# Is $(-1)^{\infty}$ an indeterminate form? [duplicate]

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?

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

• Yes.${}{}{}{}{}$ – Asaf Karagila Jul 3 '14 at 9:05
• Okay, I just wasn't sure because it wasn't mentioned anywhere. – k5f Jul 3 '14 at 9:06
• (-1)^n=±1 for any integer according to the parity. If you know that ∞ is odd or even, you can calculate (-1)^∞. – Bumblebee Jul 3 '14 at 10:18
• I wouldn't think of considering parity of $\infty$. That's interesting – k5f Jul 3 '14 at 10:36
• 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

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