On the theorem "$3$ is everywhere"

In this Numberphile video it is stated that "almost all natural numbers have the digit $3$ in their decimal representation", and a proof of this fact is proposed. A sketch of the proof follows:

Denote by $D_3$ the set of natural numbers having a digit $3$ in their decimal representation. For all $n \ge 1$, denote by $$f(n) = | D_3 \cap \{ 1, \dots , n\} |$$ it is proved that for all $n$ $$f(10^n) = 10^n- 9^n $$ holds (and this is quite clear), hence $$\lim_{n \to + \infty} \frac{f(10^n)}{10^n} = 1$$ and this concludes the proof in the video.

Now, this proof is clear and evident to me, but I think that it is incomplete, since we should prove that

$$\lim_{n \to + \infty} \frac{f(n)}{n} = 1$$

while this is not proved in the video. So my question is: how to prove this?

EDIT: Obviously, if the limit exists, then it is equal to $1$: so I am asking how to show that the last limit actually exists.

You are correct that proving $\lim_{n \to + \infty} \frac{f(10^n)}{10^n} = 1$ is not enough. I can define $g(n)=n$ if $n=10^k$ and $g(n)=0$ otherwise. I then have $\lim_{n \to + \infty} \frac{g(10^n)}{10^n} = 1$ but $\lim_{n \to + \infty} \frac{g(n)}{n}$ does not exist. We have shown that if $\lim_{n \to + \infty} \frac{f(n)}{n}$ exists, it is $1$, so all we need now is that it exists. We can use the same argument. Define $h(n)=n-f(n)$ as the number of numbers less than $n$ that are missing $3$. They show $h(10^n)=9^n$. $h(n)$ is monotonically increasing as when you go from $n$ to $n+1$ you either add $1$ or $0$ to $h$. Now for any $k$, let $m=\lfloor \log_{10}k \rfloor$ so that $10^m$ is the power of $10$ just below $k$. $\frac {f(k)}k =1-\frac {h(k)}k \gt 1-\frac {9^{m+1}}{10^m}\to 1$