How can it be sure that the $x$'s values are converging in the formal definition of a limit?

Question

In the $\delta,\epsilon$ definition of a limit : $$ \forall \varepsilon > 0,\,\exists \ \delta > 0,\,\forall x \in D,\,0 < |x - c | < \delta \ \Rightarrow \ |f(x) - L|$$

why the condition $\delta(\epsilon) \rightarrow 0 $ if $\epsilon \rightarrow 0$ is not specified? Is this condition implicit ? and is it necessary?

As far as I understand in this definition is written that when $\epsilon$ becomes smaller and smaller there's always a $\delta$'s neighbourhood for which the distance between every $f(x)$ with $x\in$ $\delta$'s neighbourhood and the limit $L$ is less than $\epsilon$ , but how can I be sure that this values of $x$ are converging to the value $c\,$ ( as it is called in the formal definition ) ? Should't be necessary something like the condition I mentioned? Or another condition ?

Answer 1

The definition clearly emphasizes that the the inequality $|f(x) - L|<\epsilon $ should hold for all values of $x$ with $0<|x-a|<\delta$. And all these values of $x$ include values of $x$ which are as close to $a$ as we please. A larger value of $\delta$ does not remove the set of values of $x$ which are near $a$. More formally if $0<\delta_1<\delta_2$ then the set of values of $x$ with $0<|x-a|<\delta_1$ is a subset of the set of values of $x$ with $0<|x-a|<\delta_2$.

Thus a large value of $\delta$ does not contradict your intuitive understanding of limit.

Answer 2

It is not necessary. Take the null function $f\colon\mathbb R\longrightarrow\mathbb R$. Hence $\lim_{x\to0}f(x)=0$. But, if you want to prove this using the $\varepsilon-\delta$ definition, you can, for each $\varepsilon>0$, take $\delta=1$. Actually, you can even take $\delta=\frac1\varepsilon$!