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.
