We want to show that
$\lim_{x \to 0}(x + a) = a$
for all $a \in \mathbb{r}$
Following a standard Epsilon-Delta proof, we initialize
$\epsilon > 0$
and we can see that
$0<\left| x \right| < \delta$
as well as
$\left|x\right|<\epsilon$
and from there, because if the inequality for Delta is true we can see from the implication in the Epsilon-Delta definition that we can set Epsilon to more than Delta.
$\delta < \epsilon$
And so we can select any Delta less than Epsilon so that the definition holds, meaning the function, when approaching zero, converges. Is that all there is to it, or should there be something more? Obviously I have left out a few steps (primarily because they are tricky to write out in LaTeX) but would you say the conclusion is satisfactory?
It is not correct to write “we can see that $0<|x|<\delta$ as well as $|x|<\varepsilon$”. How can we see such a thing when all we know about those three numbers is that $\varepsilon>0$?
What we can see is that, if $\delta=\varepsilon$, then$$|x|<\delta\implies\bigl|(x+a)-a\bigr|=|x|<\delta=\varepsilon.$$
Of course, instead of taking $\delta=\varepsilon$, you can tak any $\delta\in(0,\varepsilon]$ (as you wrote), but it is always better to proved a specific number.