Page 174 - HOW TO PROVE IT: A Structured Approach, Second Edition

P1: PIG/
0521861241c03 CB996/Velleman October 20, 2005 2:42 0 521 86124 1 Char Count= 0

160 Proofs
Since x was arbitrary, we can conclude that ∀x(0 < |x − 3| <δ →
2
2x −5x−3 − 7 <ε).

x−3
2

2x −5x−3
Therefore ∃δ> 0∀x 0 < |x − 3| <δ → − 7 <ε .

x−3

Since ε was arbitrary, it follows that ∀ε> 0∃δ> 0∀x 0 < |x − 3| <
2
δ → 2x −5x−3 − 7 <ε .

x−3
Two steps remain to be worked out. We must decide what value to assign
2

to δ , and we must fill in the proof of 2x −5x−3 − 7 <ε. We’ll work on the

x−3
second of these steps first, and in the course of working out this step it will
become clear what value we should use for δ . The givens and goal for this
second step are as follows:
Givens Goal
2
2x − 5x − 3
ε> 0 − 7 <ε

x − 3

δ = (some positive number)
0 < |x − 3| <δ
First of all, note that we have 0 < |x − 3| as a given, so x = 3 and therefore
2
the fraction 2x −5x−3 is defined. Factoring the numerator, we find that
x−3
2x − 5x − 3 (2x + 1)(x − 3)
2

− 7 = − 7
x − 3 x − 3
=|2x + 1 − 7|=|2x − 6|= 2|x − 3|.
Now we also have as a given that |x − 3| <δ ,so2|x − 3| < 2δ . Combining
2
this with the previous equation, we get 2x −5x−3 − 7 < 2δ , and our goal is

x−3
2
2x −5x−3 − 7 <ε. Thus, if we chose δ so that 2δ = ε, we’d be done. In other

x−3
words, we should let δ = ε/2. Note that since ε> 0, this is a positive number,
as required.
Solution
2
Theorem. lim 2x −5x−3 = 7.
x→3 x−3
Proof. Suppose ε> 0. Let δ = ε/2, which is also clearly positive. Let x be an
arbitrary real number, and suppose that 0 < |x − 3| <δ . Then
2
2x − 5x − 3 (2x + 1)(x − 3)

− 7 = − 7

x − 3 x − 3

=|2x + 1 − 7|=|2x − 6|
ε

= 2|x − 3| < 2δ = 2 = ε.
2