Is $\frac{x}{e^x-1}$ a monotonically decreasing function? On the first sight, the function $f(x)=\frac{x}{e^x-1}$ is a monotonically decreasing function (https://i.stack.imgur.com/ctArv.png).
But when I zoom the graph to the neighborhood of the point $(0,f(0))$, the function kind of "oscillates":
(https://i.stack.imgur.com/Wpd2T.png)
Either it's not a monotonically decreasing function, or it's an error of the calculator (desmos.com/calculator).
Formally, your question makes no sense, because $f$ is undefined at $0$. But suppose that you extend the domain of $f$ to $\mathbb R$, defining$$f(0)=\lim_{x\to0}\frac x{e^x-1}=1.$$Then, yes, $f$ is decreasing near $0$. That's so because $f(x)=\frac1{g(x)}$, where\begin{align}g(x)&amp;=\begin{cases}\frac{e^x-1}x&amp;\text{ if }x\neq0\\1&amp;\text{ otherwise}\end{cases}\\&amp;=1+\frac x2+\frac{x^2}{3!}+\cdots\end{align}Then $g'(0)=1$ and, since $g$ is increasing near $0$, $f$ is decreasing near $0$.
$$ 
f(x)=\frac{x}{e^x-1} $$
   Find its first derivative   it is zero at x=0 only and function is not defined at and x= 0 is not the vertical asymptot  ; as x approach towards 0       , y approaches towards 1 ; 
f'(x)&lt;0 for all x in its domain hence decreasing function.