Proving that: $e^{-x(1/\tau - i\xi)} \to 0$ as $x \to \infty$.

Question

I remember my friend showing me how sandwich theorem can be applied here. Unfortunately, I can't find his solution anymore and I am not familiar with sandwich theory.

Answer 1

Assume $\tau>0$ and $x,\xi \in \mathbb{R}$. Then using $\left|e^{-ia}\right|=1 $ for any $a \in \mathbb{R}$, we get $$ 0<\left|e^{-x(1/\tau - i\xi)}\right|=\left|e^{-x/\tau}\right|\times \underbrace{\left|e^{ ix\xi}\right|}_{\color{red}{=1}}\leq e^{-x/\tau} $$ giving $$ 0\leq\lim_{x \to \infty}\left|e^{-x(1/\tau - i\xi)}\right|\leq \lim_{x \to \infty}e^{-x/\tau}=0. $$