Is there a function that grows slower than the factorial function and whose derivative grows faster than the function itself?

Question

Is there a function that grows slower than the factorial function and whose derivative grows faster than the function itself? In symbols, is there a function $f(x)$ that satisfies $f(x) = o(x!)$ and $\lim_{x \to \infty} f(x)/f'(x) = 0$?

Answer 1

For any $r>0$ we have $$ \lim_{x\to\infty} \frac{r^x}{x!}=0 $$We can augment the exponent a little bit: for instance, $$ \lim_{x\to \infty} \frac{r^{x+\log(x)^2}}{x!}=0 $$But not too much: if $r>1$, for any $\varepsilon>0$ we have $$ \lim_{x\to \infty} \frac{r^{x^{1+\varepsilon}}}{x!}=\infty $$

Answer 2

Stirling's formula says $x! = e^{x \ln(x) - x + O(\ln(x))}$, so we can do $f(x) = e^{g(x)}$, where $g(x)$ is any function that grows more slowly than $x \ln(x)$, but faster than $ax$ for any constant $a$. One obvious one to use could be $g(x) = x \ln \ln x$, which makes $$f(x) = e^{x \ln \ln x},$$ for which $$f'(x) = e^{x \ln \ln x} \left( \ln \ln x + \frac{1}{\ln x} \right) = f(x)\ln \ln x + \frac{f(x)}{\ln x}$$ clearly satisfies $\lim_{x \rightarrow \infty} \frac{f'(x)}{f(x)} = \infty$.

It's easy to verify that for this choice of $f(x)$, $$\lim_{x \rightarrow \infty} \frac{f(x)}{x!} = \lim_{x \rightarrow \infty} e^{x \ln \ln x - x \ln x + x + O(\ln x)} = 0.$$