Bridge to Abstract Mathematics: Mathematical Proof and Structures

134 ELEMENTARY APPLICATIONS OF LOGIC Chapter 4

Figure 4.2 The graph of y = f (x) in Example 2.

Solution It is intuitively clear that lim,,, f (x) should not exist, since the

graph of f(see Figure 4.2) has a break in the immediate vicinity of x =

0; but how can we argue this formally in terms of epsilon and delta?

Let us try first to show that L = 1 does not satisfy the definition of limit,

or more precisely, does satisfy the negation of that definition.

We must show that there exists (i.e., we must produce) a positive E,

giving rise to a horizontal "E band" {(x, y) 1 x E R, 1 - E < y < 1 + E)

about the line y = 1 (see Figure 4.3) such that, for any 6, no matter how

small, there can always be found a value of x that is within 6 of a = 0,

but whose functional value f(x) does not lie in the given horizontal band;

that is, f(x) is not within E of 1. Let us try E = 1, SO that we are speci-

fying the horizontal E band {(x, y)l x E R and 0 < y < 2). Let an arbitrary

Figure 4.3 A value of c has been chosen small enough

so that the entire portion of the curve to the left of

x = 0 lies outside the resulting &-band.

I /

The horizontal e band

l-€<y<l+E