248 CHAPTER 5 Series and Differential Equations
Next, assume that 1> >0, and letδ > 0 be a real number such
that whenever 0<|x−a|< δ,
|f(x)−L|<3
,
3 |M|
, |g(x)−M|<3
,
3 |L|
.
(If either ofL, M= 0, ignore the corresponding fraction 3 |L|, 3 |M |)
|f(x)g(x)−LM| = |(f(x)−L)(g(x)−M) + (f(x)−L)M
+(g(x)−M)L|
≤ |(f(x)−L)(g(x)−M)|+|(f(x)−L)M|
+|(g(x)−M)L|<^2
9
+
3
+
3
<
3
+
3
+
3
=,
proving that limx→af(x)g(x) = LM.^2
As indicated above, in computing limx→af(x) we are not concerned
withf(a); in factaneed not even be in the domain off. However, in
definingcontinuityat a point, we require more:
Definition. Letf be defined in a neighborhood ofa. We say thatf
iscontinuousatx=aif
xlim→af(x) = f(a)As a simply corollary of the above theorem we may conclude that
polynomial functions are everywhere continuous.
The student will recall that the derivativeof a function is defined
in terms of a limit. We recall this important concept here.
(^2) Note that in the above proof we have repeatedly used the so-called “triangle inequality,” which
states that for real numbersa, b∈ R,|a+b| ≤ |a|+|b|. A moment’s thought reveals that this is
pretty obvious!