0
$\begingroup$

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:

enter image description here

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: enter image description here

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.:

enter image description here

)

| cite | improve this question | | | | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.