Function with all derivatives $\infty$ at $0$. I was wondering if you could construct a infinitely differentiable function where every derivative is $\infty$ at $0$. I tried some ideas like bump-function but didn't really get anywhere. Does anyone has a idea?
 A: Since infinity is not a number, this needs to be translated into limits to make sense.
Even then, there is an issue either side of $0$: if $f'(x) \to +\infty$ as $x \to 0$, then

*

*you might not be too surprised to see an example of  $f''(x) \to +\infty$ as $x \to 0^-$, since $f'(x)$ must be increasing somewhere just below $0$, and of  $f''(x) \to -\infty$ as $x \to 0^+$, since $f'(x)$ must be decreasing somewhere just above $0$;

*you would be surprised to see $f''(x) \to +\infty$ in both cases as $f'(x)$ cannot be increasing when it is already infinite.

If you do not care about the sign of the infinite derivative then you could try something like
$$f(x)=x^{1/3}$$
when $x\not =0$ and $f(0)=0$, using real cube roots of negative numbers.  It has derivatives $$f^{(n)}(x)=\frac{(-1)^{n-1}}{\left(x^{1/3}\right)^{3n-1}}\prod\limits_{k=1}^n \left(k-\frac13\right)$$ with the limit of the absolute $n$th derivative being $$\lim\limits_{x \to 0} \left|f^{(n)}(x)\right|=\infty$$ for any positive integer $n$, and both for all the odd derivatives and for negative $x$ and even derivatives you have $\lim\limits_{x \to 0}f^{(n)}(x)=+\infty$.
