The need for $\min(a,b)$ may arise because for some reason a simpler expression, such as $\delta < \epsilon/19$, doesn't always work.
In this case, there is a step in the proof that relies on the fact that
$\delta |x^2 + 2x + 4 | < 19 \delta $. How do you know this is true?
It is true because $|x^2 + 2x + 4 | < 19$.
Ah, but how do we know that is true?
The answer is, if all we know about $\delta$ is that $\delta < \epsilon/19$, then
we don't know that $|x^2 + 2x + 4 | < 19$. For example, if $\epsilon = 190$
(which is a perfectly legitimate value of $\epsilon$), then $\epsilon/19 = 10$,
so the requirement $\delta < \epsilon/19$ just means $\delta < 10$.
So $\delta$ could be, for example, $5$.
But then all we know about $x$ in that last chain of inequalities
is that $|x - 2| < 5$. For example, $x$ could be $4$, because $|4 - 2| = 2 < \delta$.
But if $x = 4$, then $|x^2 + 2x + 4 | = 28$. That's bad news, because $28 > 19$,
and now the next to last step of the proof doesn't work.
But as long as $\delta \le 1$, we know that $|x - 2| < 1$ and therefore $1 < x < 3$.
And with only those possible values for $x$, we know that $|x^2 + 2x + 4 | < 19$,
which we need for our proof. A way to insure that $\delta \le 1$ is to say
$\delta < \min(1, \mbox{something else})$. In this case the "something else" is
$\epsilon/19$, because we use the fact $\delta < \epsilon/19$ in the very last
step of the proof.
In short, $\delta < 1$ is a rule that doesn't work for very small values of $\epsilon$, and $\delta < \epsilon/19$ doesn't work for very large values of $\epsilon$, but for any $\epsilon$ one rule or the other works, so we set $\delta < \min(1,\epsilon/19)$, which enforces both rules simultaneously and always works.
Addendum:
You may well wonder how someone comes up with the idea to set a condition such as $\delta < \min(1, \epsilon/19)$. It is not an obvious guess. What might happen, however, is that you might get as far as to find that
\begin{equation}
|f(x) - f(2)| = |x - 2| \, |x^2 + 2x + 4 | = \delta \, |x^2 + 2x + 4 |.
\end{equation}
You need to show that this quantity is less than $\epsilon$. Now, if the right-hand side had come out to $\delta N$ for some constant $N$, it would be easy: just set $\delta < \epsilon / N$. We aren't so lucky here, but what we can do is this:
- Find some constant $N$.
- Somehow enforce the condition that $|x^2 + 2x + 4 | < N$.
- Require that $\delta < \epsilon/N$.
So how can we make $|x^2 + 2x + 4 | < N$? Actually, this is easy to do if we restrict $x$ to some finite range and make $N$ large enough. We already have $|x - 2| < \delta$, so if we require that $\delta < 1$, we have guaranteed
that $|x^2 + 2x + 4 | < |3^2 + 2 \cdot 3 + 4 | = 19$, so
$N = 19$ is a large enough value of $N$. But if we had instead said that $\delta < 2$, we would guarantee that $|x^2 + 2x + 4 | < 28$, so $N = 28$ would do. That is, we could put the two requirements that $\delta < 2$ and $\delta < \epsilon/28$, which together are the same as the single requirement that $\delta < \min(2,\epsilon/28)$, and this too would have worked.
So there's nothing magical about the number $1$ that makes it the "right" answer here. There's also nothing against using it. We just needed a positive number for this particular proof, and $1$ happened to be convenient to use.
Why $\delta < 1$ is not always helpful:
Suppose instead of the continuity proof posed above, we had to show continuity of $f(x) = 1/x$ at $x = 1$. Then for arbitrary $\epsilon$ we would need a $\delta$ for which we could show that if $|x - 1| < \delta$, then $|f(x) - f(1)| < \epsilon$.
But
\begin{equation}
|f(x) - f(1)| = |1/x - 1| = \left| \frac{1 - x}{x} \right| < \delta \, \left| \frac{1}{x} \right|.
\end{equation}
So if we want to do anything like the technique we did before, we need to find a constant $N$ and a way to enforce $|1/x| < N$. Unfortunately, unlike a polynomial, which only gets unbounded as $x$ goes to infinity, $1/x$ blows up as $x$ approaches zero. So not just any finite range of $x$ will do; zero has to be outside the range. The condition $\delta < 1$ tells us that $0 < x < 2$, but that's not good enough. We need the lower end of that range to be a positive number. Well, you can't say $\delta < 0$, and $\delta < 1$ doesn't work, so split the difference and say $\delta < 1/2$. Then $1/2 < x < 3/2$ and so $1/x < 2$. But in fact you could have said $\delta < 2/3$ or $\delta < 0.999$, as long as you choose a large enough $N$. For example, $\delta < \min(0.999, \epsilon/1000)$ would work for a proof of continuity.
So there's nothing magical about $\delta < 1/2$ here, either; it's just that I'd rather do the calculations for $\delta < 1/2$ than for $\delta < 0.999$, wouldn't you?
