Limits by definition: a conceptual doubt I would like to understand better $\delta - \epsilon$ proofs throughout a widespread kind of example, so please I really appriciate an anwser.
Proof that $\lim _{x \to 7} x^2 = 49$
Well, as we know:
$0 <\mid x-7 \mid < \delta$ $\implies$ $\mid x-7 \mid \mid x+7 \mid < \epsilon$
My resolution:
$\mid x-7 \mid \mid x+7 \mid < \delta \mid x+7 \mid < \epsilon$ 
(because $0 <\mid x-7 \mid < \delta$ and certainly $\mid x-7 \mid < \delta$)
Then:
$\mid x-7 \mid \mid x+7 \mid < \delta \mid x+7 \mid < \delta (\mid x\mid + \mid 7 \mid) < \epsilon$ 
Here the things became cloudy for me: 
First of all, I would like to know the real reason to suppose an $\delta = 1$. Please do not say things like "because it works". Well, I need an reason that emerges from the problem itself I mean: how can I suppose $\delta = 1$ just analysing the numbers of the problem, geometrical idea of distances etc...  from this particular limit? Why not $\delta = 2000$? 
Another point is:
Suppose $\delta = 1$.
Then we know the interval that $x$ is varying:
$\mid x-7 \mid < 1 \implies 6<x<8$
With this interval we can analyse what happens with $\mid x+7 \mid$
$6<x<8 \implies 6+7<x+7<8+7 \implies 13<x+7<15$
Here is another cloudy point:
From above we get $13 < x+7 <15$ From higher above we get: $\delta \mid x+7 \mid < \delta (\mid x\mid + \mid 7 \mid)$. Ok.
The reason that we all hear to the necessity of an bounded $x$ is : "Well $\mid x-7\mid \mid x+7\mid <\epsilon$ must have $\mid x+7\mid$ bounded in order to find the delta. Conversely, with the expression $\delta (\mid x\mid + \mid 7 \mid)$ we can see that too! Saying that $\mid x \mid$ must to be bounded in order to find the delta.
But what is the tecnical reason to bound  $\mid x+7\mid$? 
Then we must to find and upperbound. Why an Upperbound? I mean:
$6<x<8 \implies 6+7<x+7<8+7 \implies 13<x+7<15 \implies \mid x+7 \mid <15$
Why this happens $13<x+7<15 \implies \mid x+7 \mid <15 $ ?
Is it correct to say: $ 13<x+7<15 \implies 6+7<x+7<8+7$
Then: $x+7<8+7 \implies \mid x\mid + \mid 7 \mid < 8 + 7$ then, $\mid x \mid <8$?
The point is, with the 15 (why an upperbound!?) we are able to solve the problem quite easily:
$\mid x-7 \mid \mid x+7 \mid < \delta \mid x+7 \mid < \delta (\mid 8\mid + \mid 7 \mid)< < \delta 15< \epsilon \implies \delta <\frac{\epsilon}{15}$
Then, the final cloudy point:
We have to choose the $\delta = min(1, \frac{\epsilon}{15})$ I really do not grasp the meaning of this. What are the reason (geometrical,algebric etc...) for that?
Thank you! 
 A: Here is a rundown of my thoughts about how I prove $\lim_{x\to 7}x^2 = 49$.
First off, the definition of $\lim_{x\to 7}x^2 = 49$:

For any $\epsilon > 0$, there is a $\delta > 0$ such that whenever $0<|x-7|<\delta$, we have $|x^2 - 49| < \epsilon$.

This means, in terms of how we write the proof, that we're given an arbitrary $\epsilon$, and we must find a $\delta$ that works for that specific $\epsilon$. Note also, that whenever we have a $\delta$ that works, any $\delta$ which is smaller also works. This is important because it means we introduce no problems by artificially restraining ourselves with the $\delta \leq 1$ that's coming later.
$$|x^2 - 49| = |x-7|\cdot |x+7|$$
We do this factorisation because we want some kind of connection between the $\epsilon$ we're given and the $\delta$ that we're going to find / pick. We want the left-hand side to be smaller than $\epsilon$, and on the right-hand side we see $|x-7|$, which we can force to become as small as we wish by making $\delta$ small. This is good. We just need to make sure that $|x + 7|$ doesn't ruin too much for us. That means we must limit how large this term can become.
This is where declaring $\delta \leq 1$ comes in. That's just me, as the proof writer, saying loud and clear that no matter which $\epsilon$ I'm given, I will never pick a $\delta$ which is larger than $1$. As noted above, this doesn't introduce any problems logically, although it is not a very transparent move. You're not the first person puzzled by it, and you won't be the last. Not even on this site.
As long as $|x-7| \leq 1$, we must have $|x+7| \leq 15$ (in fact, we must have $13\leq x+7\leq 15$, but we don't need that level of detail). We have now limited how large this term can be, and we're ready to finish the proof, and figure out what $\delta$ should be. We take the factorisation from above, and add some explanatory text. Thus whichever $\delta$ we choose in the end, as long as it's not greater than $1$, we get the following:
$$
\overbrace{|x^2 - 49|}^{\text{We want this to be smaller than }\epsilon} = \overbrace{|x-7|}^{\text{We know this is smaller than }\delta}\cdot\underbrace{|x+7|}_{\text{We know this is smaller than }15}
$$
We want the left-hand side to be smaller than $\epsilon$, and the only thing we have control over is $\delta$. Now we see that as long as we vow never to let $\delta$ exceed $1$, then the right-hand side, and therefore the left-hand side, is always less than $15\delta$. That means that as long as we pick our $\delta$ to be at most $\epsilon/15$, then the left-hand side will stay smaller than $\epsilon$.
(Also note that $\epsilon$ itself didn't really enter into our proof until this late. Everything we've done up till now is to show that we can get control over $|x^2 - 49|$ by using $\delta$, and only once we have that do we bring in the arbitrary $\epsilon$ and say which $\delta$ we should use.)
This concludes the proof. As long as we keep our promise that we never pick a $\delta > 1$, anything below $\epsilon/15$ works. I will actually choose to pick exactly $\epsilon/15$, unless $\epsilon > 15$, in which case I have to pick $1$. This is written mathematically as $\boxed{\delta = \min\left(1, \frac\epsilon{15}\right)}$
