About the rigorous $(\epsilon,\delta)$ definition of limit I have questions about the complete and rigorous $(\epsilon,\delta)$ definition of two sided limits. The definition of the two sided limit in http://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit says that for a function $f(x)$ which is defined on an open interval containing $c$, possibly excluding $c$, it is $\lim_{x \to c} f(x)=L$ iff there is some $\delta > 0$ for every $\epsilon > 0$ such that for every $x$ which provides $0 < |x - c| < \delta$ it is $|f(x) - L| < \epsilon$. 
My first question is about the "open interval" which is stated at the beginning. If we assume that $f(x)$ is defined on an open interval $I$ which contains $c$, shouldn't the definition state that "for every $x \in I$ which provides $0 < |x - c| < \delta$" instead of just saying "for every $x$ which provides $0 < |x - c| < \delta$" in order to limit $x$ with interval $I$ of our selection?
My second question is about the interpretation of the above definition. Let's say we the $f(x)$ as in the following sketch:

Here, the function is continuous, defined on an open interval $(A,B)$ with $a \in (A,B), c \in (A,B)$ and $b \in (A,B)$ and $|a-c|>|c-b|$. A valid selection for $\delta $ is $\delta = |c-b|$ such that for every $x$ with $0 < |x-c| < |c-b|$ it is $|f(x) - f(c)| <\epsilon$.
But if we have the following $f(x)$ instead:

which is the same as the first one, but the the domain of $f$ is $(A,a)$ this time and it is $|c-b| > |a-c|$. My question is, in this case, can $|b-c|$ again be a valid $\delta$ value, since again for every $x$ which provides $0 < |x-c| < |b-c|$ we have $|f(x) - f(c)| < \epsilon$. I ask this because I used to think this definition always for functions defined on the whole real line and this makes one to believe that we should have intervals of the same length around $c$ which provides $|f(x)-L|$ where $L$ is a general limit value. But in the second sketch, this is not the case.
 A: *

*Absolutely yes. We only require the function $ f $ to be defined on the set $ I $ and so it doesn't make sense to talk about $ f(x) $ for any $ x $ outside of $ I $.

*For the second function and the given $ \epsilon $, $ \delta = \left| c - b \right| $ is indeed a valid choice of $ \delta $ and this follows from adding "$ x \in I $" to the definition. In practice, however, it might be convenient to choose a smaller $ \delta $ for some functions because then you don't need to consider what happens at the boundary of $ I $.
The two functions you've drawn have different domains and so are considered to be different functions. This means that when dealing with the second function you don't need to worry about what the "rest of the function" looks like (and of course you could extend the second function to one which is and looks different from the first).
A: Values of $\delta$ comparable to the distance from $c$ to the endpoints of $I$ are irrelevant and do not affect the property of continuity of the function. In fact, it could be incorporated in the definition that $\delta$ should be small enough in this sense. It wouldn't make any difference in fact if we require $\delta$ to be less than a millionth of the distance to the boundary points.
To make this point even clearer, it is helpful to keep in mind that the quantifier definition of continuity that you are seeing is a (somewhat windy) paraphrase of Cauchy's original definition of continuity of $f$, which said that if $x$ is infinitely close to $c$ then $f(x)$ is infinitely close to $f(c)$ (assuming of course that $f$ is defined at $c$). 
Another way of putting it is that if $x-c$ is infinitesimal then $f(x)-f(c)$ is infinitesimal, and you don't need quantifiers.
Once you understand the intuition behind this concept it will be easier to understand the technicalities of the epsilon, delta definition. 
In this sense, only "small" values of $\delta$ are relevant, even though the official definition does not say anything about this.
At the end of the 19th century people were nervous about infinitesimals and, unable to justify them rigorously, they replaced Cauchy's original definition by the one you are seeing. However, 60 years ago infinitesimals were finally justified rigorously, so you don't have to hide Cauchy's definition anymore (except from your professor :-) ).
