Bridge to Abstract Mathematics: Mathematical Proof and Structures

198 METHODS OF MATHEMATICAL PROOF, PART II Chapter 6

In most theorems about general properties of limits the hypotheses in-

clude a statement about the existence and value of one or more limits. In

such cases the epsilon-delta proof of a conclusion about limits will often

use in a crucial way the technique of specialization, introduced in Article 5.3.

EXAMPLE 12 Assume that lim,,, f (x) = L, and lim,,, g(x) = L,. Prove

that lirn,,, (f + g)(x) = L, + L,.

Solution Let E > 0 be given; we fix this E here at the outset and note that

we work with it throughout the remainder of the proof. We must produce

6 > 0 such that whenever x is a real number satisfying 0 < Ix - a( < 6,

then (f + g)(x) satisfies l(f + g)(x) - (L, + L,)I < E. Now our hy-

potheses will tell us something about the quantities (f (x) - L, I and

Ig(x) - L,I, and we note for future reference that the quantity we wish

to make less than E can be related to these two quantities by the inequality

Now precisely how can we make use of our hypotheses? Since

lirn,,, f(x) = L,, then corresponding to any positive real number, and

in particular to our given E, divided by 2, there is a positive 6, having

the property that whenever 0 < Ix - a1 < dl, then 1 f(x) - L,I < ~/2.

Similarly, there exists 6, > 0 such that whenever 0 < lx - a1 < a,, then

Ig(x) - L,I < ~/2. Now if we can choose 6 so small as to make both

these inequalities concerning f and g true at the same time, we will be

very close to our final goal, due to expression (1). How shall we choose

6? Recalling the argument in the first part of Example 9, we let

6 = min {a1, 6,). Then if x satisfies 0 < Ix - a1 < 6, it will automatically

satisfy both 0 < Ix - a1 < 6, and 0 < Ix - a1 < d2, so that

I(f + g)(x) - (L, + L,)I = I(fW - Ll) + (g(x) - L2)I

5 IfW - LlI + Ig(x) - L2I

.-- < &/2 + &/2 = E, as desired.
Note the role of specialization in the preceding proof. We have set up
the proof that (f + g)(x) tends to L, + L,, as x tends to a, by fixing a value
of E at the outset. Our two hypotheses, involving f and g separately, in-
dicate that something works for any positive real number, and so we can
conclude that this something works for the particular positive number ~/2.
Proofs of other limit theorems (e.g., Exercises 15, 18) call for similar appli-
cations of the specialization technique.
Students often ask why 42 is used in this proof and how they can ever
be expected to think of such a "trick" in writing their own proofs. The
first question can be answered on two levels. We use ~/2 (1) because there
is no reason we shouldn't (we violate no rules in doing so) and (2) because