Asymptotic equivalence and little-o relation In my course resources (calculus I), there is a relation between asymptotic equivalence and little-o without a proof. I just tried to give my own proof but I would know if it's actually correct, or I've made some mistakes. Just to uniform the notation:
Asymptotic equivalence between f and g for $x \to \overline{x}$:
$$ f \sim_{\overline{x}} g \implies \lim_{x \to \overline{x}} \frac{f(x)}{g(x)}=1$$ 
Little o  for $x \to \overline{x}$:
$$ f(x)=o(g(x)) \iff \lim_{x \to \overline{x}} \frac{f(x)}{g(x)}=0$$ 
The relation say that properties a),b) are equivalent:
$$
a)\quad \quad f(x) = g(x)+\tilde{g}(x) \quad \text{ with:} \quad \tilde{g}(x)=o(g(x)) $$
$$
b)\quad \quad f(x)\sim_{\overline{x}}g(x) 
$$
This is my "proof":
$$
\tilde{g}(x)=o(g(x)) \iff \lim_{x \to \overline{x}} \frac{\tilde{g}(x)}{g(x)}=0
$$
Intuitively this means that $\tilde{g}$ grows much more slowly than g, so I can "omit"  $\tilde{g}$ for this reason, the remaining equation is:
$$
f(x)=g(x) \quad \text{ for} \quad x \to \overline{x}
$$
Since the 2 quantities are equals in a neighbour of $\overline{x}$, I can conclude:
$$
 \lim_{x \to \overline{x}} \frac{f(x)}{g(x)}=1  \implies f \sim_{\overline{x}} g
$$
I know that for prooving an equality I must proove in both directions, but I just want to know if the main concept behind the proof is right. I'm pretty sure that I've made a some technical mistakes.
 A: Note that we'll use $o(x)$ notation instead of $o_{x\to\overline{x}}(x)$ for convenience, but remember this notation is always related to a point.


*

*Let's work $a\Rightarrow b$
You have $$\text{ }f(x) = g(x) + o(g(x))$$
So you can subsitute easily , and here we suppose $g(\overline{x})\neq 0$:
$$\lim_{x\to\overline{x} }{f(x)\over g(x)} = \lim_{x\to\overline{x} }{g(x) + o(g(x))\over g(x)} = 1+0 = 1$$
 Which gives you $b$


*

*Now $b \Rightarrow a$
You have :
$$f(x) \sim_{\overline x} g(x)$$
Which is by definition :
$$ f(x)-g(x) = o(g(x)) $$
EDIT : Using your definition (notice that this is an equivalence, ie : it goes both ways), if $g(\overline{x})\ne 0$ :
$$f(x)\sim_{\overline{x}}g(x) \Leftrightarrow \lim_{x\to\overline{x}}{f(x)\over g(x)}=1$$
But we know that : 
$$ (1) \forall x, {g(x)\over g(x)} = 1$$
$$ (2) \lim_{x\to\overline{x}} {o(g(x))\over g(x)} = 0$$
So using $(1)+(2)$ we have :
$$\lim_{x\to\overline{x}} {g(x)+o(g(x))\over g(x)} = 1 + 0 = 1 = \lim_{x\to\overline{x}}{f(x)\over g(x)}$$
And now we can conclude : 
$$f(x) = g(x)+o(g(x))$$
Remember that $o(g(x))$ is really $o_{x\to\overline{x}}(g(x))$ and is only "negligeable" in regard to $g$ in the neighborhood of $\overline{x}$, so $f$ can be a lot different than $g$ on any other point.
