Epsilon-Delta Confusion I don't understand the epsilon delta definition of a limit 
"According to the formal definition above, a limit statement is correct if and only if confining  x to d units of c will inevitably confine f(x) to epsilon units of L."
So, if we can confine x to infinitely small delta units of c, we can confine f(x) to infitely small epsilon units of L.  Like, constricting f(x) to L ?
Is that a right way of explaining what the general idea of the epsilon-delta method is ?
 A: The definition I learned is
$ lim_{x \rightarrow y}{f(x)} = c $ if and only if for every $ \epsilon > 0 $, there is a $ \delta > 0 $ such that
$ 0 < |x - y| < \delta$ implies $ | f(x) - c | < \epsilon $ for all $x$
This is still a little bit technical, lets see what it means. I think it's easiest to read
$ 0 < |x - y| < \delta $ as "the distance between $x$ and $y$ is between $0$ and $\delta$".
If we use this interpretation, the definition becomes:
$lim_{x \rightarrow y}{f(x)} = c$ if and only if for every $ \epsilon > 0 $, there is a $\delta > 0$ such that if the distance between $x$ and $y$ is between $0$ and $\delta$, the distance between $f(x)$ and $c$ is less than $\epsilon$.
This basically means that $f(x)$ gets arbitrarily close to $c$ (and I think this expression is still used sometimes as a more informal definition) without necessarily becoming $c$ (as James S. Cook pointed out). Suppose $f(x)$ does not get arbitrarily close to $c$, i.e., there is some constant $d$ such that $f(x)$ will stay away from $c$ with at least distance $d$.
Then we can show the limit definition does not hold:
take $\epsilon = d$. Now there doesn't exist a $\delta$ such that 
$0 < |x - y| < \delta \rightarrow |f(x) - c| < \epsilon $
(because we just assumed that $f(x)$ would never get closer to $c$ than distance $d$, and remember that this happening is equivalent to $|f(x) - c| < \epsilon = d$).
What may be helpful too is a (trivial) proof using the limit definition. Usually in these proofs, you take an $epsilon$, and return a $\delta$ for which you have proved that $ 0 < |x - y| < \delta \rightarrow | f(x) - c | < \epsilon $ (when dealing with continuous functions you usually just use $|x - y| < \delta$ in your proof).
I prove that
$\lim_{x \rightarrow y} x = y$ (or, equivalently $f(x) = x$ and $c = y$ in the original formulation).
Take $\delta = \epsilon$. Then $ |x - y| < \delta \rightarrow | f(x) - y | = | x - y | < \delta = \epsilon $. So the condition that there exists some $\delta$ for which $ |x - y| < \delta \rightarrow | f(x) - c | < \epsilon $ holds is true, as I've just shown. More advanced proof usually use a similar logic, but the expression for $\delta$ and working out the $| f(x) - c | < \epsilon$ can become quite hard. This is why other theorems are often used (for continuous functions, $ lim_{x \rightarrow y} f(x) = f(y) $, and compositions, products, sums of continuous functions are again continuous, which can help you out very often).
Also, the definition requires that something must hold for every $\epsilon$. Sometimes, teachers explain this as a game: you can choose $\epsilon$ freely, and you can give a procedure to show that something (there exists a $\delta$ such that...) will hold, then we can say $lim_{x \rightarrow y} f(x) = c$. So, if you win this game by giving such a procedure, you basically got the recipe for a proof!
If you cannot win, it suffices to give a single $\epsilon$ such that $f(x)$ will never get closer to $c$ then distance $\epsilon$.
The wikipedia page on the ($\epsilon$, $\delta$)-definition is pretty good. Also, the picture from there may be helpful when you try to visualize $\delta$ and $\epsilon$ as distances.

A: Let $y \in \mathbb{R}$ and $f$ a function whose domain is defined for points near $y$. We say $\lim_{x \rightarrow y} f(x)=L \in \mathbb{R}$ iff for each $\epsilon >0$ there exists $\delta >0$ such that if $x \in \mathbb{R}$ and $0 < |x-y| < \delta$ then $|f(x)-L|<\epsilon$.
Notice the $0<|x-y|$ means we do not consider the value $x=y$ for the limiting process. Now, it is often the case that $L=f(y)$ and the limit is calculated by mere evaluation of the function $f$; this is the case of a function which is continuous at $y$. In fact, this is the definition of continuity at $y$ for $f$. We say $f$ is continuous at $x=y$ iff $\lim_{x \rightarrow y}f(x) = f(y)$. Unless $y$ is an endpoint of the domain or an isolated point then some fine print applies, but let us focus on the essential point here.
One important example (there are others, but certainly this class of examples is unavoidable in the study of calculus!) of the exclusion of $x=y$ in the limit is found in difference quotients: if the limit 
$$  \lim_{x \rightarrow y} \frac{f(x)-f(y)}{x-y} $$
exists then we call it $f'(y)$ the derivative of $f$ at $y$. It is never reasonable to just plug $x=y$ into the difference quotient because it necessitates division by zero. So, the $0$ in the definition "$0 < |x-y|$" is of critical importance because it is the subtle point which separates taking a limit from just doing algebra. Although, I will admit, the majority of the limits we encounter in first semester calculus are just that; algebra. But, conceptually, the exclusion of the limit point from the limit is not optional in my view. It is the point.
(of course, again, for continuous expressions there is no difference, but continuity cannot be expected in many cases)
