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A little bit of context. I was given a problem which went like if $X_n$ is normally distributed with mean $a_n$ and is converging in distribution to $X$, then $a_n\to a$ for some $a\in\mathbb R$ and $X$ is normally distributed. The question is quite doable using characteristic functions, I guessed, until I ended up with the problem if both limits $$\lim_{n\to\infty}\cos(a_nt) \ \ \ \text{ and } \lim_{n\to\infty} \sin(a_nt)$$ for all $t\in\mathbb R$ then the limit $$\lim_{n\to\infty} a_n$$ exists as well and is finite. If I assume $a_n$ is bounded, then I don't have any problem proving the statement. But if $a_n$ is unbounded, then there is, without loss of generality, a subsequence $a_{n_k}\to\infty$. Therefore my question which is a bit general:

Question. Let $f:\mathbb R\to\mathbb R$ be a continuous function and $a_n$ be a sequence diverging to $\infty$, if $$\lim_{n\to\infty} f(a_nt)$$ exists for all $t\in\mathbb R$, do we have $\lim_{x\to\infty} f(x)$ exists as well?

The result is well-known if $a_n=n$. If we have proven this statement then my original claim is easy to prove. I attempted to mimic the proof in this post, but I stopped where they say $(f(nmx))_{n\in\mathbb N}$ is a subsequence of $(f(nx))_{n\in\mathbb N}$ since in my particular case I don't have $(f(a_na_mx))_{n\in\mathbb N}$ is a subsequence of $(f(a_nx))_{n\in\mathbb N}$. However deep inside, I believe there is a possibility for mimicing.

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For each positive real $x$, there is a unique integer $r=r(x)$ such that $1\le2^rx<2$. Let $f(x)=\sin(2\pi(2^rx-1))$. Then $f$ is continuous, $f(x)=f(2x)=f(4x)=\cdots$ for all positive $x$ (so $\lim_n f(a_nx)$ exists for $a_n=2^{n-1}$, $n=1,2,3,\dots$), but $\lim_{x\to\infty}f(x)$ does not exist.

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    $\begingroup$ $r(x)$ is not continuous. Is it obvious that $f$ is continuous? $\endgroup$ – Kavi Rama Murthy Feb 12 '19 at 12:19
  • $\begingroup$ Yes, it is obvious. All we're doing here is putting a single period of the sine function on each of $[1,2),[2,4),[4,8),\dots$. $\endgroup$ – Gerry Myerson Feb 12 '19 at 20:46
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Let $g(x)=\begin{cases}2x&0\le x\le\frac12\\ 2(1-x)&\frac12\le x\le 1\\0&\text{otherwise}\end{cases}$, a bump of height $1$ supported on $[0,1]$.

For $a_n=2^n$, consider the following function: $$f(x)=g(x-2)+g(x-6)+g(x-14)+\cdots+g(x+2-2^n)+\cdots$$ For any $x$, at most one element of the sequence $2^nx$ is in one of those intervals $(2^k-2,2^k-1)$ where $f$ is nonzero. As such, $\lim_{n\to\infty}f(2^n x)=0$ for all $x$. On the other hand, $\lim_{x\to\infty}f(x)$ doesn't exist, since there are peaks with value $1$ no matter how far out we go.

So no, the theorem does not extend to arbitrary sequences.

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