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?
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
$$ \delta^2 \le \delta \implies 9 \delta^2 \le 9 \delta \implies 9 \delta^2 + 60 \delta \le 69 \delta \le 69 \times \dfrac{\epsilon}{69} $$