First off, $\frac{x}{x-1} > 0$ iff $x < 0$ or $x > 1$, so we can't take the natural logarithm if $x\in[0,1]$. My answer addresses the inequality for real-valued $x$, as in the original post (proving it for $x > 1$ and disproving it for $x < 0$). Now $ e\le \left(\frac{x}{x-1}\right)^{x}= \left(1-\frac1x\right)^{-x} \to e $ already guarantees the limiting behavior for us. Depending on the sign of $x$, this inequality becomes $ e^{-1/x} \le 1-\frac1x \qquad(x < 0) $ $ e^{-1/x} \ge 1-\frac1x \qquad(x > 1) $ Setting $t=\frac1x\lt1$ and using the Taylor series, this translates to $ \eqalign{ 1-t &\le e^{-t} = \sum_{n=0}^\infty\frac{(-1)^n t^n}{n!} \\ &\le 1-t+\frac{t^2}2+\frac{t^3}6+\cdots } $ for $t \in (0,1)$, which is patently true, while for negative values of $t$, the reversed inequality is patently false (which we can easily check in the original by trying $x=-1$ since $2 < e$). Therefore I would suggest adding the stipulation that $x > 0$ (necessarily) or actually, $x > 1$.