A couple of proofs

I'm going through some examples of applications of the various theorems in my calculus book. I understand the easier examples, but I'm having a hard time following the finer details of some of the more complex ones.

The definition of convergent sequanses:

"The sequence {an} converges to a number a, if there for every real number (epsilon)>0 exists a real number N which makes |an-a|<(epsilon) for every n>(or equal to)N. In which case we write:

lim(n-->(infinite))an = a"

Example application:

"Let {an} and {bn} be two convergent sequenses with lim(n-->(infinite))an=A and lim(n-->(infinite))bn=B. Prove that lim(n-->(infinite))(an*bn)=A*B."

My book does this as follows:

"Given (epsilon)>0, we'll have to find an N that will make |an*bn - AB| < (epsilon) when n>(or is equal to)N. It is difficult to measure the distance |an*bn - AB| directly, as both an and bn varies with n. The solution is given by seperating the expression into two parts, by adding and subracting an*B. The triangle inequality gives:

|anbn - AB| = |(anbn - anB + anB - AB)| <(or equals) |anbn-anB| + |anB - AB|= |an| * |bn - B| + |B| * |an - A|

If we can show that we can make each part, |an| * |bn - B| and |B| * |an - A|, smaller than (epsilon)/2 by giving n a high enough value, then the original statement is proven."

All this makes perfect sense to me, it's the last part I'm having trouble with. (Small sidenote: What are the rules for factorizing absolute values?)

"Starting with the easiest part, |B| * |an - A|: Since lim(n-->(infinite))an=A, there has to exist an N1 which makes |an - A| < (epsilon) / (2 * |B|) when n>(or is equal to)N1. Which in turn makes |B| * |an - A| < (epsilon)/2."

I'm still following.

"The part |an|*|bn-B|is a little more complicated because |an| is variable."

Now the above I don't get.

"Since lim(n-->(infinite))an=A, we do know that there exists an N2 which makes |an| < |A| + 1 when n >(or equals)N2."

Now I'm totally lost. Why does it have to be less than |A| + 1? I thought |an| could never reach |A| + 0 even, and that that would be true for all values of n. Is this just some random value which could have been whatever? Continuing:

"Since lim(n-->(infinite))bn=B, there has to be an N3 which makes |bn-B| < (epsilon)/(2*(|A|+1)) when n>(or equals)N3."

Got it.

"If we choose n larger or equal to both N2 and N3, it gives us:

|an|*|bn-B| < (|A| + 1) * (epsilon)/2(|A|+1) = (epsilon)/2.

If we let N be the largest of the numbers N1, N2 and N3, we se that if n>(or equals)N:

|anbn - AB| <(or equals) |an| * |bn - B| + |B| * |an - A| < (epsilon)/2 + (epsilon)/2 = epsilon.

This completes the proof."

The above also makes sense, if I accept that |an| < |A| + 1. So what I'm wondering is basically how they made that claim. And also, the factorization with absolute numbers-thing. If you have |AB + AC|, do you have to write |A|*|B+C|, or does A * |B+C| count? Why?

Another small detail of a similiar example:

I'm supposed to show how small h has to be if |h|*|12+2h| is to be smaller than (epsilon). From the book:

"We see that when h is small, (12+2h) roughly equals 12. If we're being safe, we can state that 12+2h is less than 14 when |h| is less than 1. If we at the same time make sure that the other |h| is less than (epsilon)/14, the product |h|*|12+2h| will be less than (epsilon)."

What, are they using two different numbers for the two different h's?

"We therefore choose (delta) = min{(epsilon/14), 1} (which means (delta) equal to whichever number is less). "

(Delta) is the number h has to be less than. So now they're using one value for h.

"This means that |h|*|12+2h| < (epsilon)/14 * (12+2*1) = epsilon"

But they used two different values for the two different h's anyway? Or am I missing something?

Anyway, though I understand the arguments, there's some holes in my understanding of them. If anyone could clarify it would be greatly appreciated!