# Indeterminate form - 1^∞

1. May 10, 2012

### 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.

2. May 10, 2012

### micromass

Staff Emeritus
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!!

Last edited by a moderator: May 10, 2012
3. May 10, 2012

### Infinitum

Aha!! That clears it up.

Thank you.