Let $a_n$ be a sequence of positive real numbers so that for each $t>0$ the ratio $\dfrac{a_{[nt]}}{a_n}\to x_t$ as $n$ goes to infinity, where $x_t$ is a finite positive number for each $t>0$. Show that $\dfrac{a_{[[nt]t]}}{a_{[nt^2]}}\to1$ as $n$ tends to infinity.
Observe that $x_1=1$. I think I need to show the required limit is this but although intuitively clear it doesn’t seem to yield.
Let us write $x(t) = x_t$ for readability. Then we claim that
Assuming this claim, the desired limit immediately follows from $a_{[[nt]t]}/a_{[nt^2]} \to x(t)^2/x(t^2)$. Also, this claim allows us to establish a stronger claim that $x(t) = t^{\alpha}$ for some $\alpha \in \mathbb{R}$.
Proof of Claim. We first consider the case where $t = p/q$ is rational with $p, q \in \mathbb{Z}_+$. Then
$$ x(st) = \lim_{n\to\infty} \frac{a_{[stqn]}}{a_{qn}} = \lim_{n\to\infty} \frac{a_{[spn]}}{a_{pn}} \cdot \frac{a_{[tqn]}}{a_{qn}} = x(s)x(t) \tag{1} $$
Next we consider arbitrary $s, t > 0$. In view of $\text{(1)}$, it suffices to assume that $s$ is sufficiently small so that $s < \min\{ \frac{1}{2t}, 1 \}$ holds.
Now we seek for a condition on $n$ such that $[tsn] = [t[sn]]$ holds. To this end, write $m' = [sn]$ and $r' = sn-[sn]$. Then this condition is equivalent to
$$ r' < \frac{1 - (tm' - [tm'])}{t}. \tag{2} $$
We note that $1 - (tm - [tm]) \geq \frac{1}{2} $ holds for infinitely many $m \in \mathbb{Z}_+$. (This claim is non-trivial only when $t$ is irrational, where the claim follows by the density of $\{tm-[tm] : m\in\mathbb{Z}_+\}$ in $[0, 1]$.) Also, for each such $m$, we can pick $n$ such that $[sn] = m$ and $sn-[sn] < \frac{1}{2t}$. Now the upshot is that we can find a subsequence $(n_j)_{j=1}^{\infty}$ satisfying $[tsn_j] = [t[sn_j]]$ for all $j$. Then
$$ x(st) = \lim_{j\to\infty} \frac{a_{[st n_j]}}{a_{n_j}} = \lim_{j\to\infty} \frac{a_{[s n_j]}}{a_{n_j}} \cdot \frac{a_{[t[s n_j]]}}{a_{[s n_j]}} = x(s)x(t). $$
Therefore the desired claim follows. ////