Strengthening the notion of 'limit equivalence' The notation $f \sim g$ is used to signify that two functions are asymptotically equivalent, either at $± \infty$ or near some real number $c$. 
For example, near $0$ we have $\sin x \sim x$. The formal definition is $$f \sim g \ \text{at} \ c \ \text{iff} \ \lim_{x \to c} \frac{f(x)}{g(x)} = 1$$
where $c \in \mathbb{R} \cup \{-\infty, +\infty\}$.
However, this notion is somewhat imprecise, in my opinion. For 'equivalence' of functions, an important property we ought to have (in my opinion) is for any sufficiently well behaved (e.g., $C^{\infty}$ or even just $C^{1}$) function $h$, $f \sim g$ implies $h \circ f \sim h \circ g$. In other words, our notion of 'equivalence' should be such that composition should preserve 'equivalence'. 
However, this isn't necessarily true with the standard definition of equivalence. In particular, if $h$ varies or grows rapidly near $c$ (again, $|c|$ may be infinity), then the composition with $h$ may 'exacerbate' the otherwise small (relative) difference between $f$ and $g$. An example of this phenomenon is how $\sin x \sim x$ as $x \to 0$ but $$\lim_{x \to 0}\frac{\exp\left(-\exp\left(\frac{1}{(\sin x)^2}\right)\right)}{\exp\left(-\exp\left(\frac{1}{x^2}\right)\right)} = 0$$
Another example, this time as $x \to \infty$, is how $x^2 \sim x^2 + x$ but $\exp(x^2 + x)$ dominates $\exp(x^2)$. 

My question is, is there a stronger definition of 'asymptotic equivalence' which does not have this apparent weakness? In other words, can we strengthen the definition of $f \sim g$ such that $f \sim g$ implies $h \circ f \sim h \circ g$ for any smooth $h:\mathbb{R} \to \mathbb{R}$.  

It could be as strong as needed, but not so strong that the notion is rendered useless. For example, we could 'redefine' $f \sim g$ so that $f \sim g$ iff $f \equiv g$ in some neighbourhood of $c$ [or, in the case that $c=+\infty$ (resp. $-\infty$), for sufficiently large positive (resp. negative) $x$ we have $f \equiv g$] but this would be trivial and uninteresting. 
A valid answer to this question is that this is impossible. In other words, if $f$ and $g$ have the property that $f \sim g$ and $h \circ f \sim h \circ g$ for all smooth $h$, then $f \equiv g$ near $c$. 
Note that a related question is here. 
 A: 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.
A: Asymptotic equivalence at some finite $c\in\mathbb{R}$ actually behaves rather nicely. First, let us generalize the notion to metric spaces. Given metric spaces $X,Y$, a point $\alpha\in X$, and functions $f,g:X\to Y$, we define $\{f,g\}_\alpha$ to mean:
$$\forall{\epsilon>0}\quad\exists{\delta>0}\quad\forall{x\in X}\quad d(x,\alpha)<\delta\implies d\big(f(x),g(x)\big)\leq\epsilon d\big(f(x),f(\alpha)\big)$$
$\{f,g\}_\alpha$ is meant to capture the notion of $f$ and $g$ approximating one another near $\alpha$. When $X=Y=\mathbb{R}$, this definition is almost equivalent to the limit definition you use in your question, except that it avoids division by zero problems. Since the main result of this post relies on many properties, we state most of them without proof. $\{\cdot,\cdot\}_\alpha$ is an equivalence relation on functions $X\to Y$. Assuming $\{f,g\}_\alpha$, we must have $f(\alpha)=g(\alpha)$. Additionally, $f$ is continuous at $\alpha$ if and only if $g$ is continuous at $\alpha$.
Right Composition: Let $X,Y,Z$ be metric spaces and $\alpha\in X$. Suppose we have $\phi:X\to Y$ and $f,g:Y\to Z$ where $\phi$ is continuous at $\alpha$ and $\{f,g\}_{\phi(\alpha)}$. Then we can conclude $\{f\circ\phi,g\circ\phi\}_\alpha$.
Let $Y$ be either $\mathbb{R}$ or $\mathbb{C}$ for the remainder of this post. Recall a function $f:Y\to Y$ is analytic at $y\in Y$ whenever there exists a power series for $f$ with positive radius of convergence centered at $y$. Assume $f$ is analytic at $y$. Then either $f$ is constant in a neighborhood of $y$, or there exists unique non-zero $\lambda\in Y$ and $n\in\mathbb{N}$ such that $\{f,p\}_y$ where $p(z)=f(y)+\lambda(z-y)^n$. Think about this as taking the constant and the first non-zero term of the power series of $f$ around $y$.
Lemma: Suppose we have $f,g:X\to Y$ where $\{f,g\}_\alpha$. Let $p:Y\to Y$ be any function of the form $p(z)=\gamma+\lambda\big(z-f(\alpha)\big)^n$ for some $\gamma,\lambda\in Y$ and $n\in\mathbb{N}$. Then we can conclude $\{p\circ f,p\circ g\}_\alpha$.
