Real Analysis Limit Problem with Continuity and Existence of Limit [duplicate]

I was working on this problem and I cannot manage to solve it.

Let $f : \mathbb{R} \to \mathbb{R}$ be continuous. For all $\alpha > 0$, the sequence $f(\alpha), f(2\alpha), f(3\alpha), \ldots \to 0$. Show that $\lim\limits_{x\to\infty} f(x) = 0$.

Here are my thoughts:

1. Immediately I thought this was trivial because the limit is unique. But, I believe the limit is unique, if it exists. Therefore, I do not think this approach works.

2. Next was maybe some liminf or limsup thing and show they are equal. However, just doing this doesn't make use of continuity. I can probably think of some sort of weird noncontinuous function where the rest of the conditions of the problem hold. For example, for each $\alpha$ and fixed $\varepsilon$, we can set the $n$ which eventually bounds the function $|f|$ by $\varepsilon$ further and further away.

3. Maybe something with the continuity. All I could say was that theres a small ball $(\alpha-small, a+small)$ such that there is some $n$ where the function is trapped under $\varepsilon$ in that small ball.

Help would be appreciated. I do not know how to use continuity to solve this problem.