review.

Don't explain for me, I'll do some research...

The proof of this uses just a simple common technique. I'll use the method of finding the limit of ln (n^(1/n)) as n approaches infinity, and raising e to the power of that limit.

limit of ln (n^(1/n)) as n approaches infinity

= limit of 1/n ln( n) as n approaches infinity = limit of (ln n)/n as n approaches infinity ( = ∞/∞) since this is an indeterminant form, we use L' hopitals rule, 1/n / 1 = 1/n. As n approaches infinity 1/n = 0.

e^0 = 1 so the limit of the the nth root of n as n approaches infinity is one! Thats just awsome! Check it on a calculator if you don't believe me.

]]>