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 $\delta$——in fact, 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, and it can be to the contrary even in some case. 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 $\delta\to 0$.