Strengthening the notion of 'limit equivalence'

Suppose we're in the case $f\sim g$ at $\infty,$ where $\lim_{x\to \infty} f(x) = \lim_{x\to \infty} g(x)=\infty.$ Assume that for every $R>0$ there exists $x>R$ such that $f(x)\ne g(x).$ Then WLOG we can choose $x_n\to \infty$ such that $f(x_n)< g(x_n)$ for all $n.$ Because $f(x_n),g(x_n)\to \infty,$ we can pass to a subsequence, which I'll continue to denote by $x_n,$ such that

$$f(x_1) < g(x_1) < f(x_2) < g(x_2) < \cdots.$$

We can then construct $h\in C^\infty(\mathbb R)$ such that $h(f(x_n)) = 1, h(g(x_n))=2$ for all $n.$ I won't say too much about how to do this at this point; please feel free to ask questions.

It follows that $h\circ f, h \circ g$ are not asymptotically equivalent at $\infty.$ We were able to design this $h$ assuming that for every $R>0$ there exists $x>R$ such that $f(x)\ne g(x).$ Thus if $h$ preserves asymptotic equivalence, there must be some $R>$ for which the above fails, i.e., $f\equiv g$ on $(R,\infty).$

The assumption $f\sim g$ at $\infty,$ where $\lim_{x\to \infty} f(x) = \lim_{x\to \infty} g(x)=\infty,$ is just one case, but I think the same $f\equiv g$ conclusion will hold in the other cases.