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.

Each limit has a proper way to be approached. Normally, we assume that $0 < |x - a| < \delta_{\varepsilon}$, where $\delta_{\varepsilon}$ is a function of $\varepsilon$ which we do not know yet. So the idea is to use this relation in order to get the desired expression.
At the given case, we can proceed as follows (it is not the unique way):
\begin{align*}
|(x^{2} + 9) - 10| & = |x^{2} - 1|\\\\
& = |x - 1||x + 1|\\\\
& \leq |x - 1|(|x - 1| + 2)\\\\
& < \delta_{\varepsilon}(\delta_{\varepsilon} + 2) := \varepsilon
\end{align*}
whence we get that $\delta_{\varepsilon} = -1 + \sqrt{1 + \varepsilon} > 0$.
Hopefully this helps!
