evaluating limits

Problem: Identify a value of delta that
corresponds to epsilon = 0.01 given
lim_x->-1 (x^2 + 3) = 4 according to
the definition of limits.

|f(x) - L| < Epsilon
|(x^2 + 3) - 4| < .01
|x^2 - 1| < .01
|x + 1||x - 1| < .01

Objective: get the above
in the form: |x - a| < delta

Below is quoted from my book:

"Begin by assuming delta is less
than 1. Which is reasonable
because it should represent
an extremely small distance."

Why is that? Why can't it represent
an extremely large distance?

"If delta < 1, then |x + 1| < 1
which means that -1 < x + 1 < 1."


"Subtract 2 from each of those
expressions to get
-3 < x - 1 < -1. Therefore,
|x - 1| < 3"

This part, I just don't follow at all.
Why subtract 2? What drives us
to do this? If we start out with
-1 < x + 1 < 1, to say that we have
to subtract 2 seems arbitrary.



"Your goal is to produce the
expression |x - a|
in the middle of the compound
expression. Recall that |x - 1| < 3
and substitute 3 into the inequality.

|x - 1||x + 1| < 0.01
3|x + 1| < 0.01
|x + 1| < 0.01/3

Therefore, delta = .01/3"

I don't follow why we use
|x - 1| < 3 or where it comes from.
If we had:
-1 < x + 1 < 1, algebra does not
say to pull a -2 from thin air and apply
it to the expression or to magically
produce |x - 1| < 3. I see a whole lot
of fluff here and too little reason being
applied. I would like to reduce the fluff
and increase the reasoning. For example,
sure I can recognize that |x - 1| < 3 because
originally we had |x - 1||x + 1| < .01
but by that same token, I could just as well
say |x - 1| < 4, |x - 1| < 100, |x - 1| < 10^100
etc, what makes 3 so special?

Also, he continues from stating that
|x - 1| < 3 to substituting
3 for |x - 1|, again, I'm left wondering
why?


Any clarification appreciated.


--
conrad

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