For the purposes of limits, the precise dependence of $\delta $ on $\epsilon$ is simply not important. The proofs do not require any knowledge of that relation. As said, $\delta (\epsilon)$ usually tends to $0$ as $\epsilon$ tends to $0$, but this is not always the case. To make this precise, let $\delta(\epsilon)$ be the largest $\delta$ corresponding to $\epsilon$ in the definition of continuity at $x$ for a function $f$. Thus, $\delta(-)$ is a function whose domain is $(0,\infty )$ and whose range is $(0,\infty]$, and it is monotonically non-decreasing. If $f$ is a constant function, then $f(\epsilon)=\infty $ for all $\epsilon>0$, showing that indeed $\lim_{\epsilon\to 0}\delta(\epsilon)$ need not be $0$.
The definition of limit captures the following: $\lim_{x\to a}f(x)=L$ means that for any prescribed distance $\epsilon>0$, there exists some upper bound for distances $\delta$ such that if $x\ne a$ is within $\delta $ units from $a$, then $f(x)$ is guaranteed to be within $\epsilon$ units from the limit $L$.
Remark: The function $\delta(-)$ above is known as a modulus of continuity for $f$. Functions whose moduli of continuity have certain properties (e.g., are concave) are of importance.
You can use the delta epsilon method of proving this (which I assume is what you want) by the following.
Right-handed limit (when $x>a$)
There exists $\delta$ such that if $x-a < \delta$, $|f(x)-L|<\epsilon$
Left-handed limit (when $x<a$)
There exists $\delta$ such that if $a - x < \delta$, $|f(x)-L|<\epsilon$
The two-sided limit is defined as :
There exists $\delta$ such that if $|x-a| < \delta$, $|f(x)-L|<\epsilon$
We can split up the absolute value around $x-a$ above on the two sided limit in the following way:
If $x-a>0$, or equivalently if $x>a$
$|x-a| < \delta$
Becomes
$x-a < \delta$
The same as the right sided limit definition.
If $x-a<0$, or equivalently if $x<a$
$|x-a| < \delta$
Becomes
$-(x-a) < \delta$
$a - x < \delta$
The same as the left sided limit definition.
This means that, in order for the two sided limit to exist, both the right handed and left handed limits must exist.
Best Answer
If $f\colon A\subset\mathbb{R}\to \mathbb{R}$, you have that $\lim_{x\to c}f(x)=l$ iff $\lim_{x\to c^{-}}f(x)=\lim_{x\to c^{+}}f(x)$.
Now, you defined $\lim_{x\to {c^{-}}}f(x)=l$ as: $(\forall \varepsilon>0)( \exists \delta_{1}>0)(\forall x\in A)(-\delta_{1}<x-c<0\Rightarrow |f(x)-l|<\varepsilon)$
analogously you define the limit to the right, and with these two definitions you show that $\lim_{x\to c}f(x)=l$