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).
 A: 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)&=\begin{cases}\frac{e^x-1}x&\text{ if }x\neq0\\1&\text{ otherwise}\end{cases}\\&=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 $0$.
A: No, the function is really decreasing ! Indeed, if you differentiate the function, you have : $$f'(x)=-\left(\frac{e^x(x-1)+1}{(e^x-1)^2} \right)\le 0. $$
And the function is continuous when $x=0$. Indeed : $$e^x-1\sim x $$ when $x$ is at the neighbourhood of $0$, so : $$f(x)\sim\frac{x}{x}\sim 1, $$ and by continuity you have : $f(0)=1$.
There must be an error when your calculator draws the function, espacially when $x$ is at the neighbourhood of $0$.
A: Interpret the function as $1$ when $x=0$. Claim: $(1-x)e^{x} \leq 1$ for all $x$.  Use the fact that $e^{-x} \geq 1-x$ for all real $x$ to see that the inequality holds.  The claim is proved and now you can check that the derivative of the given function is $\leq 0$ so it is decreasing on the entire real line.  
A: $$ 
f(x)=\frac{x}{e^x-1} $$
   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)<0 for all x in its domain hence decreasing function.
