The result is probably that it is undefined but we can say that $\lim\limits_{n \to \infty} (-1)^n \in \{-1, 1\}$. Right? Could we continue the calculation by considering separate cases with $1$ and $-1$ ? This is studied in the case of sequences therefor $n$ is always an integer.
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4Infinity is not a number and cannot be used in an algebraic expression. Therefore $(-1)^\infty$ is meaningless until explicitly defined. – Jam Feb 01 '20 at 18:36
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1See Question 260876: What is infinity?. – Jam Feb 01 '20 at 18:37
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3Does this answer your question? Why is $1^{\infty}$ considered to be an indeterminate form – Brian61354270 Feb 01 '20 at 18:39
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2Instead consider $\lim_{n \rightarrow \infty}(-1)^n$. – Peter Szilas Feb 01 '20 at 18:39
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@Jonathcraft: "... we can say that $(-1)^\infty\in{-1,1}$. Right?" ... You're assuming "$\infty$" is not merely a number, but an integer. Stepping back from that, you might want to assert instead that $\left|(-1)^\infty\right|=1$. (See @ Mourad's comment.) – Blue Feb 01 '20 at 18:43
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1Expand it to $(-1)^x=\cos(\pi x)+i\sin(\pi x)$. – Mourad Feb 01 '20 at 18:44
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1Re your last line: How can you simultaneously replace a single symbol by two different values? – Allawonder Feb 01 '20 at 18:55
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@Blue I assume "$\infty$" to be an integer because I want to study this in the context of sequences. – BrockenDuck Feb 01 '20 at 18:56
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@Jonathcraft: You should include all such context in the body of the question. (Comments are easily overlooked.) – Blue Feb 01 '20 at 18:57
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I edited my question, I was a bit lazy on the notation :) – BrockenDuck Feb 01 '20 at 18:57
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@Mourad I edited my question because $x$ is always an integer – BrockenDuck Feb 01 '20 at 19:00
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1The limit of the sequence $1, -1, 1, -1, \ldots$ is undefined, so it isn't proper to make any statement that treats "$\lim_{n\to\infty} (-1)^n$" as a number. However, in certain contexts, it is proper to state that $+1$ and $-1$ are each "limit points" of the sequence. (The Wikipedia entry gives the example $(-1)^n\frac{n}{n+1}$, which has no limit, but two limit points (namely, $+1$ and $-1$).) It's a subtle, but significant, distinction. – Blue Feb 01 '20 at 19:24
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1The limit itself already has a definition, and using that definition, $\lim_{n\to\infty}(-1)^n$ doesn't exist. That said, there is the concept of "the set of sub sequential limits" which is a useful generalization of the limit studied in its own right, and seems to be exactly what you are talking about. – Brian Moehring Feb 01 '20 at 19:26