The answer to your question is that _Yes, it captures the intuitive understanding of $f(x)\to L$ if $x\to x_0$_ (except that technically one requires $0<|x-x_0|<\delta$ rather than $|x-x_0|<\delta$ unless you’re expecting $f$ to be possibly continuous at $x_0$); however, for the enumerated questions (which probably you seek answers to better understand your main question),

(i) No, $\delta$ is not arbitrary but the supremum of all possible values of $\delta$ is _dependent_ on $\epsilon$; the statement _such that whenever $0<|x-x_0|<\delta$_ simply means that $|x-x_0|$ can be _arbitrarily smaller than_ $\delta$——and I suppose this is what you mean when you ask “... $\delta$ is arbitrary?” The very construct of the definition does not demand any lower positive bound on $\delta$——i.e., the infimum or all possible values of $\delta$ is $0$ itself.

(ii) No, one **does not need** to show that $\epsilon$ becomes arbitrarily small as $\delta$ becomes arbitrarily small. To better see this, take the function 
$$f(x)=
\left\lbrace
\begin{array}{ll}
0&\mbox{if $x\ne 0$}\\
1&\mbox{if $x=0$}
\end{array}
\right.
$$
It is immediately clear that _for every $\epsilon>0$ and **any** $\delta>0$, we have that $|f(x)-0|<\epsilon$ whenever $0<|x-0|<\delta$_; thus, by the definition, we have that $\lim_{x\to 0}f(x)=0$. Clearly, in this particular case, **any** positively-valued $\delta$ (as a function of $\epsilon$) works——that is, for each $\epsilon$, the supremum of all ‘satisfiable’ values of $\delta$ is $\infty$; thus, if you take $\delta:=2^{-\epsilon}$ say, then obviously $\epsilon\to\infty$ as $\delta\to 0$.