Understanding algebra used in $\varepsilon$-$\delta$ definition I would like to preface my questions by saying that I understand how the $\epsilon$-$\delta$ defintion of a limit works. The proof seeks to formalise the intution about a graph approaching a certain $f(x)$ value, by saying that there exists an $f(x)$ $\forall x$ between $x=a$ (limiting value) and an abritrarily chosen $x$ value in a region that is less than $\delta$ units away from $a$, but greater than $0$. Therefore, when we get as differentially close to $x=a$ as we want to and we can still find f(x) values in that region, then $\lim_{x \to a}f(x)$ truly exists.
However, the algebra always stumps me for all functions other than linear ones, where we have to limit certain factors to find an initial delta. Take for example the following:$$\mathit{Prove} \lim_{x \to 1}(x^2+9) = 10$$
This is what process I would use to prove the limit:$$\forall \epsilon \gt 0, \exists \delta : 0< |x-a|<\delta \implies |f(x)-a|<\epsilon$$ $$\therefore |x^2+9-10|<\epsilon$$ $$|x^2-1|<\epsilon$$ $$|x-1||x+1|<\epsilon$$
At this point, I get stumped. I know to assume $|x-1| \lt 1 $, but I am not sure why we are allowed to do that in the first place. So, I just assumed it was for convenience purposes since $\epsilon$ cannot be expressed in terms of $x$. But, obviously that is not an explanation of why specifically it is done.
Continuing: $$|x-1| \lt 1 \implies 0\lt x \lt 2$$ $$1\lt x+1 \lt 3 \implies |x+1|\lt3$$ $$\therefore |x-1|\cdot3 \lt \epsilon \implies |x-1|\lt \frac{\epsilon}{3}$$
So, take $\delta$ = min{1,$\frac{\epsilon}{3}$}
$$|x-1|<\delta \implies |x-1||x+1| \lt \delta |x+1| = 3 \cdot\frac{\epsilon}{3}=\epsilon$$
This last part too confuses me. I do not understand why we use the $min$ notation here and what purpose it serves in the final part where we actually prove the limit exists.

It's not so much a question of whether we're "permitted" to assume that $|x-1| < 1$, as much as we're looking for a narrow enough $\delta$-window so that we put $f$ into the appropriate $\varepsilon$-window. The $\delta$-window can be tighter than strictly necessary, so even if we don't need to make $x$ closer to $1$ than the interval $(0, 2)$, it's OK if we do.
For instance, if $\varepsilon = 100$, it turns out that we can make $\delta = 9$. But we don't have to make it that high; $\delta = 5$ will work, as will $\delta = \pi$, as will $\delta = 1$. The only important thing is that we find a $\delta > 0$ that will work—it has to be strictly positive.
Doing so may serve, among other things, to make the reasoning easier. We might find that if we assume that $|x-1| < 1$, then we can easily conclude that a $\varepsilon/3$-window around $x = 1$ works just fine, for any $\varepsilon \leq 3$. Then, for any $\varepsilon > 3$, we know that a $1$-window around $x = 1$ will work just fine. That is why it is appropriate to write, for example, $\delta = \min(1, \varepsilon/3)$.
