4.3 THE LIMIT CONCEPT (OPTIONAL) 141What then would be the situation in the limit problem we proposed earlier?
In particular, would we still have lim,,, f(x) = 11, according to this "defini-
tion"? Here is the reason we would not! Let E be any positive real number
less than or equal to 9 (note that 9 is the gap between y = 2 and y = 11).
Then, no matter how small we choose 6, the particular value x = 3 of x satis-
fies lx - 3) < 6, since 13 - 3) = 0 c 6, whild at the same time 1 f(x) - LJ =
1 f(3) - LI = 12 - 11 1 = 9 2 E. This shows that L = 11 fails to satisfy this
(incorrect) version of the definition.
The correct epsilon-delta definition of limit requires, in order for L to
equal lirn,,, f(x), that we be able to make all corresponding values of f(x)
(other than f(a) itself if it exists) as close as we please to L, that is, within
distance E of L, by considering only values of x sufficiently close to a, that
is, within S of a (but again, not equal to a). If we omit the parenthetical parts
of the preceding sentence, the example of the previous paragraph shows
that limit problems we have designated "Type 11" would instead be of Type
111, since no value of L could satisfy the epsilon-delta definition. Indeed, if
we allow the value off at to influence the value of lirn,,, f(x), by using
the preceding incorrect "definition" of limit, we are not only doing away
with the category of "Type I1 limitythe most important category because
of its role in the definition of derivative), but are simultaneously removing
much of the subtlety that makes the limit concept interesting.
Finally, why does the correct epsilon-delta definition of limit yield the
rule of thumb that "L = the y component of the missing point" in a Type
I1 example? If our characterization of the "0 < (x - a(" part of the defi-
nition as mandating that the value of f(a) is irrelevant to the value of L
is accepted, then the following is evident. Suppose g is a function continu-
ous at each point of some open interval I containing a. Further suppose
ifx~I,x#a
. By definition of
either undefined or not equal to g(a), if x = a
continuity at a, lim,,, g(x) = g(a) It stands to reason that, since f agrees
with g everywhere except at a, and since lirn,,, f(x) is completely deter-
mined by the values off (x) at points x close to a, and not at all by f (a), then
lirn,,, f (x) should equal lirn,,, g(x); that is, should equal g(a). But g(a) is
precisely the "missing point" in the graph off as shown in Figure 4.8.
EXAMPLE 5 Compute lim,, , [(x2 - 5~ + 6)/@ - 3)]
Solution f(x)=[(x2-5x+6)/(x-3)]isundefinedatx=3,butifx#3,
then f(x) = [(x2 - 5x + 6)/(x - 3)] = x - 2, where we denote the latter
function by g(x). Now g, being a linear polynomial function, is continuous
everywhere, in particular, lirn,,, g(x) = g(3) = 3 - 2 = 1. Since f agrees
with g everywhere except at a = 3, so that the graph of f is simply a
line with the point (3, 1) missing, then lirn,,, f(x) = lirn,,, g(x) = 1.
0