Prove that this function is bounded. I am trying to prove that this function is bounded for all $x$ $$\frac{e^{-ix}-1}{x}$$

I found that :

$|\frac{e^{-ix}-1}{x}| \le \frac{|cos(x)|}{|x|}+\frac{|sin(x)|}{|x|}+\frac{1}{|x|}$

So when $x \rightarrow (-)\infty $, the function goes to $0$

and in $x=0$, we have that : $\lim_{x \to 0} |\frac{e^{-ix}-1}{x}| = \lim_{x \to 0} |\frac{e^{-ix}-e^(i0)}{x-0}| = (e^{-ix})'(0) = 0$

and by continuity of the function in $]0,\infty[$ it cannot diverge between $0$ and $(-) \infty$, so it must be bounded

Is it correct ?
I'll write $f(x)= \frac{e^{-ix}-1}{x}$, $x \in \mathbb{R} \backslash \{0\}$. As you've noted, a bound like
$$|\frac{e^{-ix}-1}{x}| \leq \frac{|e^{ix}|+1}{|x|} = \frac{2}{|x|}$$
implies that $f$ is bounded at infinity. In particular, say, $|f(x)| \leq 2$ for $|x| \geq 1$.
In addition, you've effectively noted that 
$$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{e^{-ix}-e^{-i0}}{x-0} = -ie^{-ix}|_{x=0} = -i,$$
and this implies that $\exists \epsilon &gt; 0$ such that $|x| &lt; \epsilon \implies |f(x)+i| &lt; 1$, which by the triangle inequality implies $|f(x)| \leq 2$. 

So, $|f(x)| \leq 2$ for $|x|&lt;\epsilon$ and $|x|&gt;1$. Meanwhile, the set $A=\{x \in \mathbb{R} \backslash \{0\} : \epsilon \leq |x| \leq 1\}$ is compact, so continuity of $x \mapsto |f(x)|$ implies that it achieves its maximum value $M \geq 0$ on $A$. Hence, $\forall x \in \mathbb{R} \backslash \{0\}$ we have the bound $|f(x)| \leq M+2$.

A similar argument yields that any continuous function on $\mathbb{R}^n \backslash \{p\}$ for which the limits as $x \to \infty$ and $x \to p$ exist is bounded. More generally, the existence of the $x \to \infty$ limit may be relaxed to $f$ being bounded outside of a sufficiently large sphere.
Hint: $$ \left|e^{ix} - 1\right|\le \min(2, |x|) $$