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.
 A: 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, $\delta = \min(1, \varepsilon/3)$.
A: 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!
A: In order to prove that $\lim_{x\to a}f(x)=g$  we have to show that the value $\Delta(x)=|f(x)-g|$ is arbitrarily small provided $|x-a|$ is sufficiently small. Here we take any $\varepsilon$ and we have to show the existence of such $\delta>0$ that $\Delta(x)=|x-1||x+1|<\varepsilon$ provided $|x-1|<\delta$.
As I said, our goal is to find some $\delta$, not necessarily the best (the largest) one. We know that if some $\delta$ is good then smaller is also good. Therefore we can always restric our search for $\delta$ to the set $(0,1)$, that is we can always assume that $\delta<1$ and look what it implies. In our case this implies that $0<x<2$. This makes the second factor $|x+1|$ bounded by $3$. This makes our task a lot easier.
This technique is what I call the method of small and bounded factors. It consists in finding in each summand the factor that is small and a factor that is bounded. For example in our case $|x-1||x+1|$ the first factor is small (it can be arbitrarily small for $x$ close to $1$) and the second factor is bounded (by 3). The idea is that if $a$ is smaller than $\delta$ and $b$ is smaller than a fixed value $M$ then if we take $\delta \leq \varepsilon/M$ then $ab<\varepsilon$. Of course this doesn't give you the optimal (largest) possible value of $\delta$ but let you avoid problems with nonlinear inequalities.
To be more precise that I mean by problems with nonlinear inequalities, let's try another approach: $$\Delta(x)=|x-1|\cdot|x+1|\leq |x-1|\cdot||x-1|+2|.$$ If $|x-1|<\delta$ then $\Delta(x)<\delta(\delta+2)=\delta^2+2\delta$. This is a nonlinear inequality and we wouldn't like to be forced to solve such inequalities (in more difficult cases they can be much more complicated). Instead we factorise $\Delta(x)$ and divide our problem of finding $\delta$ into two parts. First we force $\delta$ to be just not greater than $1$. Then the second factor $\delta+2\leq 3$. Now, we see that $\Delta(x)<\delta\cdot 3$. In order to get $\Delta(x)<3$ it's sufficient to consider $\delta\leq \varepsilon/3$. So in order to make both the inequalities
$$\Delta(x)<3\delta\leq \varepsilon$$ we can take any $\delta$ that satisfies both the inequalities $\delta\leq 1$ and $\delta\leq \varepsilon/3$. Now we see that it sufficies to take the minimum of these two.

I would like to show this method in a bit more sofisticated example. We'll show that $$\lim_{x\to 1}\frac{x+5}{4-x^2}=2.$$ Let's assume that $|x-1|<\delta$ where $\delta$ is a constant that will be defined later. Let's calculate $\Delta$:
$$\Delta(x)=\left|\frac{(x-1)(2x+3)}{(2-x)(2+x)}\right| < \delta\cdot\frac{|2x+3|}{|2-x||2+x|}.$$ We managed to extract the small and the bounded factor. In order to make $\Delta$ small we need to find the upper bound of $|2x+3|$ and lower bound of $|2-x|$ and $|2+x|$.

*

*First we assume that $\delta\leq 1$. Then $|2x+3|=|2(x-1)+5|< 2\delta+5<7$.

*Now we have to be far away from $2$ and $-2$. Therefore we assume that $\delta\leq 1/2$. Then
$$|2-x|=|1-(x-1)|\geq 1-|x-1|>1-\delta\geq 1/2,\\ 
|2+x|=|3+(x-1)|\geq 3-|x-1|>3-\delta\geq 5/2.$$ These bounds are well seen on the axis.

Now we see that $$\Delta(x)<\delta\cdot \frac{7}{\frac 12\cdot \frac 52}=\frac{28}{5}\delta.$$ Therefore $\Delta(x)<\varepsilon$ if $\delta\leq \frac 5{28}\varepsilon$.
The last thing to do is to find what restrictions do we put on $\delta$ so that all the inequalities hold. There are:
$$\delta\leq 1,\quad \delta\leq \frac 12,\quad \delta\leq \frac 5{28}\varepsilon.$$
Therefore we can put $$\delta:=\min\left\{1,\frac 12,\frac 5{28}\varepsilon\right\} = \min\left\{\frac 12,\frac 5{28}\varepsilon\right\}.$$
