As for a rigorous proof, we note that
\\ $For large enough$ \ x \ $note that$ \\ \\ 0 < \frac{x}{e^x-1} < \frac{1}{x} \\ \\ $The upper bound can be proven with calculus, as with the lower bound$ \\ \\ $Hence by taking$ \ x \ $to infinity$ \ 0 < \lim_{x \to \infty} \frac{x}{e^x-1} < \lim_{x \to...