Page 173 - HOW TO PROVE IT: A Structured Approach, Second Edition

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More Examples of Proofs 159
are not prime. These are completely unnecessary and are only included to make
the proof easier to read.

For readers who are familiar with the definition of limits from calculus, we
give one more example, showing how proofs involving limits can be worked
out using the techniques in this chapter. Readers who are not familiar with this
definition should skip this example.
2
2x − 5x − 3
Example 3.7.4. Show that lim = 7.
x→3 x − 3
Scratch work

According to the definition of limits, our goal means that for every positive
number ε there is a positive number δ such that if x is any number such that 0 <
2

|x − 3| <δ, then 2x −5x−3 − 7 <ε. Translating this into logical symbols, we

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

x − 3
We therefore start by letting ε be an arbitrary positive number and then try to
find a positive number δ for which we can prove

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

x − 3

The scratch work involved in finding δ will not appear in the proof, of course.
In the final proof we’ll just write “Let δ = (some positive number)” and then
proceed to prove
2x − 5x − 3
2
∀x 0 < |x − 3| <δ → − 7 <ε .

x − 3
Before working out the value of δ , let’s figure out what the rest of the proof
will look like. Based on the form of the goal at this point, we should proceed by
2

letting x be arbitrary, assuming 0 < |x − 3| <δ , and then proving 2x −5x−3 −
x−3

7 <ε. Thus, the entire proof will have the following form:

Let ε be an arbitrary positive number.
Let δ = (some positive number).
Let x be arbitrary.
Suppose 0 < |x − 3| <δ .
2

[Proof of 2x −5x−3 − 7 <ε goes here.]

x−3
2
2x −5x−3

Therefore 0 < |x − 3| <δ → − 7 <ε.

x−3