Why do we impose the constraint $\delta\le 1$ when proving the continuity of $x^3$ using the definition of continuity? 
$3)$ $f(x)=x^3+1$ is continuous at $x=2$. Let $\epsilon\gt0$ be given. Choose $\delta\gt0$ such that $$\delta\lt \color{red}{\min\{1,\tfrac\epsilon{19}\}}.$$ Then for $|x-2|\lt\epsilon$ we have $$\eqalign{|f(x)-f(2)| & = |x^3+1-9| = |x^3-8|\\
& = |x-2|\,|x^3+2x+4|\lt\delta|x^2+2x+4| \\ &\lt 19\delta\lt\epsilon.
}$$

Hey sorry about the picture, its of an example in a lecture slide.
Just a quick question is I understand how they get the answer (in red)  if they are told the constraint that delta must be less than 1, but how are you meant to know that?
Also how do you know you need a min{x,y} for a general question of this nature?
 A: 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?
A: is this from Ian Wood's lectures? haha!
Anyway, with regards to choosing $\delta < 1$, I think that's just a "safe assumption" and can generally be applied to most problems. If this is from the lectures I'm thinking of, it was fine in the class test! You assume that $|x-2| < 1$ to be able to get a bound on $2x$ and $x^2$.
As for a min{..,..} solution, you want to get $|something|$ = $|x-2||something else|$. One part of the solution comes from the assumption $|x-2| < 1$ earlier on, whereas if you cant immediately get a bound on the $|something else|$ (eg its a polynomial) you know you'll need to put a bound on it, giving another solution.
