Bridge to Abstract Mathematics: Mathematical Proof and Structures

4.3 THE LIMIT CONCEPT (OPTIONAL) 139

Variations of the arguments used in Examples 2 and 3 are needed to

prove that limits do not exist in other cases, as in Example 4.

, EXAMPLE 4 Use an epsilon-delta argument to show that 1 # lim,,, h(x),

1, x rational

where h(x) =

• 1, x irrational
  Solution Note first from Figure 4.7 that the graph of h consists of parts
  of the two horizontal lines y = 1 and y = - 1, each line, however, having
  infinitely many "missing points." The two pieces of the graph of this
  function near x = 0 have a. gap of 2 separating them. Thus, to prove
  1 # lirn,,, h(x), we may let E = $ (we could in fact choose any specific
  value of E less than or equal to 2), thus determining a horizontal E band
  about the line y = 1, extending from -3 to $, measured along the y
  axis. Let 6 > 0 be arbitrary. Can we find x between -6 and 6 whose
  corresponding f(x) is not between -3 and $? Yes! Simply let x be an
  irrational number between -6 and 6 (Note: It can be proved that there
  is an irrational number between any two reals). Then f(x) = - 1 $
  (-+,$). a

An argument similar to the preceding one can be used to show that

lirn,,, h(x) does not exist and, more generally, that lirn,,, h(x) does not

exist for any real number a.

As indicated earlier, we will not focus on epsilon-delta proofs that L =

lirn,,, f(x) until Article 6.1, but a few words about them now may be ap-

propriate. This type of proof differs from the "negative" epsilon-delta ar-

guments we've seen in Examples 2, 3, and 4 in that we may not specify a

Figure 4.7 Graph pertaining to Example 4.

1, x rational

The function h(x) =

{ - 1, x irrational

/--- cannot really be graphed. Its graph consists
of parts of two parallel lines, each with injinitely
many missing points. Every vertical line x = k
meets either the upper portion (k rational) or
the lower portion (k -irrational), but not both.

/

x irrational

y=l

7""""'

y=-l