Recently, I have been looking at complex limits, The most famous being $e^{ix}$=$\lim\limits_{n \rightarrow \infty}{(1+{ix\over n})^n}$. An example would be that when $x = \pi$ we know that the answer will be -1. But how do you determine this due to the fact that you can always $+1$ which will determine the outcome.

I am fully aware that you are able to do this via the $i\cdot \sin(a \ln b) +\cos(a\ln b)$ however, how can you prove this via a limit, because if you test it on a calculator, most of the time you'll end up with some imaginary part.

Specifically I have been looking at the representation of $\sin x={ie^{-ix}\over 2}-{ie^{ix}\over 2}$. Everyone would be safe to assume that $\sin x$ is always real, but when you apply a limit then how can you determine if it is only real or imaginary and real?


marked as duplicate by Simply Beautiful Art, Robert Wolfe, Claude Leibovici limits Sep 2 '17 at 8:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @SimplyBeautifulArt, that's not even close. $\endgroup$ – Paul Sep 1 '17 at 13:22
  • $\begingroup$ I don't see how I have duplicated these posts? $\endgroup$ – Morgan Sep 1 '17 at 13:40
  • $\begingroup$ @Paul Have you viewed the link? It literally follows through on Yves Daoust's answer. $\endgroup$ – Simply Beautiful Art Sep 1 '17 at 13:41
  • $\begingroup$ Particularly, see this answer. $\endgroup$ – Simply Beautiful Art Sep 1 '17 at 13:42
  • $\begingroup$ @SimplyBeautifulArt the question is certainly different, but you are right that the answer does seem to cover both questions. $\endgroup$ – Paul Sep 1 '17 at 13:52

Using the polar form, you can rewrite the expression as $$\left(\sqrt{1+\frac{x^2}{n^2}}\right)^n\text{cis}\left(n\arctan\frac xn\right).$$

It tends to $1\cdot\text{cis }x$.


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