Understanding 1^∞ as an Indeterminate Form

[Infinitum]

I am sure that x^{\infty} as x\to1 is an indeterminate form.

But can someone please explain how 1^{\infty} is indeterminate? I always thought it is equal to 1.

 

[micromass]

We choose 1^\infty to be indeterminate. The reason we do this, is because of the limit

\lim_{n\rightarrow +\infty} \left(1+\frac{1}{n}\right)^n

This will converge to "1^\infty" in a certain sense, but it will not converge to 1. The reason that it does not converge to 1 is that 1+\frac{1}{n} converges to 1 too slow and that n converges to infinity too fast.

Another explanation is the following, say that f(n) converges to 1 and that g(n) converges to infinity, then

\lim_{n\rightarrow +\infty} f(n)^{g(n)} = \lim_{n\rightarrow +\infty} e^{g(n) log(f(n))}=e^{\lim_{n\rightarrow \infty} g(n) log(f(n))}

But as f(n) converges to 1, we have that log(f(n)) converges to 0. So g(n)log(f(n)) converges to "0\cdot (+\infty)" and this is a known indeterminate form.

So if we want to define 1^\infty and if we want to let it behave like we want it to, then we would have to define 0\cdot (+\infty) a well!

 

[Infinitum]

Aha! That clears it up.

Thank you.