Under what conditions asymptotic equivalence can hold after composition? Let define the equivalence relation $$f\sim_\alpha g:\iff\lim_{x\to\alpha}\frac{f(x)}{g(x)}=1$$
Suppose we have two functions $f$ and $g$ and a continuous function $h$. Under what conditions we can says that $f\sim_\alpha g\implies(h\circ f)\sim_\alpha(h\circ g)$?
This question comes from this example: let $f(x):=\frac{\sqrt{x^2+x+1}}x$, then $f\sim_\infty g$ for $g(x):=1$ but $\ln f\not\sim_\infty \ln g$.
If there is some textbook or reference about this kind of questions related to the above equivalence relation I will be glad to know it.
 A: First of all, your example is kind of silly since $\ln g$ is identically $0$ and so $\ln f/\ln g$ is simply never defined.  Obviously, if $h(x)$ can be $0$, you can get similar silly counterexamples by letting $g$ be such that $h(g(x))$ is always $0$.  However, you can get a less silly example for $h(x)=\ln x$ by choosing $g$ to approach $1$ very fast without actually being equal to $1$, so $\ln g$ will be much closer to $0$ than $\ln f$.
With this in mind, let me assume that $h(x)$ is always positive, or at least is always positive on the images of $f$ and $g$ in a neighborhood of $\alpha$.  I will also assume for convenience that $f$ and $g$ are positive in some neighborhood of $\alpha$, so we may consider $h$ as a function $A\to(0,\infty)$ for some $A\subseteq(0,\infty)$ containing the images of $f$ and $g$ near $\alpha$ (this is no real loss of generality, since if $f\sim_\alpha g$ then $f$ and $g$ have the same sign in some neighborhood of $\alpha$).  Define a function $h'$ by $h'(x)=\log(h(\exp x))$.  Then I claim $f\sim_\alpha g$ always implies $h\circ f\sim_\alpha h\circ g$ iff $h'$ is uniformly continuous on its domain $\{x:\exp x\in A\}$. 
Indeed, suppose $h'$ is uniformly continuous.  Suppose $f\sim_\alpha g$, and fix $\epsilon>0$.  Choose $\delta_0>0$ such that if $|x|<\delta_0$, $|\exp(x)-1|<\epsilon$.  Choose $\delta_1>0$ such that if $|x-y|<\delta_1$, then $|h'(x)-h'(y)|<\delta_0$.  Choose $\delta_2>0$ such that if $|x-1|<\delta_2$, $|\log x|<\delta_1$.  Finally, let $U$ be a neighborhood of $\alpha$ such that if $x\in U$, then $\left|\frac{f(x)}{g(x)}-1\right|<\delta_2.$
Now if $x\in U$, we have $$\left|\frac{f(x)}{g(x)}-1\right|<\delta_2$$ which implies $$|\log f(x)-\log g(x)|=\left|\log\frac{f(x)}{g(x)}\right|<\delta_1$$ which implies $$|\log(h(f(x)))-\log(h(g(x)))|=|h'(\log f(x))-h'(\log g(x))|<\delta_0$$ which implies $$\left|\frac{h(f(x))}{h(g(x))}-1\right|=|\exp(\log(h(f(x)))-\log(h(g(x))))-1|<\epsilon.$$ Since $\epsilon>0$ was arbitrary, this means $h\circ f\sim_\alpha h\circ g$.
Conversely, suppose $h'$ is not uniformly continuous.  Then for some $\epsilon>0$, there exist sequences $(x_n)$ and $(y_n)$ such that $|x_n-y_n|\to 0$ but $|h'(x_n)-h'(y_n)|>\epsilon$ for all $n$.  Choosing $f$ and $g$ such that $\log f(x)=x_n$ and $\log g(x)=y_n$ when $1/n\leq |x-\alpha|<1/(n-1)$, we will then have $f\sim_\alpha g$ but $h\circ f\not\sim_\alpha h\circ g$.
A: Concerning the example you mention (composition on the left by the log), what is true is this:

If $f\sim_a g$ and if, in some neighbourhood of $a$, $g$ doesn't approach the value $1$, i.e. explicitly, if $\dfrac1{\ln\circ g}$ is bounded, then 
  $$ \ln\circ f\sim_a\ln\circ g. $$

