Lingustic representation of a number

Question

Let $f(n)$ be the length of a number when written out. For example, $f(5)=4$, because "Five" has 5 letters in it. Now, it is quite obvious that in English, $4$ is the greatest number such that $f(x) \geq x$.

However, in a video he conjectures that in any (non-trivial) linguistic number system there is an $x$ for which $f(x) \geq x$ and for no number bigger than $x$ is this true. I have tested in a variety of different languages and this holds true.

I realize this isn't a very rigorous stating of a conjecture, but I was wondering if there is any way to mathematically formalize his conjecture, or even prove it.

Answer 1

In any reasonable language, the number of letters required to express a number should correspond "linearly" to the number of digits needed. That is, if $d(x)$ is the number of digits in the decimal representation of a number, then we should certainly have $$ f(x) \leq C d(x) $$ for some sufficiently big number $C$ (for example, this works in English easily with $C = 100$). However, it is notable that $$ d(x) \leq \log_{10}(x) $$ So, we know that we will always have $f(x) \leq C \log_{10}(x)$. As such, it's enough to show that:

For any value of $C > 0$, there exists an $r > 0$ such that whenever $x > r$, we have $C \log_{10}(x) < x$.

That is,

For any value of $C > 0$, there exists an $r > 0$ such that whenever $x > r$, we have $\frac{\log_{10}(x)}{x} < \frac 1C$.

In fact, proving that this is the case is precisely the same as proving that $$ \lim_{x \to \infty} \frac{\log_{10}(x)}{x} = 0 $$ which is easy to do using L'Hôpital's rule, for example.