Question
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?
Answer
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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