Is $(-1)^{\infty}$ an indeterminate form?

Question

We know that $\lim_{n\to\infty}(-1)^n=(-1)^{\infty}$ doesn't exist. Now take $\lim_{n\to\infty}(-1)^{2n}=(-1)^{\infty}$. This limit exists, because $\lim_{n\to\infty}(-1)^{2n}=\lim_{n\to\infty}((-1)^2)^n=1$. Does this mean that $(-1)^{\infty}$ is an indeterminate form?

Answer 1

Not necessarily because the concept "indeterminate form" does not have a widely accepted formal definition; in many books this is simply presented as a list of examples. Your example is not in that list, so in those books, it would not be an "indeterminate form". However, if you try to write a definition that is not merely a list, then your example likely could be an indeterminate form, depending of course on how the definition is written.

Answer 2

I suppose so ... even $1^\infty$ is an indeterminate form, and you are just adding a question about the sign.

Answer 3

Maybe over the integers, but over the reals - which is what's more often meant when people talk about indeterminate forms - it'll always either diverge or go to zero, since for values where the base is sufficiently close to $-1$, further approaching the limit enough to increase the exponent by $1$ (which you can always do since the exponent approaches infinity) will reverse the sign. So is that an indeterminate form? I'd say no.