In the many epsilon-delta proofs I have seen for the various limit rules, most include a step where they define new epsilons from the one provided then go on to state that "There exists $\delta_1$ for every $\epsilon_1$ such that for every $x$ the expression...". An example of such a step is this, found here http://www.milefoot.com/math/calculus/limits/GenericLimitLawProofs04.htm:
Now, I am fine with this step, however, the proofs usually then go on to choose a delta for the proof they are working, often defining it as the minimum of the already existent deltas. And, from this delta they then go on to state a list of implications. Such an example is this:
What I fail to understand is whether the list of implications are true regardless of the definition of a delta (i.e. $\delta$) for the proof, because as I see it they would be true regardless of the new delta since they are merely statements extracted from what we are provided with (i.e.:
)
The conclusion "$|f(x)-L|<\epsilon_2$ and $|g(x)-M|<\epsilon_2$" could not be inferred for a bad choice of $\delta$. If $\delta>\delta_1$ it might happen that $|f(x)-L|\ge \epsilon_2$, and if $\delta>\delta_2$ it might happen that $|g(x)-M|\ge \epsilon_2$. You might pick an arbitrary number $\delta$ that fulfills the three conditions $\delta>0$, $\delta\le \delta_1$, and $\delta\le \delta_2$. But for such a pick to succeed, we need to know that a suitable such number exists at all. Instead of contemplating whether such a $\delta$ exists, the proof simply exhibits such a $\delta$ concretely, namely $\min\{\delta_1,\delta_2\}$, where the check of the three conditions is straightforward.
Recall that we start from two statements of the form $\forall \epsilon\exists \delta\ldots$ and want to show a third statement of the same form. The $\epsilon$'s and $\delta$'s of these statements are unrelatetd. For the proof we must start from an arbitrary $\epsilon$, then make use what the given statements give us (they give us just "some" $\delta$, and each may give us a different one), and from this we have to construct yet another $\delta$ that works for the target claim.