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This question already has an answer here:

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?

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marked as duplicate by projectilemotion, Daniel W. Farlow, Juniven, Shailesh, Leucippus Feb 20 '17 at 2:59

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    $\begingroup$ Yes.${}{}{}{}{}$ $\endgroup$ – Asaf Karagila Jul 3 '14 at 9:05
  • $\begingroup$ Okay, I just wasn't sure because it wasn't mentioned anywhere. $\endgroup$ – k5f Jul 3 '14 at 9:06
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    $\begingroup$ (-1)^n=±1 for any integer according to the parity. If you know that ∞ is odd or even, you can calculate (-1)^∞. $\endgroup$ – Bumblebee Jul 3 '14 at 10:18
  • $\begingroup$ I wouldn't think of considering parity of $\infty$. That's interesting $\endgroup$ – k5f Jul 3 '14 at 10:36
  • $\begingroup$ Has nothing to do with the "parity" of $\infty$, just picture $(-1)^{1/2}$, is not even defined. That is more a problem in the definition of the function than an indetermination. $\endgroup$ – Smurf Feb 19 '17 at 13:26
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I suppose so ... even $1^\infty$ is an indeterminate form, and you are just adding a question about the sign.

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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.

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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.

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