Does $\frac{a_n}{x_n} \approx L$ for sufficiently large $n$ $\implies a_n \approx Lx_n$ for sufficiently large $n$?

Question

Given two strictly increasing sequences $a_n$ and $x_n$, is it true that: $$\lim_{n\to\infty} \frac{a_n}{x_n} = L \implies a_n = L*x_n$$ For sufficiently large $n$? It seems intuitive, given one can simply write state that $\frac{a_n}{x_n}=L$ for sufficiently large $n$, but how might a formal proof of this go? I think it would begin something like: $$L-\epsilon \leq \frac{a_n}{x_n} \leq L + \epsilon$$ For $n>0$ and arbitrary $\epsilon \geq 0$. Then, we have: $$x_nL-x_n\epsilon \leq a_n \leq x_nL + x_n\epsilon$$ Letting $\epsilon \to 0$: $$x_nL \leq a_n \leq x_nL \implies a_n=x_nL$$ Is this proof sound? If not, what might be the salvage?

Answer 1

No, it is not true. For example, with $a_n=1-\frac{1}{n}$ and $x_n=1-\frac{2}{n}$, both sequences are strictly increasing and you have $L=1$, but there is no $n$ for which $a_n=x_n$.