Explanation For The Non Existence Of A Limit The definition of limit (for sequence) is:
$\forall \epsilon>0,\, \exists n_{\epsilon}\, \forall N>n_{\epsilon}: |x_N-L|<\epsilon $ 
which means that for every $\epsilon$ we will choose there is $n_{\epsilon}$ that for every $N>n_{\epsilon}$ the elements of there sequence will be close to the limit in a distance that is smaller than $\epsilon$ .
the opposite definition is $\exists \epsilon>0,\, \forall n_{\epsilon}\, \exists N>n_{\epsilon}: |x_N-L|\geq\epsilon $

What is the explanation for the non existence of a limit?
 A: The first definition you listed (for a convergent sequence) makes use of the fact that $x_n$ gets as close as you want to $L$. No matter how big or small of a "bubble" of radius $\varepsilon$ you put around $L$, you can always find an index $N$ so that every element with index greater than $N$ is in that "bubble." The natural intuition is to then say that $x_n$ converges to $L$.
The negation of that statement does just the opposite. It says that no matter which element you start with in the sequence, there will be some size of bubble such that at least one element past your starting element will be outside that bubble. That is to say, $x_n$ doesn't get close enough to $L$ as $n \to \infty$; it strays away just enough. 
For a very contrived example, think of the sequence $\left\{\frac{1}{n} \right\}_{n=1}^\infty$. It converges to zero by definition. If we alter the sequence such that: $$\left\{a_n = \frac{1}{n} \space \text{if} \space 10 \nmid n, \space a_n = 1 \space \text{else} \right\}_{n=1}^\infty$$ Then the sequence will not converge, because it keeps "bumping" up to $1$ every tenth element, hence not converging to zero as $n \to \infty$.
A: That is not the definition of the non-existence of a limit.
The statement:
$$\forall \epsilon > 0 : \exists N: \text{etc}$$
is disproven with a counterexample:
$$\exists \epsilon > 0 : \neg \exists N: \text{etc}$$
The meaning is, if you choose $\epsilon$ small enough, then you can't make $x_n$ $\epsilon$-close to the $L$.
A: Let's try to parse this out in words:
$$\forall\,\epsilon>0, \exists\,N\in\Bbb N, \forall\,n\ge N\quad |x_n-L|<\epsilon.$$
This means for any positive distance $(\epsilon)$, and $n$ large enough $(n\ge N)$, such that $x_n$ will be closer to L than 
$\epsilon$.
So the negation of this statement is that there is a positive distance $\epsilon$ such that no matter how far in the sequence we go, there will be an $x_n$ which is no closer to $L$ than $\epsilon$. Translating this into the logical notation gives
$$\exists\,\epsilon>0,\forall N\in\Bbb n,\exists n\ge N |x_n-L|\ge\epsilon.$$
Hope that helps.
