Is the function $ \exp(2\pi i /\epsilon) $ equivalent to zero?

Question

Is the function

$$ f\left(\frac{x}{\epsilon}\right)=\exp\left(\frac{2\pi ix}{\epsilon}\right)=g(x) $$

equivalent to zero ? in the limit $ \epsilon \to 0 $ ?

If I take the derivative $$\frac{ g(x+\epsilon)-g(x)}{\epsilon} $$ is $0$ because the function $g(x)$ has a period 'epsilon'

Also if I take the integral of $g(x)$ is $0$ almost everywhere

So is this function equivalent to $0$ ??

Answer 1

$|g(x)|=1$ for all real numbers $x$ and all $\epsilon >0$. How can $g$ be equivalent to $0$?

Answer 2

You cannot take the derivative with the same $\epsilon$ as that in the function.

$$\lim_{\epsilon\to0}\frac{e^{2i\pi\epsilon/\epsilon}-1}\epsilon\ne \lim_{\epsilon\to0}\lim_{\eta\to0}\frac{e^{2i\pi \eta/\epsilon}-1}\eta.$$