Limit of e with imaginary number Important part:
$\lim\limits_{x\to\infty} e^{-ix} - e^{-x} $
This is suppose to approximate to "$1$" but the way I see it we have $0 - 0$ ...

Expand the term with the imaginary exponent using Euler's formula:
$$\lim_{x\to\infty} \left[e^{-ix} - e^{-x}\right] = \lim_{x\to\infty} \left[\cos(-x) + i\sin(-x) - e^{-x}\right]=\lim_{x\to\infty} \cos(x) - i\lim_{x\to\infty} \sin(x)$$
which does not converge to either 0 or 1, but instead oscillates around the unit circle.