Notice we do not not require any sort of continuity condition on $f$ or $g$. The proof of this lemma is the most difficult part, so we will go through it carefully. If $\lambda$ or $n$ is zero, then this is trivial, so assume both are non-zero. Then the lemma reduces to the case where $\gamma=f(\alpha)=0$ and $\lambda=1$. We want to show:
$$\forall{\epsilon>0}\quad\exists{\delta>0}\quad\forall{x\in X}\quad d(x,\alpha)<\delta\implies\left|f^n(x)-g^n(x)\right|\leq\epsilon\left|f^n(x)\right|$$
Take any $\epsilon>0$. Define $\kappa:=\frac{\epsilon}{n(1+\epsilon)}$. Notice $0<\kappa<\frac{1}{n}$. Since $\{f,g\}_\alpha$, we can find $\delta>0$ such that $\left|f(x)-g(x)\right|\leq\kappa\left|f(x)\right|$ for all $x\in B_\delta(\alpha)$. Fix some $x\in B_\delta(\alpha)$. For the sake of having concise notation, let $a:=f(x)$ and $b:=g(x)$. We want to prove $\left|a^n-b^n\right|\leq\epsilon\left|a^n\right|$.
$$\left|a^n-b^n\right|=\left|a-b\right|\left|\sum_{k=1}^{n}a^{k-1}b^{n-k}\right|\leq\kappa|a|\left|\sum_{k=1}^{n}a^{k-1}b^{n-k}\right|\leq\kappa\sum_{k=1}^{n}\left|a^{k}b^{n-k}\right|\leq$$
$$\kappa\sum_{k=1}^{n}\left|a^{k}a^{n-k}\right|+\kappa\sum_{k=1}^{n}\left|a^{k}\big(b^{n-k}-a^{n-k}\big)\right|=n\kappa|a^n|+\kappa\sum_{k=1}^{n-1}\left|a^{k}\big(b^{n-k}-a^{n-k}\big)\right|=$$
$$n\kappa|a^n|+\kappa|b-a|\sum_{k=1}^{n-1}\left|a^k\sum_{\ell=1}^{n-k}a^{\ell-1}b^{n-k-\ell}\right|\leq n\kappa|a^n|+\kappa^2|a|\sum_{k=1}^{n-1}\left|a^k\sum_{\ell=1}^{n-k}a^{\ell-1}b^{n-k-\ell}\right|\leq$$
$$n\kappa|a^n|+\kappa^2\sum_{k=1}^{n-1}\sum_{\ell=1}^{n-k}\left|a^{k+\ell}b^{n-(k+\ell)}\right|=n\kappa|a^n|+\kappa^2\sum_{k=2}^{n}(k-1)\left|a^{k}b^{n-k}\right|\leq$$
$$n\kappa|a^n|+n\kappa^2\sum_{k=1}^{n}\left|a^{k}b^{n-k}\right|$$
Defining $\beta:=\sum_{k=1}^{n}\left|a^{k}b^{n-k}\right|$, here are the relevant pieces of the above chain of inequalities:
$$\left|a^n-b^n\right|\leq\kappa\beta\leq n\kappa|a^n|+n\kappa^2\beta$$
We can plug this inequality into itself to get:
$$\left|a^n-b^n\right|\leq n\kappa|a^n|+n\kappa(\kappa\beta)\leq n\kappa|a^n|+n\kappa\left( n\kappa|a^n|+n\kappa^2\beta\right)= n\kappa|a^n|+(n\kappa)^2|a^n|+(n\kappa)^2(\kappa\beta)$$
Using induction, we get a geometric series and have the following for all $m\in\mathbb{N}$
$$\left|a^n-b^n\right|\leq|a^n|\left(\frac{n\kappa+(n\kappa)^{m+1}}{1-n\kappa}\right)+(n\kappa)^m\kappa\beta$$
Since $0<n\kappa<1$, we may let $m\to\infty$ to obtain the desired result:
$$\left|a^n-b^n\right|\leq|a^n|\left(\frac{n\kappa}{1-n\kappa}\right)=\epsilon|a^n|$$
Thus the lemma is proven. Now we can prove a theorem regarding composition.
Theorem: Let $X$ be a metric space and let $Y$ be either $\mathbb{R}$ or $\mathbb{C}$. Suppose $f,g:X\to Y$ and $\alpha\in X$ where $\{f,g\}_\alpha$ and $f$ is continuous at $\alpha$. Suppose $\phi,\psi:Y\to Y$ where $\{\phi,\psi\}_{f(\alpha)}$ and $\phi$ is analytic at $f(\alpha)$. Then we can conclude $\{\phi\circ f,\psi\circ g\}_\alpha$.
Since $\phi$ is analytic at $f(\alpha)$, there exists $\lambda\in Y$ and $n\in\mathbb{N}$ such that $\{\phi,p\}_{f(\alpha)}$ where $p(z)=\phi\circ f(\alpha)+\lambda\left(z-f(\alpha)\right)^n$ (if $\phi$ is constant in a neighborhood of $f(\alpha)$, just take $\lambda=n=0$). Since $f$ is continuous at $\alpha$, $g$ must be continuous at $\alpha$. Then we can right compose by $f$ and $g$ to get $\{\phi\circ f,p\circ f\}_\alpha$ and $\{\phi\circ g,p\circ g\}_\alpha$. From the lemma, we know $\{p\circ f,p\circ g\}_\alpha$. By the transitive property, we must have $\{\phi\circ f,\phi\circ g\}_\alpha$. Now we right compose by $g$ to get $\{\phi\circ g,\psi\circ g\}_\alpha$. Then the transitive property implies $\{\phi\circ f,\psi\circ g\}_\alpha$ as desired.
