DEF.$$ \lim_{x \rightarrow x_0} f(x)=L \Leftrightarrow \forall \varepsilon >0: \exists \delta >0: 0<\left|x-x_0 \right|< \delta \Longrightarrow \left|f(x)-L \right|<\varepsilon $$
Why doesn't the definition have any other requirements for $\delta $, for example that when $\varepsilon$ decreases, $\delta $ decreases as well?
If we for example found $\delta $ s.t. $$ \delta=\delta (\varepsilon)=\begin{cases} \varepsilon+1, \ \ \ &\mathrm{if}\ 0<\varepsilon \le 2\\ \varepsilon-2, \ \ \ &\mathrm{if}\ \varepsilon > 2. \end{cases} $$
and $$ \forall \varepsilon>0: 0 < \left|x-x_0 \right|<\delta (\varepsilon) \Longrightarrow \left| f(x)-L \right|<\varepsilon,$$ would it still imply that $ \lim_{x \rightarrow x_0} f(x)=L $?
And the same question for limits at infinity:
DEF. $$ \lim_{x \rightarrow \infty} f(x)=L \Leftrightarrow \forall \varepsilon>0: \exists M>0: \ x>M \Longrightarrow \left|f(x)-L \right|<\varepsilon $$
If we found $ M = M(\varepsilon) $ s.t. M actually decreases as $\varepsilon$ decreases, would it still imply, that $ \lim_{x \rightarrow \infty} f(x)=L $?
If some $\delta$ works for a given $\epsilon$, then all smaller $\delta$'s will work as well. Assume that $\delta(\epsilon)\le\epsilon$ works in all cases (for instance with $f(x)=x$), then $\delta(\epsilon)=\epsilon \dfrac{\cos\epsilon+2}3$ is equally valid.
If there is no reason to enforce a condition, do not enforce it, that could make some proofs more complicated than necessary.