# 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 ?

• Try expanding $e^{ix}$, but you should show some of your own work and then ask about specifically what you are confused about. – Nalt Jun 3 at 14:46
• The derivative of $e^{-ix}$ evaluated at zero is $-i$, not $0$. You should add a bit to your argument explaining why continuity implies it cannot "diverge" between $0$ and $\pm \infty$ given the limits you've computed. – jawheele Jun 3 at 16:04

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 $$\exists \epsilon > 0$$ such that $$|x| < \epsilon \implies |f(x)+i| < 1$$, which by the triangle inequality implies $$|f(x)| \leq 2$$.
So, $$|f(x)| \leq 2$$ for $$|x|<\epsilon$$ and $$|x|>1$$. 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: $$\left|e^{ix} - 1\right|\le \min(2, |x|)$$