What is the best way to explain setting a restriction on $\delta$ in $\epsilon$-$\delta$ proofs? I'm trying to prepare a somewhat informal lesson striving to provide an intuitive understanding of why for some limit proofs, we have to set an upper bound on $\delta$.
For example, here's part of the preliminary analysis of the proof I'm walking through:

Problem: Prove that $\displaystyle \lim_{x\to3} 9x^2+6x+1=100$.
  $$\begin{align*}
\left|9x^2+6x+1-100\right|&=\left|9x^2+6x-99\right|\\
&=3\left|3x^2+2x-33\right|\\
&=3|(3x+11)(x-3)|\\
&=3|3x+11||x-3|
\end{align*}$$
  Now if we let $\delta\le1$, we have
  $$|x-3|<1\implies-1<x-3<1\implies17<3x+11<23\implies|3x+11|<23$$
  $$\begin{align*}
\left|9x^2+6x+1-100\right|&=3|3x+11||x-3|\\
&<69|x-3|\\
&<\epsilon
\end{align*}$$
  which means $\delta=\min\left\{1,\dfrac{\epsilon}{69}\right\}$ is sufficient.

What's the best way to illustrate why we set $\delta\le1$ in an intuitive way, and why it is sometimes necessary to use a different upper bound?
 A: I am no teacher but this is just how I wish I was first taught this stuff. 
I was taught that the goal in these proofs was to come up with a $\delta$ when we are given with any $\epsilon$ whatever. 
That is, some antagonist is throwing an arbitrary positive quantity $\epsilon$ at us, and it was our job to produce a $\delta$ in response. 
Since we are the ones coming up with $\delta$ we can make it as small as we wish because it is inside our jurisdiction. 
As for the need of the upper bound the following inequalities are great for a first sight. you can impress upon the student that you are trying to make $\left|9x^2+6x+1-100\right|$ small by making $|x - 3|$ small. Then rewrite,  
$$ 3|3x+11||x-3| = 9 \left|{ (x -3) + \frac{20}{3}}\right  | \left|{x - 3}\right| \le 9|x - 3|^2 + 60 |x - 3|$$
If $\delta \le 1$ then 
$$ \delta^2 \le \delta \implies  9 \delta^2 \le 9 \delta \implies 9 \delta^2 + 60 \delta \le 69 \delta \le 69  \times \dfrac{\epsilon}{69} $$ 
A: It should be clear that it suffices to consider only small $\epsilon$. Therefore, it is often easier to grasp if one replaces the 

Let $\epsilon>0$. ... Let $\delta=\min\{1,\frac\epsilon{69}\}$.

with 

Let $\epsilon>0$. ... We may assume without loss of generality that $\epsilon<69$ and let $\delta=\frac\epsilon{69}$.

A: We want to define what it means for a function to be continuous at a point using the mathematically rigorous definition known as the $\varepsilon - \delta$ approach. But before we jump into that, it might be helpful to start with something simpler.
Let a function $f: \mathbb R \to \mathbb R$ be given and suppose that $f(0) = 0$. Then is said to be well-behaved and nice at the origin $(0, 0)$ if a number $a \gt 0$ and a number $\delta \gt 0$ can be found such that the graph of the function $f$ restricted to the closed interval $[-\delta, \; +\delta]$ satisfies
$\tag 1  -a |x| \le f(x) \le a |x|$
At the mathisfun site you can slide the value of $a$ around. Here are some pictures (free draw some function graphs and set some $\delta$'s):

and (missing some of the 'containment region', but OK if $\delta = .5$, say).

Now this definition is stronger that the usual $\varepsilon - \delta$ continuity at a point definition. 
Without any further ado, let me draw some $\varepsilon - \delta$ rectangles on our graph paper (the function 'containment regions'), and after some discussion the formal presentation will be given.
etc.
