Bridge to Abstract Mathematics: Mathematical Proof and Structures

138 ELEMENTARY APPLICATIONS OF LOGIC Chapter 4

Solution This problem generalizes a part of Example 2, in which we dealt

with the particular case a = 1. To prove that 0 # lirn,,, f(x), we must

start by specifying a positive E. The key is how to choose this E; that

choice quite clearly depends on a. Recalling the fourth paragraph of the

solution to Example 2, we see that any specific positive value of E less

than or equal to a will work. If we let E = a, for instance, then for any

positive 6, no matter how small, there can always be found a value of

x within 6 of 0, measured along the x axis, whose corresponding f(x) is

a distance greater than E (or a) away from 0, measured along the y axis.

In fact, any x on either side of x = 0, between -6 and 6 will have this

property.

Let us consider some implications of Example 3. The number 0 cannot

2x + a,

serve as lim,,, f (x), where f (x) = '"},for anyo > 0. Fur-

x-a, x<O

thermore, there is a specific connection between a and the largest value of

E that will work in the epsilon-delta proof, namely, they're equal. If a =

i, we have a "gap" of 1 between the two pieces of the graph, with 0 halfway

between, and E = (or any smaller positive value of E) serves to show that

0 # lirn,,, f(x) in this case. If a = i, the gap is 4 and E = $ can be used.

If a = the gap is very small, namely, 2 x but still exists, and

E = lod5 (or any specific positive number smaller than lod5) can be used

to prove that 0 is not the limit. The same argument, with E = a, works no

matter how small a is, and thus no matter how small the gap is between

the two pieces of the curve, as long as a is positive. But suppose we now

let a = O? We observe that, geometrically, the gap between the two pieces

of the curve has now been closed completely, since we are now dealing with

the function g(x) = {2:: 1. What has become of our argument

that 0 # lirn,,, g(x)? Can we still find a positive E that works to prove this?

As seen previously, we must have 0 < E I a, but now a = 0. The largest E

we could use has been shrunk down to zero; that is, no positive E can be

found to prove that 0 # lirn,,, g(x). What we are saying can be symbolized,

in general (remember that a = 0 and L = 0 in our current example):

Geometrically, we know that lirn,,, g(x) exists and equals zero in this

example, since there is no longer any break in the curve in the immediate

vicinity of x = 0. This corresponds to the fact that the preceding symbolized

statement is logically equivalent to

where the latter is precisely the definition of lirn,,, g(x) = L.

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