Help me solve $\lim_{n \to \infty } \frac{e^{in} - e^{-in}}{n} $ for my exam preparation

Question

I'm trying to solve or determine if this complex succession exist: $$\lim_{n \to \infty } \frac{e^{in} - e^{-in}}{n} $$

What I first realized is that the top part of the equation kind of looks like $sen(n) $,

$$\sin(n) = \frac{e^{in} - e^{-in}}{2i} $$

The only part that is missing is $ \frac{1}{2i}$. So maybe, to solve it I have to use something related to that.

The other idea that I had was, maybe I need to divide everything by $n$ in order to "get rid of" the denominador but I can not see what that is going to help me later.

Thank you for your time

Answer 1

You do not need the $\sin$ because $|e^{ix}|=1$ for all $x \in \mathbb{R}$.

Hence

$$\left|\frac{e^{in} - e^{-in}}{n} \right|\leq \frac{|e^{in}| + |e^{-in}|}{n} = \frac{2}{n}\stackrel{n\to \infty}{\longrightarrow}0$$

Answer 2

We have that

$$\frac{e^{in} - e^{-in}}{n}=\frac{e^{in} - e^{-in}}{2i}\frac{2i}{n}=2i\frac{\sin n}{n}$$

or also since $z-\bar z=2i \Im(z)$

$$\frac{e^{in} - e^{-in}}{n}=\frac{2i\Im(e^{in})}{n}$$

and then

$$\left|\frac{e^{in} - e^{-in}}{n}\right|\le \frac2n\to 0$$