epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

92 The Nuts and Bolts of Proof, Third Edition
quantities and accepted and used more vague statements of the form: "The
value of/(x) gets as close as we want to Lfor all values of x sufficiently close
to c." On one hand, the formal definition just presented
removes the subjectivity of words such as "close" and "sufficiently close,"
while, on the other hand, for beginners in calculus and real analysis it seems
to remove the intuitive meaning as well. But details are of paramount
importance when trying to reach absolute precision. How can one make the
transition from intuitiveness to rigor without missing the important points?
Let us try to do so very carefully.
The statement "The value of/(x) gets as close as we want to L" can be
reworded as "The difference between the value of/(x) and Lgets as small
as we want"; that is, "|/(x) - L\ gets as small as we want." This means that
the value of |/(x) - L\ has to be smaller than every positive number s we can
think of. So, the value of s cannot be handpicked.
Because the value of/(x) depends on x, |/(x) - L\ will be smaller than e
only for values of x suitably close to c. One question still remains: "How
close to c should x be?" This is the same as asking "How small (large) does
|x - c\ need to be?"
Therefore, to prove that lim/(x) = L, we need to prove the existence of
an estimate for |x - c| so that all the values of x that satisfy that estimate will
make \f{x) — L\ smaller than the given s. So, the proof of fact that the
number Lis indeed the correct limit is an existence proof. (See Existence
Theorems section.)
EXAMPLE 1. Prove that lim(3x — 5) = L
Proof: According to the definition of limit, one must prove that for every
given positive number s there exists a positive number 8 such that if x is a
value that satisfies the requirement |x — 2|<5, then |(3x — 5) — l|<^.
The existence of 6 will be proved exphcitly, by giving a formula for it that
depends on s and possibly on 2, the value to which x is close.
Let us start by performing some algebraic steps, and then we will consider
the geometric meaning of the result. The statement |(3x — 5) - l| <£ can be
rewritten as:
or
|3x — 6\ < s
3|x — 2\<£.
This inequality will hold true only when |x - 2| < -.
Therefore, in this case we have 8 = s/3, and because s is positive, 8 is
positive as well.