Why upperbound $|x-a|$ by 1 in the proof of continuity? In most (all?) proofs of continuity of polynomials ($x^2, x^3$, etc), for example in Max Rosenlicht's book (http://www.math.pitt.edu/~frank/pittanal2121.pdf, page 97), the usual trick is to get to the expression 
$$
|x-a||x+a|
$$ 
and then bound $|x-a|$ by 1, which is then used to bound $|x+a|$ and then obtain $\delta = \min \{1, \frac{\epsilon}{2a +1} \}$. 
This baffles me even after numerous attempts. My questions are:
1) Why 1? What will change if I chose a different value? 
2) Why do we need this other 'smaller' bound at all? 
I realize the question is not particularly challenging, but afte numerous attempts I still can't get my head around it.  
 A: In general, the $\delta$ is given in such a way that can give us enough conditions to get $|f(x)-L|<\epsilon$. 
More generally, the method (or strategy) is:
1) To transform an expression like $|f(x)-L|$ into a expression like $|x-a||g(x)|$ where $|g(x)|\leq M$ for all $x$, 
2) Then use the inequality $$|x-a||g(x)|\leq M|x-a|<\epsilon$$ (because the case with $M|x-a|<\epsilon$ is an easy one!). 
So, in your example, note that your $\delta$ given is such that $\delta = min \{1, \dfrac{\epsilon}{2|a|+1} \}$. Let's analize this one. 
It has two "parameters", one is 1, and another is $\dfrac{\epsilon}{2|a|+1}$.
The first parameter 1 is neccesary for bounding |g(x)| and the other "parameter" is necessary for get the $\epsilon$.
Let's take an example. (It´s the classical one, but I'm going to explain you the tricks)
Suppose that you are asked to proof that $$\lim_{x \to a}{x^2}=a^2$$
Then we start by making a preliminar analysis, ie. 
$|x^2-a^2|=|(x-a)(x+a)|=|x-a||x+a|$ and note that this has the form of $|x-a||g(x)|$ (with g(x) = x+a). Now we want to find an M>0 such that $|g(x)|=|x+a| \leq M$
To achieve this, let´s say that $|x-a|<1$ (It can be another positive value, but the easiest positive value is nearly always 1). Then we are going to construct our g(x) from here, i.e.
$|x-a|<1$ implies that $-1<x-a<1$, then $-1+2a<x+a<1+2a$, therefore $|x+a|<1+2|a|$ (you can check the details). AND NOTE that the M that we are looking for is $|x+a|=|g(x)|\leq M=1+2|a|$. Also note that a is fixed, so M is a fixed constant.
So we already have that $$|x-a||x+a|=|x-a||g(x)| \leq |x-a| M = |x-a|(1+2|a|)$$. (Note that our chosen 1, let us bound g(x))
NOW WE WANT THAT LESS THAN EPSILON. 
So the plan is simple, we should make $|x-a|<\dfrac{\epsilon}{1+2|a|}$. 
Next, we have $$|x-a||x+a|=|x-a||g(x)| \leq |x-a| M = |x-a|(1+2|a|)<\dfrac{\epsilon}{1+2|a|}(1+2|a|)=\epsilon$$
That's why our $\delta$ should be $\delta = min \{1, \dfrac{\epsilon}{2|a|+1} \}$. The $1$ bounds our $|g(x)|$ and the $\dfrac{\epsilon}{1+2|a|}$ makes the cancelation posible to get the $\epsilon$. You can say that the $\delta$ has enough conditions (in this case, just two conditions) to solve the problem.
Now the proof will be:
PROOF: Let $\epsilon >0$. Let's take $\delta = min\{1,\dfrac{\epsilon}{2|a|+1} \}$.
If $|x-a|<\delta$, then $|x-a|<1$ and $|x-a|<\dfrac{\epsilon}{2|a|+1}$, therefore $-1<x-a<1$, so $-1+2a<x+a<1+2a$, and we can say that $|x+a|<1 + 2|a|$. (you can check the little details)
Now, note that $$|x^2-a^2|=|(x-a)(x+a)|=|x+a||x-a|<(1+2|a|)\left(\dfrac{\epsilon}{1+2|a|}\right)= \epsilon$$
I hope this may help.
A: 1) You can bound $\delta$ by any factor (most times), it does not have to be $1$. We take the number $1$ because we're such lazy people and $1$ is nice: $1 \cdot {\rm anything} = {\rm anything}$. However, you must be careful when dealing with functions like $1/x$. It is not safe to bound $\delta$ by $1$, if $a = 1/2$, say, because the interval $(-1/2,1/2)$ contains $x = 0$, for which the function is not defined. Meaning, the bound is too weak. For polynomials like $x^2$, this does not happen.
2) I, particularly, don't know another efficient strategy for proving continuity and limits by $\epsilon$'s and $\delta$'s. The idea of continuity is that the function doesn't varies too much near the point in question, if the function is continuous, if $x$ doesn't goes too far from $a$, then $f(x)$ won't go too far from $f(a)$. Formalizing this "$x$ doesn't goes too far from $a$" precisely is exactly placing a bound in $\delta$ (and hence in $|x-a|$).
I have made two examples of proofs by epsilons and deltas here, just like the book you're using, and I invite you to rewrite them using $2$ instead of $1$ for a bound, to get the feeling of it. I hope my answer helps you, even a little bit. And good question, by the way.
