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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1Yes.${}{}{}{}{}$ – Asaf Karagila Jul 03 '14 at 09:05
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Okay, I just wasn't sure because it wasn't mentioned anywhere. – k5f Jul 03 '14 at 09:06
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1(-1)^n=±1 for any integer according to the parity. If you know that ∞ is odd or even, you can calculate (-1)^∞. – Bumblebee Jul 03 '14 at 10:18
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I wouldn't think of considering parity of $\infty$. That's interesting – k5f Jul 03 '14 at 10:36
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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. – Smurf Feb 19 '17 at 13:26
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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.
Mark
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I suppose so ... even $1^\infty$ is an indeterminate form, and you are just adding a question about the sign.
GEdgar
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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.
user361424
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