Real Analysis: Why do we take an arbitrary delta value to prove that a function is continuous I was watching this video showing how to prove if a function is continuous and if it is not.
I understand the steps except for one in 3:25 when he takes $ \delta = 1 $, and then he takes
$$ \delta = \min \left \{ 1 , \frac \varepsilon { \max \{ | 2 a + 1 | , | 2 a - 1 | \} } \right \} \text . $$
What I'm having trouble understanding is why did he use $ \delta = 1 $ specifically, and why did he take that minimum with those two values. Especially, why $ \frac \varepsilon { \max \{ | 2 a + 1 | , | 2 a - 1 | \} } $?
Everything laid out perfectly at the end to get $ < \varepsilon $. So I would like to know why and how he made those choices.
I'm sorry for not formatting correctly, I am new in this stack.
Thanks in advance
 A: The general $\delta$-choosing strategy in a continuity proof is to identify a chain of inequalities that would complete the proof if they all hold, then determine which $\delta$ satisfy them simultaneously. These will always be make-$\delta$-small constraints, so can be combined with a $\min$ wrapper. It's wrong to think of this as, "first we take this value for $\delta$, then we change it later after thinking about it some more". It's more accurate to say we assume bounds on $\delta$, making them stricter as need be so each step in our logic works. Fortunately, old steps will still work just fine, because the assumption on $\delta$ is getting stronger.
To take the example at hand, if $|x-a|<\delta$ then the reasoning$$|x^2-a^2|=|x-a||x+a|<\delta|\delta+2a|\le\frac{\varepsilon}{\max\{|2a+1|,\,|2a-1|\}}|\delta+2a|\le\varepsilon$$serves our needs provided that$$\delta\le\frac{\varepsilon}{\max\{|2a+1|,\,|2a-1|\}},\,\delta\le1,$$which are respectively used in the last two inequalities. In particular, the first constraint's denominator is chosen by "peaking at the end" so to speak, since by the triangle inequality (which this strategy will invariably use without even stopping to mention it)$$|\delta|\le1\implies|\delta+2a|\le\max\{|2a+1|,\,|2a-1|\}.$$In the discovery stage, one is likely to instead write$$|x^2-a^2|=|x-a||x+a|<\delta|\delta+2a|\le\frac{\varepsilon}{M}|\delta+2a|\le\varepsilon,$$and decide what $M$ is later. The first constraint is then $\delta\le\frac{\varepsilon}{M}$, whatever that means. But since there's a $|\delta+2a|$ factor afterwards, which we can't expect to make arbitrarily small, we define $M$ as some finite upper bound on it we can obtain. As a rule of thumb, "mystery factors" like this (not a technical term!) get set by such "large" bounds on factors such as e.g. $|\delta+2a|$. But as is so often the case in mathematics, the discovery process is often hidden when the proof is cleaned up for presentation.
