Prove that $f(x)=\frac x{e^x-1}$ for $x\neq 0$, $f(0)=1$, is $C^\infty$

Question

How to prove that $f : x \mapsto \frac{x}{\exp(x)-1}, \ x\neq 0$, and $1$ if $x=0$ is $\mathcal{C}^{\infty}(\mathbb{R})$.

For $x\neq 0$ , $f'(x)=\frac{\exp(x)-1-x\exp(x)}{(\exp(x)-1)^2}$, so we can extend $f'$ to $0$ taking : $f'(0)=-\frac{1}{2}$. Then we can conclude that $f \in \mathcal{C}^1(\mathbb{R})$.

Now maybe an induction could work but the calculation of the successive derivatives seems tricky.

Thanks in advance !

Answer 1

A possible approach is the following. Consider the function $g=1/f$, and use the Taylor expansion of the exponential function. The remark the derivatives of $g$ at zero are never equal to zero, and conclude that $f$ is infinitely often differentiable at zero.

Answer 2

HINT.-The limit of $f(x)$ when $x\to0$ exists and is in fact equal to $1$ so one can well define $f(0)=1$ but the second derivative of $f(x)$ is equal to $$e^x\left(x+\frac{2}{(e^x-1)^3}\right)$$ and the limit of $\dfrac{2e^x}{(e^x-1)^3}$ does not exists for $x\to0$ (we have $-\infty$ when $x\to0^-$ and $+\infty$ when $x\to0^-$)