$1^\infty=1$ or is indeterminate?

Question

$$\underbrace{1\cdot1\cdot1\cdot1\cdot1\cdot1\cdot1\cdots1\cdot1}_{\infty}=1$$ because of $1^\infty$ is indeterminate?$$$$ And $$\lim_{x\to\infty}1^x=\lim_{x\to\infty}1=1?$$

Answer 1

In calculus, when we talk about indeterminate forms, it's usually something like this:

Suppose $\lim_{x\to\infty} f(x) = 1$ and $\lim_{x\to\infty} g(x) = \infty$. Then what is $\lim_{x\to\infty} f(x)^{g(x)}$? In this case, yes, the form is indeterminate.

Consider the logarithm: $\log f(x)^{g(x)} = g(x)\log f(x)$. We've said that $g(x)\to\infty$, and since $f(x)\to 1$, then we have $\log f(x)\to 0$. Therefore, the logarithm of our "$1^\infty$" form is a $\infty \cdot 0$ form, which is definitely indeterminate.

Answer 2

The answer depends on what exactly is meant by $1^\infty $. The most natural way is to define it as a limit: $1^\infty =\lim_{n\to \infty}1^n $, and this limit is clearly equal to $1$. So, if you accep this definition of $1^\infty $, then it follows that $1^\infty =1$. Of course, you may choose to define this concept otherwise, and obtain, perhaps, different answers.