calculus - If $lim_{n to +infty} f(alpha n)$ exists , does this imply $lim_{x to +infty} f(x)$ exists?

Let $f:\mathbb{R} \to \mathbb{R}$. Suppose $f$ has the following property: for all $\alpha >0$, the sequence $\{f(\alpha n)\}_{n \geq 0}$ converges either to a finite or infinite extended real number as $n \to +\infty$. In other words, we assume

$$\forall \alpha \in \mathbb{R}_+,\lim_{n \to +\infty} f(\alpha n) = L(\alpha)$$ where $L(\alpha)$ is a value depending on $\alpha$. Note that $L(\alpha)$ may be $\pm \infty$.

Does it follow that $$\lim_{x \in \mathbb{R}, x \to +\infty} f(x)$$ exists as a finite or infinite extended real number?

Note that if the above claim is true, then $L(\alpha)$ is in fact independent of $\alpha$. However, we do not assume this.

We can also play around with this to get stronger and weaker statements. Does it hold if we just assume $a \in \mathbb{Q}_+$? Does it hold if we assume $f$ is continuous?

I haven't made too much progress. However, this is quite similar to a classic little theorem which states that for continuous $f$, if for all $\alpha>0$, $f(\alpha n)\xrightarrow[n \in \mathbb{N}, n\to+\infty]{} 0$ then $f(x)\xrightarrow[x \in \mathbb{R}, x\to+\infty]{} 0$ which is proven with the Baire category theorem. However, my 'conjecture' only assumes the existence of the limit (rather than giving an explicit value, like $0$) and also does not assume continuity. I can't quite modify that proof to make it work here.

Answer

It is true if we further assume that $f$ is continuous!

First we shall use the following Lemma:

Lemma. Let $I_n=[a_n,b_n]\subset(0,\infty)$ be an increasing sequence of nontrivial closed intervals in the sense that
$$
a_n $$Then the set$$
S=\Big\{x\in(0,\infty) : \text{$nx\in \bigcup_{k\in\mathbb N} I_k$
for infinitely many $n\in\mathbb N$}\Big\},
$$is dense in $(0,\infty)$. In particular, if $\mathbb N=K\cup L$, with $|K|=|L|=\aleph_0$ and $K\cap L=\varnothing$, then$$
S=\Big\{x\in(0,\infty) :
\text{$mx\in \bigcup_{k\in K} I_k$ and $nx\in\bigcup_{\ell\in L} I_\ell$
for infinitely many $m$ and infinitely many $\,n\in\mathbb N$}\Big\},
$$
is also dense in $(0,\infty)$.

We postpone the proof of this fact. Clearly, for every $\,x_0\in(0,\infty)$
$$\liminf_{x\to\infty}\, f(x) \le \lim_{n\to\infty}\,f(nx_0)\le \limsup_{x\to\infty}\, f(x).$$
Hence, if $\lim_{x\to\infty}\,f(x)$ does NOT exist, then we can pick
$\,A,B \in\mathbb R$, such that
$$\liminf_{x\to\infty}\, f(x) < A < B < \limsup_{x\to\infty}\, f(x).$$
Due to the continuity of $f$ it is possible to define intervals $I_n=[a_n,b_n]$ and $J_n=[c_n,d_n]$, $n\in\mathbb N$, such that $\,a_n
and
$$
f\,\big|_{I_k} B,
\quad \text{for all $n\in\mathbb N$}.
$$
Due the Lemma, there exists a dense set of points $x$, with the property
that, for infinitely many $n$'s the multiple $nx$ belongs to $\bigcup_{k\in\mathbb N} I_k$, and for infinitely many $n$'s the multiple $nx$ belongs to $\bigcup_{k\in\mathbb N} J_k$.

This in turn implies that $\lim_{n\to\infty}f(nx)$ does not exist.

Proof of the Lemma. Let $[c,d]\subset(0,\infty)$. We shall prove that
there exists an $x\in [c,d]$, such that for infinitely many $n$ the multiple
$nx$ belongs to an interval of the form $I_k$, $\,k\in K$, and for
infinitely many $n$ the multiple $nx$ belongs to an interval of the form
$I_\ell$, $\ell\in L$. This is based on the observation that, if $a_n$ is
sufficiently large, then
$$[c',d']=\frac{1}{N}[a_n,b_n]\cap[c,d]\quad\text{is a nontrivial interval},$$
for $N=\lfloor a_n/d\rfloor+1$. This allows us to recursively define a sequence of nontrivial

closed intervals $J_n=[c_n,d_n]$, where $[c_0,d_0]=[c,d]$, $[c_1,d_1]=[c',d']$, and we pick
the $a_n$'s, so that the first one is in $K$, the next in $L$, the next in $K$ and so on. In
this way we have a sequence of closed intervals
$$J_0\supset J_1\supset\cdots\supset J_n\supset J_{n+1}\supset\cdots.$$
Clearly $S=\bigcap_{n\in\mathbb N} J_n\ne\varnothing$, and for each $x\in S$, infinitely
many multiples belong to $I_k$'s, $k\in K$, while infinitely
many multiples belong to $I_\ell$'s, $\ell\in L$.