$\alpha$ is unique if $f(x) \leq \alpha \leq g(x)$ for all $x$ and $\lim_{x\to a} ( g(x)-f(x)) = 0$ Let's say we have a function $f$ and another one as $g$, both are functions of, say, $x$. Let $\alpha$ be the number which lies between $f$ and $g$ for every $x$, that is 
$$
∀x ( f(x) \leq \alpha \leq g(x) )$$
It is a given condition that $$\lim_{x\to a} ( g(x)-f(x)) = 0$$
Now, I want to know what does it mean to say "$\alpha$ is the only number in between $f$ and $g$ in the limiting process as $x$ goes to $a$". My understanding speaks like this: The given condition that difference between $g$ and $f$ reduces to zero as $x$ goes to $a$ rigorously means that we can make $g(x)$ as close to $f(x)$ as we desire, and since $\alpha$ is always going to lie between them, a stage would come when $\alpha$ will be the only number in between them. 
But the flaw in my understanding is that it violates the elementary statement "Between any two numbers there lies infinitely many numbers, no matter how they close they are". So, no matter how close we can make $g$ to $f$ there always gonna lie infinitely many numbers not just $\alpha$. 
 A: Suppose that there are to distinct such numbers, $\alpha$ and $\beta$. You can assume, without loss of generality, that $\alpha<\beta$. Then, for each $x$, $g(x)-f(x)\geqslant\beta-\alpha>0$, and therefore you will not have$$\lim_{x\to\infty}f(x)-g(x)=0.$$Yes, between any two distinct real numbers, there are infinitely many real numbers. So, for a fixed real number $x$, if $g(x)>f(x)$, there are infinitely many real numbers between them. But you cannot deduce from this that there are infinitely many real numbers $\alpha$ such that $f(x)\leqslant\alpha\leqslant g(x)$ for every number $x$.
A: First Approach
I understand your remark.
1-Indeed for two given, fixedreal number $f(x) $ and $g(x) $, there exists infinitely number between them.
2-Yet, here we are dealing with a sequence $${f(x) - g(x)}_{x \in V(a)} $$ i. e. all values of the difference between f and g with $x$ is a neighborhood of $a$. 
What is the key is the limit
Here the uniqueness comes from the limit. 
Indeed if $f(x)-g(x) \to 0$ when $x\to a$.
 Supposing that $f$ and $g$ converge and $f \leq\alpha \leq g(x) $ for all $x$. 
Then  because $f$ and $g$ are supposed to converge and 
$$f(x)-g(x) \to 0$$
$f$  and $g$ have the same limit say $\lambda$
Thus taking the limit in the inequality  $f(x) \leq\alpha \leq g(x) $ 
We finally have $$\lambda\leq \alpha \leq \lambda$$
i. e. 
Then $$\alpha=\lim_a f=\lim_a g$$ hence its uniqueness. 
Second approach
More intuitively you know you can approach $f$ to $g$ as close as you want thanks to 
$f(x)-g(x) \to 0$
Imagine there are two numbers that want to take the role of $\alpha$, say $\alpha$ and $\beta $ with $
\alpha<\beta$. 
Then $f$ is always under $\alpha$ and $g$ is always over $\beta$. 
$$ f\leq \alpha<\beta \leq g$$
Imaging, between  $\alpha$ and $\beta$ you get a vacuum, like a nonmansland, for $f$ and $g$ so they can't meet i. e. you can't have
$f(x)-g(x) \to 0$. 
Rigorously it means that $f<\dfrac{ \alpha +\beta} {2}<\dfrac{\alpha+3\beta} {4}<g$ taking the arithmetic average between $ \alpha$ and $\beta$ and therefore between $b$ and $\dfrac{ \alpha+\beta} {2}$.
So $$\forall x, |f(x) - g(x) |>\dfrac{\beta- \alpha} {4}  $$
which is contradictory with :
$$f(x)-g(x) \to 0$$. 
