# How to determine if a $\lim\limits_{n \rightarrow \infty}{(1+{ix\over n})^n}$ would be complex [duplicate]

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?

## marked as duplicate by Simply Beautiful Art, Robert Wolfe, Claude Leibovici limits StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 2 '17 at 8:15

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 $1\cdot\text{cis }x$.