Limits: why $f(x)$ can be equal to $L$ and $x$ can't be equal to $c$ The definition of the limit states that limit of $f(x)$ when $x$ approaches $c$ is $L$ iff  for every $\epsilon > 0$ there exists $\delta > 0$ such that $|f(x) - L | < \epsilon$ and $0 < |x - c| < δ )$.
This states that $f(x)$ can reach $L ( L- ε < f(x) < L + \epsilon)$, while $x$ cannot reach $c ( 0 < |x-c|)$.
The informal definition says that limit means the value the function approaches as the input approaches some value. (They use the same word.) Why can one reach its corespondent value ($L$) while the other can't ($x$ to equal $c$)? Why f(x) can equal L , and x can't equal c. What is the intuitive answer to this question?
 A: The definition is not quite what you wrote. It should be:

For all $\varepsilon > 0$, there exists $\delta > 0$ such that, if $0 < |x - c| < \delta$, then $|f(x) - L| < \varepsilon$.

Putting the $0 < $ next to the $|x - c| < \delta$ actually makes the definition weaker, in the sense that it's easier to satisfy. By putting the $0 < $ there, it means that we don't care what happens when $x = c$. The if...then statement imposes something to be true whenever $0 < |x - c| < \delta$ is true. When $0 < |x - c| < \delta$ is false (e.g. if $x = c$), the premise is false, so the conclusion $|f(x) - L| < \varepsilon$ can be true or false.
If we were to get rid of the $0 < $, we actually recover the concept of continuity, i.e. we don't just get a limit, we actually force the limit $L$ to be $f(c)$. Why? Because we can always choose $x = c$, and knowing that
$$|x - c| < \delta \implies |f(x) - L| < \varepsilon$$
implies that, in the case of $x = c$,
$$0 < \delta \implies |f(c) - L| < \varepsilon.$$
Note that $0 < \delta$ is always true, hence the conclusion $|f(c) - L| < \varepsilon$ must also be always true, regardless of $\varepsilon$. But, there's only one number $L$ so that $|f(c) - L| < \varepsilon$ for all $\varepsilon > 0$, and that's $L = f(c)$.
On the other hand, if we added the $0 <$ to the conclusion $|f(x) - L| < \varepsilon$, this becomes a stronger definition, in the sense that it's harder to satisfy. Not only do we require that, on the interval $(c - \delta, c + \delta)$, the function to be $\varepsilon$-close to $c$, we also require it to be different from $c$. This would exclude constant functions from having limits, as on every $\delta$-neighbourhood, they take their limit as a function value at every point!
A: The idea is that if, for instance,$$f(x)=\begin{cases}1&\text{ if }x=0\\2&\text{ otherwise,}\end{cases}$$then $\lim_{x\to0}f(x)=2$. The fact that $f(0)=1$ is not relevant here. What matters is that when $x$ is close to $0$ and distinct from $0$, then $f(x)$ is close to $2$. And in this case it is actually equal to $2$. So, if we had imposed that $0<\bigl\lvert f(x)-L\bigr\rvert$, the limit would not exist.
A: Indeed by the definition of limit 
$$\lim_{x\rightarrow c} f(x) = L \iff (\forall \varepsilon >0\, \exists \delta > 0: \forall x\in D\quad \color{red}{0<\vert x-c\vert <\delta} \implies \vert f(x)-L\vert <\varepsilon $$
it suffices that $f(x)$ approches $L$ as $x$ approaches $c$, as for example
$$\lim_{x\to 0} \frac{\sin x}{x}=1$$
for which we can define any value for $f(0)$ or also assume $f(x)$ not defined at $x=0$.
