TITLE: What exactly is a modulus of continuity?
QUESTION [8 upvotes]: This is my first post on here, so forgive me if I am ignorant of certain customs.
I am currently reading Courant & John's Introduction to Calculus and Analysis Volume I. Unfortunately, I have stumbled upon an ambiguous notation on page 41.
It begins:

Our definition of continuity of the function $f(x)$ at $x_0$ requires that for every degree of precision $\epsilon > 0$ there exists quantities $\delta \gt 0$ (so-called moduli of continuity) such that $|f(x)-f(x_0)| < \epsilon$ for all $x$ in the domain of $f$ for which $|x-x_0| < \delta$.

He then goes on to say that (for "regular continuity"):

In general, as our example show, this $\delta = \delta(\epsilon)$ depends not only on $\epsilon$ but also on the value of $x_0$.

What exactly is meant by a modulus of continuity and how is it related to $\delta(\epsilon)$?
Edit - Upon further inspection I have noted that the "modulus of continuity" depends on both $x_0$  and $\epsilon$ for "regular continuity", however only depends on $\epsilon$ for uniform continuity. I believe that some of the confusion has been cleared by the notation of $\delta(\epsilon)$. However, he states that moduli of continuity are simply quantities $\delta > 0$. Is this "it", or is there further discussion to be had?

REPLY [3 votes]: The definition of continuity of a function $f$ at $x_0$ is: For any $\varepsilon >0$ there exists $\delta >0$ such that $|f(x)-f(x_0)|<\varepsilon $ provided that $|x-x_0|<\delta $. 
We thus say that for every $\varepsilon $ there is a corresponding $\delta $, in the sense that the above statement holds. The thing to notice is that
1) The precise value of $\delta$ may depend on both $\varepsilon $ and the point $x_0$. 
2) For any given $\varepsilon >0 $ and $x_0$, there may be many choices for $\delta$ as above. 
Any choice of $\delta$ for a given $\varepsilon $ and $x_0$ is called a modulus of continuity for $f$ at $x_0$ corresponding to $\varepsilon $. Moduli of continuity are not unique.