Help with Epsilon-Delta proof for limits I tried to teach myself these types of proofs.  I understand the reasoning behind it very well, but I have trouble understanding specific parts  when simplifying inequalities.  Let me give an example:
Say I wanted to prove the following:
$$\lim_{x\to1}(x^2+3)=4$$
I start by supposing: given $ε>0$, I want to find $δ>0$ such that
$$0<|x-1|<δ => |(x^2+3)-4|<ε$$ 
I start by simplifying the RHS to find an expression that relates $ε$ to $δ$:
$$|x^2-1|<ε$$
$$=>|(x-1)(x+1)|<ε$$
$$=>|x-1||x+1|<ε$$
At this point I was stuck and did research on how I should proceed.  Apparently we can restrict $δ$ to only be at most $1$ unit away from $a$.  Since we are dealing with the limit of $f(x)$ as ${x\to a}$, it is reasonable to restrict our "radius" around $a$ this way.  I sort of understand this reasoning, although a more detailed explanation would be appreciated.  Anyway, this implies (in my case):
$$|x-1|<δ≤1$$
$$=>|x-1|<1$$
$$=>0<x<2$$
$$=>1<x+1<3$$
$$=>1<|x+1|<3$$
This means that the min value of |x+1| is larger than 1 and the max value is smaller than 3.  However, here comes the part where I get stuck: according to multiple solutions I found online, it is reasonable to say that:
$$|x-1||x+1|<3|x-1|$$
Now the part above I totally understand, but the following part I do not.  Apparently, it is a logical step to deduce the following:

$$|x-1||x+1|<3|x-1|<ε$$
$$=>3|x-1|<ε$$

How can we logically deduce that $$3|x-1|$$ is smaller than the given $ε$?  In my reasoning, if $$3|x-1|$$ is larger than $$|x+1||x-1|$$ it does not necessarily mean that the former is also smaller than $ε$.  For instance, if $3<5$ and $6>3$, then it does NOT mean that $6<5$, obviously.  In my opinion, it is correct to deduce the following:
$$|x-1|<|x+1||x+1|$$ 
(Using the same reasoning as earlier)
$$=>|x-1|<|x+1||x-1|<ε$$
$$=>|x-1|<ε$$
This, to me, is pretty clear.  Any expression smaller than the middle one is logically smaller than the third.  Therefore, I set ε=δ.  But then I'm stuck again.  In the proofs I looked at online, they say to set $$δ=min{(1,ε)}$$, to pick the smaller value of the two.  Why is that?
To sum up, I would appreciate any feedback on my work, especially an explanation on why the part between ****(...)**** is a valid deduction and why we pick the smallest value for $δ$? Thanks!  
 A: "How can we logically conclude that $3|x-1|<\epsilon$?"
We don't. We conclude that as long as $\delta<1$ , then $|x+1|\cdot |x-1|<3|x-1|$. We want $|x+1|\cdot |x-1|<\epsilon$, so if we can somehow make sure that $3|x-1|<\epsilon$, then we are good.
So now we are free to set our $\delta$ to make sure that the last inequality above holds. Choosing  any $\delta<\epsilon/3$ should work nicely, since if $|x-1|<\delta$, then
$$
|x^2-1|=|x+1|\cdot|x-1|< 3|x-1|<3\delta < \epsilon
$$
But remember that the middle step here also required $\delta<1$, so actually we have to pick $\delta<\min(1,\epsilon/3)$.
A: We get to choose our $\delta$, hence to get the conclusion that 
$$3|x-1|<\epsilon$$
I can choose my $\delta$ to be $$\delta=\min(1, \epsilon/3)$$
Hence, For $|x-1| <\delta$,
$$|(x+1)(x-1)|<3|x-1|<3\delta\leq 3(\epsilon/3)=\epsilon$$
$|x+1|<3$ is due to $\delta \leq 1$, $3\delta \leq \epsilon$ is due to we choose $\delta \leq \epsilon/3$.
Since I want $\delta$ to be both smaller than $1$ and $\epsilon/3$, I choose the minimal of both quantities.
A: If you start from $\epsilon$  (as you have done and as it's correct in general, if we have not an intuition about $\delta$), than you have to solve the inequality:
$$
|x^2-1|<\epsilon
$$
this is equivalent to the two systems:
$$
\begin{cases}
x^2-1\ge0\\
x^2<1+\epsilon
\end{cases}
\quad \lor \quad
\begin{cases}
x^2-1<0\\
x^2>1-\epsilon
\end{cases}
$$
With the condition $x>0$ (since we have $x \to 1$), the solution of this is:
$$
[1,\sqrt{1+\epsilon}) \cup(\sqrt{1-\epsilon},1)=(\sqrt{1-\epsilon},\sqrt{1+\epsilon})
$$
so you can chose $\delta= \min\{1-\sqrt{1-\epsilon},\sqrt{1+\epsilon}-1\}$
