- #1

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- TL;DR Summary
- Does a limit exist? I believe no!

##i^\frac{1}{n}## has n roots. If one is not careful, the limit as ##n \to \infty## is 1. Simple proof: ##i=e^\frac{\pi i}{2}## or ##i^\frac{1}{n}=e^\frac{\pi i}{2n} \to e^0=1##.

This does not take into account the n roots, since ##i=e^{(\pi i)(2k+\frac{1}{2})}##.. Here ##\frac{k}{n} ## can have any value with ##1\le k \le n## for the n roots. Therefore given any ##0\lt x \le 1##, it is easy to construct a sequence of ##\frac{k}{n}\to x## so any point on the unit circle will be the limit of a sub-sequence.

Am I missing something?

This does not take into account the n roots, since ##i=e^{(\pi i)(2k+\frac{1}{2})}##.. Here ##\frac{k}{n} ## can have any value with ##1\le k \le n## for the n roots. Therefore given any ##0\lt x \le 1##, it is easy to construct a sequence of ##\frac{k}{n}\to x## so any point on the unit circle will be the limit of a sub-sequence.

Am I missing something?

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