# Infinite differentiability of a function with a removable discontinuity

How would I prove that $\frac x{e^x-1}$ is infinitely differentiable? (This question came up since the No 1 answer in Maclaurin series for $\frac{x}{e^x-1}$ states that the function is infinitely differentiable, which - at least to me - isn't obvious at the first glance. Both the function and all its derivatives have a power of $e^x-1$ in the denominator which causes a (removable) discontinuity in $x=0$, but I suppose the answer can't be to look the removability of this discontinuity for all derivatives, right?)

• What the poster really meant was that $f(x)=\{ \frac{x}{e^x-1}$ if $x\neq 0$, $1$ if $x=0$ is infinitely differentiable. In proving the differentiability you'd have to look at 1-sided limits in the definition of the derivative. May 26 '16 at 17:33
• @mathematician I know, but still - how do I prove that this function is infinitely differentiable? May 26 '16 at 17:34
• Inductively. The only point of concern is $x=0.$ Now take derivative at nonzero $x$'s and take derivative at zero using its definition as the usual quotient. If limit of $f'(x)$ as $x$ goes to zero is equal to $f'(o)$ the $f'$ is continuous, and that proves $f$ is $C^1.$ Then you must see a pattern emerging for second, third, ... derivatives. May 26 '16 at 17:57

Hint: Look at $g(x) = (e^x-1)/x$ instead. Then $g$ equals a power series that converges on all of $\mathbb R,$ hence is $C^\infty.$ Show that $g$ is never $0.$ Therefore …
Note that, if $f(x)$ is infinitely often differentiable and $f(0) \neq 0$ then $\frac{1}{f(x)}$ is infinitely often differentiable in $x=0$. This follows by induction, since you can show by induction that the $n$-th derivative of $\frac{1}{f(x)}$ is of the form $\frac{p(f, f^\prime, ..., f^{(n)})}{f^{2n}}$ with a polynomial $p$ of $n+1$ variables.
Apply this to $f(x)= \frac{e^x-1}{x}$