How does the epsilon-delta definition define a limit? I understand what the epsilon-delta definition is saying in regards to the distance from a point c and the distance from your limit, but I don't understand how this defines a limit.  Any help is appreciated.
Thanks!
 A: In expressing the idea that
$$\displaystyle\lim_{x\to a}f(x)=L,$$
we often use intuitive phrases such as "$f(x)$ approaches $L$ as $x$ approaches $a$" or "$f(x)$ gets closer and closer to $L$ as $x$ closer and closer to $a$".  While these phrases do capture some aspects of limits, the problem with them is that they are imprecise and, in some sense, inaccurate.  Hence the need for the $\epsilon$-$\delta$ definition.
Consider, for example, the statement
$$\displaystyle\lim_{x\to0}\sin x=0.$$
It certainly is not always the case that $\sin x$ is getting closer to $0$ as $x$ is getting closer to $0.$  If you start with $x=\pi,$ for example, $\sin x$ equals $0$ there, but moves away from $0$ for a while as $x$ moves towards $0.$  You might say, well OK, but when $x$ is sufficiently close to $0,$ say $\lvert x\rvert<\pi/2,$ then $\sin x$ always moves towards $0$ as $x$ gets closer to $0.$  This is true, and illustrates why the precise definition of limit doesn't require that $f(x)$ always be getting closer to $L$ as $x$ gets closer to $a,$ but only that there exist some window around $a$ within which $f(x)$ is close and getting closer to $L.$  This is the role of $\delta$ in the $\epsilon$-$\delta$ definition: $\delta$ tells you the size a window around $a$ that will ensure that $f(x)$ is close and getting closer to $L$ as $x$ gets closer to $a.$
But that still isn't quite accurate, as the following example reveals.  Consider the function $g(x)=x\sin(1/x).$  You've probably seen graphs of $\sin(1/x)$ and know that it oscillates back and forth between $-1$ and $1$ infinitely many times as $x$ approaches $0,$ and that the number of oscillations between $x$ and $0$ remains infinite no matter how close $x$ gets to $0.$  In fact $\sin(1/x)$ has no limit as $x$ approaches $0$ for precisely this reason.  But $g(x)$ is different because of the factor $x.$  Instead of oscillating between $-1$ and $1,$ the oscillations are contained within an envelope, and furthermore, that envelope shrinks to zero width as $x$ approaches $0.$

It is intuitively reasonable in this example that
$$\lim_{x\to 0}g(x)=0,$$
and we need a definition of limit that reflects that.  The problem is that it is not literally true that $g(x)$ is getting closer and closer to $0$ as $x$ gets closer and closer to $0:$ even when your $\delta$-width window is very small, there are infinitely many $x$ within that window where $g(x)=0$ but then moves away from zero again.  So in general we don't require that $f(x)$ always move toward $L$ as $x$ moves toward $a.$  Instead, we require that any movement of $f(x)$ away from $L$ be contained within some envelope of shrinking width.  This is the role of $\epsilon$ in the $\epsilon$-$\delta$ definition: we require that eventually $f(x)$ get and stay within some window of width $\epsilon$ around $L,$ and that this be true for arbitrarily small $\epsilon.$
Putting it all together, $\lim_{x\to a}f(x)=L$ means that, for arbitrarily small $\epsilon,$ $f(x)$ stays within a distance of $\epsilon$ of $L$ provided that $x$ is sufficiently close to $a.$  By "sufficiently close", we mean within some distance $\delta$ of $a,$ where $\delta$ is allowed to depend on $\epsilon.$
Note that we don't require that $f$ be defined at $x=a,$ or, if defined, that it take the limiting value.  In fact, one of the main uses for limits in introductory calculus is to understand what is happening to $f$ close to points where $f$ is undefined.  So we have to exclude the point $x=a$ itself from the definition.  Therefore we require that, for every $\epsilon>0,$ there exist a $\delta>0$ such that $\lvert f(x)-L\rvert<\epsilon$ whenever $\lvert x-a\rvert<\delta$ except possibly at $x=a.$
A: To add to what user99680 said with an example-
The delta epsilon definition of a declaration of the form "if P(F, L, X) is true, then L is the limit of F at X".   It is not a universal equivalence, so it leaves many cases undefined.
One of the most prominent examples is that of an infinite series.  An infinite series is defined as :
$$\sum_{k=0}^{\infty} f(k) = \lim_{n \rightarrow \infty} \sum_{k=0}^{n} f(k)$$
In the case that $f$ is divergent, that leaves the limit undefined.  I have a more detailed explanation in this answer, but the in short the undefined nature of the limit leads to the freedom to define things like $1 + 2 + 3 + 4 \dots = {-1 \over 12}$.
