Bridge to Abstract Mathematics: Mathematical Proof and Structures

6.1 CONCLUSIONS INVOLVING V, FOLLOWED BY 3 197

EXAMPLE 10 Prove that lim,,, (4x + 7) = 19.

Solution Note that f(x) = 4x + 7, a = 3, L = 19 in this problem. Let E > 0
be given. We must produce 6 > 0 such that whenever 0 < Ix - 31 < 6,
then 1 f(x) - LI = 14x + 7 - 191 = 14x - 121 < E. Let us look for a rela-
tionship between lx - a1 and 1 f(x) - LI; in this case it's easy to spot since
14x - 121 = 41x - 31 for every real number x. Hence we simply let 6 = &/4.
Note that if 0 < (x - 31 < 6, then 1 f(x) - 191 = 14x - 121 = 41x - 31 <
46 = 4(~/4) = E, as desired.

Note that if f(x) is a linear function y = Mx + B, we can always use

IMI as the positive constant k satisfying 1 f(x) - LI 1 klx - a(. (Verify this.)

Furthermore, this inequality is valid for all real x so that we do not need

to deal with the neighborhood N(x; 6,) referred to prior to Example 10.

Hence the choice of 6 in terms of E in any epsilon-delta proof involving a

linear function is particularly simple: Just take delta to be epsilon divided

by the absolute value of the slope!

EXAMPLE 11 Prove that lim,,, x3 = 125.

Solution Let E > 0 be given. Our goal is to produce 6 > 0 such that when-

ever x satisfies 0 < Ix - 51 < 6, then f (x) satisfies (f (x) - 1251 < &; that is,

1x3 - 1251 = (x - 51 1x2 + 5x + 251 < E. Now your first impulse, based

on the solution to Example 10, might be to assert that 6 can be taken

to equal &/Ix2 + 5x + 251. The problem with this suggestion is subtle,

but crucial. The definition of limit contains the sequence of quantifiers

(VE > 0)(36 > O)(Vx), not (VE > O)(Vx)(36 > 0). That is, the 6 we are

charged to find may depend on E only, not on E and x! Hence we must

take the approach outlined before Example 10. Namely, we must ask

whether there is some neighborhood N(5; 6,) of 5 and a positive constant

M such that lx2 + 5x + 251 s M whenever 0 < Ix - 5) < 6,. Let us try

6, = 1; that is, suppose that 0 < Ix - 51 < 1. Then - 1 s x - 5 < 1 and

x = 5 # 0, so that 4 1 x < 6 and x # 5. Since x 5 6, then we may con-

clude 1x2 + 5x + 251 < 36 + 30 + 25 = 91. That is, 1x2 + 5x + 251 < 91.

Thus our choice of 6 is dictated by the rule given just prior to Ex-

ample 10, with 6, = 1 and k = 91; let 6 = min (1,~/91}. Then if

0 < Ix - 5) < 6, we have both 0 < Ix - 51 < 1 so that lx2 + 5x + 251 5

91, and 0 < lx - 51 < e/91. Hence

A final version of this proof begins: "Let E > 0. We must find 6 > 0 such

that if 0 i (x - 51 < 6, then Ix3 - 1251 < E. If 0 < Ix - 51 < 1, then

(x2 + 5x + 251 s 91. Let b = min (1, ~/91).... "