Is there any surjective property in $\epsilon-\delta$ relation in the definition of limit? We know that surjectivity is defined as follows:
Definition (Surjective Function): Let $f:A\subset\mathbb{R}\rightarrow B\subset\mathbb{R}$. $\forall y\in B$, $\exists x\in A$ such that $y=f(x)$, then $f$ is called a surjective function.
My inquiry rise while defining the limit:
Definition (Limit): Let $f:A\subset\mathbb{R}\rightarrow B\subset\mathbb{R}$. The following statements are equivalent.


*

*$\forall \epsilon>0$, $\exists\delta=\delta(\epsilon)>0$ whenever $0<|x-x_0|<\delta$ then $|f(x)-L|<\epsilon$.

*$\lim_{x\rightarrow x_0} f(x)=L$.


My question is: Is the relation $\delta=\delta(\epsilon)$ surjective? If not, is there any condition which makes it surjective?
 A: Well, your question is a bit unclear, and as often is the case with such questions, the answer is a resound "Maybe!"

Interpretation one: functions.
The problem is that $\delta$ isn't a function of $\epsilon$. Can we fix this?
The answer is yes: look for the biggest possible $\delta$. Namely, if $f$ is continuous and $x_0$ is a point in the domain of $f$, we define $\delta_{x_0}(\epsilon)$ as $$\sup\{\delta: \forall x(0<\vert x-x_0\vert<\delta\implies \vert f(x)-L\vert<\epsilon)\}$$ (where $L=\lim_{x\rightarrow x_0}f(x)$). 

Exercise: show that $\delta_{x_0}(\epsilon)$ actually "works" for $\epsilon$ - that is, $$\forall x(0<\vert x-x_0\vert<\delta_{x_0}(\epsilon)\implies \vert f(x)-L\vert<\epsilon).$$

Now for each $x_0$, $\delta_{x_0}$ is a map from $\mathbb{R}_{>0}\cup\{\infty\}$ to $\mathbb{R}_{>0}\cup\{\infty\}$ (exercise: think about why it's convenient to have "$\infty$" here), and we can ask if this map is surjective. Unfortunately, the answer is no: e.g. if $f(x)=0$ for all $x$, then $\delta_{x_0}(\epsilon)=\infty$ for all $x_0$ and all $\epsilon$.

Interpretation two: relations.
Another approach is to talk about surjective total relations. A total relation $R$ between $A$ and $B$ is basically a multivalued function: for each $a\in A$, there is at least one $b\in B$ with $R(a, b)$. There is a natural "delta-relation" given by $\Delta_{x_0}(a, b)$ iff $$\forall x(0<\vert x-x_0\vert<b\implies \vert f(x)-L\vert<a)$$ (in words, $\delta=b$ "works for" $\epsilon=a$). If $f$ is defined on all of $\mathbb{R}$, then it is then true that "every $\delta$ works for some $\epsilon$", and this is a good exercise (HINT: extreme value theorem).
Note: here I'm restricting to finite $\epsilon$ and $\delta$. It's a good exercise to think about how the previous claim needs to change depending how infinite values of $\epsilon$ and $\delta$ are allowed in the picture.

Note, in both interpretations, the dependence on the point $x_0$. This can't be done away with; in general, the "$\epsilon-\delta$ structure" of a function at one point is very different than at a different point. But see here for more details.
