I was curious in showing that $\left(1+\frac{ix}{n}\right)^n$ goes to $e^{ix}$ as n goes to infinity. I believe this is right as included it is included numerous times in here: Proof of Euler's formula that doesn't use differentiation?. I have:
$e^{ix}=\left(\left(1+\frac{1}{n}\right)^n\right)^{ix}=\left(1+\frac{1}{n}\right)^{ixn}$ then name ixn as k so we have: $\left(1+\frac{ix}{ixn}\right)^{ixn}=\left(1+\frac{ix}{k}\right)^k$. As per the definition of e, n approaches infinity, and so k (which equals ixn) approaches $i\infty$.
My problem is that k is going to $i\infty$, not just $\infty$ as in the original. Did I make an error or am I misunderstanding something?
Thanks and regards!
Edit: My question is where did I make a mistake? I am aware there may not be one and that it may be salvageable, but I don't know how.
$$a_n=\left(1+\frac{ix}{n}\right)^n\implies \log(a_n)=n\log\left(1+\frac{ix}{n}\right)$$ Now, by Taylor $$\log(a_n)=n\left(\frac{i x}{n}+\frac{x^2}{2 n^2}+O\left(\frac{1}{n^3}\right)\right)=i x+\frac{x^2}{2 n}+O\left(\frac{1}{n^2}\right)$$ $$a_n=e^{\log(a_n)}=e^{ix}\left(1+\frac{x^2}{2 n}\right)+O\left(\frac{1}{n^2}\right)$$