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

Special Kinds of Theorems 93
Thus, when |x - 2|< f it will follow that \(3x - 5) - l\<6. This proves
that lim(3x - 5) = 1. •
The construction of the proof presented in Example 1 is very dehcate, and
it depends on the choice of the correct number as the limit. The algebraic
steps performed cannot be replicated to show the existence of the number 8,
if the limit was incorrectly chosen. We leave it to the reader to see how the
proof breaks down if one tries to modify it to show that lim(3x — 5) is some
number other than 1.
Before proceeding further, let us look behind the algebraic approach and
examine the geometric meaning of our findings. We will do so by choosing
some values for s and studying the detailed consequences of our choices.
This will provide examples (which are useful but can never replace a proof),
and it will also provide a visual approach to the concept of limit, which some
people find quite helpful.
Consider s = 4.5. The conclusion just obtained states that
|(3x-5) - l|< 4.5 if we use values of x that satisfy the inequality
|x - 2| < 1.5 (i.e., values of x such that -1.5<x--2<1.5or more explicitly
0.5<x<3.5).
Consider the graph the function f{x) = 3x-5 corresponding to the
interval (0.5, 3.5) (see Figure 1.)
It can be observed that all the corresponding values of/(x) fall between
-3.5 and 5.5—that is, in less than 4.5 units from the value L = 1, as shown in
Figure 2.
If we now consider 6 = 0.75, we see that in order to have
|(3x — 5) - l| < 0.75 we cannot use again all the values of x in the interval
(0.5, 3.5). In this case, because we want the values of/(x) to be closer than
0.75 to 1 (much closer than when we chose s = 4.5), we might need to choose
X much closer to 2. Indeed, we will need \x — 2| < 0.75/3 = 0.25.
Figure 1
-6 J