What does this use of little-o notation mean? I am currently going through the proof of Prime Number Theorem, as given in Hardy and Wright, and in it they define the following constant:
$$\alpha = \limsup_\limits{x \to \infty} \left|V(x) \right|$$
It then states:
$$|V(x)| \leq \alpha + o(1)$$
The book explains that $f(x)=o(g(x))$ means that $f(x)/g(x) \to 0$ as $x \to \infty$, and likewise $f(x)=o(1)$ simply means that $f(x) \to 0$ as $x \to \infty$. It also states that the use of $o(g(x))$ in an equation simply means "some unspecified function f(x) such that f(x)=o(g(x))". This definition makes sense to me when the equation is an equality, however I don't know how to interpret this statement, which is an inequality. 
Does it mean that there is a specific function $f(x)$ such that $f(x)=o(1)$ and $|V(x)| \leq \alpha + f(x)$?
Does it mean that for all functions $f(x)$ such that $f(x)=o(1)$, the inequality $|V(x)| \leq \alpha + f(x)$ remains true (at least for sufficiently large $x$)?
 A: The whole point of the big-oh and little-oh notation is to provide a way to write down that a particular function is negligible/is of the same order as another simpler function in a rigorous way. It allows us to deal with possibly disgusting and unpredictable functions by comparing them to much simpler functions of the same order. 
The point is that once we've written $f(x)=o(g(x))$, we no longer care what $f(x)$ actually is; just that it is in a sense "smaller than $g(x)$". 
With this in mind, whilst the way to read it would probably be $$|V(x)|\le \alpha +\text{a function that is equal to }o(1),$$
the take home message is 
$$|V(x)|\le \alpha +\text{something which tends to $0$ as $x$ tends to $\infty$,}$$
or perhaps in this context
$$|V(x)|\le \alpha +\text{an error term that is negligible for large $x$.}$$

EDIT as per the comments. 
If any function $f(x)\to \alpha$ as $ x\to \infty$, then by definition, $f(x)-\alpha=o(1)$, so $$f(x)= \alpha +o(1).$$
In the case of $\limsup$, we have$$\alpha =\limsup_{x\to\infty}f(x) =\lim_{x\to\infty}\big(\sup_{y\ge x}f(y)\big),$$
So for any $x$ we have $$f(x)\le\sup_{y\ge x}f(y)=\alpha + o(1).$$
