226 Chapter 5 • Continuous Functions5.1 Continuity of a Function at a Point
Definition 5.1.1 (Continuous Function at a Point) Suppose f: 'D(j) ---+
IR, and x 0 E 'D(f). Then f is continuous at xo if
Ve> 0, :l 8 > 0 3 Vx E 'D(f), Ix - xol < 8 ==;. lf(x) - f(xo)I < c.
If f is not continuous at x 0 then we say that f is discontinuous at xo.
Notes: (1) In Definition 5.1.1, xo must be in 'D(f) but need not be a cluster
point of 'D(f). Thus, lim f(x) need not exist, even when f is continuos at x 0.
X-+Xo
(See Exercises 1 and 3.)
(2) In case xo is a cluster point of 'D(f), Definition 5.1.1 is equivalent to
conditions (a)-(c) given above (See Exercise 2). In fact, we usually combine
(a)-(c) into one statement, and say:
If x 0 is a cluster point of 'D(f), then f is continuous at x 0 iff
lim f(x) = f(xo).
X--+Xo(3) Suppose xo is a cluster point of'D(j). Calculating lim f(x) is trivial if
x-+xo
f is continuous at xa. Continuity off at x 0 means that to calculate lim f(x)
X--+Xo
we merely "plug in" x = x 0. In a sense, this suggests that limits are not of much
interest in studying functions that are known to be continuous at a point; limits
will often be of more interest at points where a function is discontinuous. But
this definition also suggests that we can use limits to determine whether a
function is continuous at a point.
Students frequently ask for an example that shows how to use Definition
5.1.l to prove that a function is continuous at a point. For them we include the
following example. You will see right away that there is nothing more to this
example than showing that lim f(x ) = f(2). In truth, however, this example
X --> 2
serves another purpose: to show the difference between ordinary continuity and
another type of continuity known as uniform continuity, to be introduced in
Section 5.4. The details will be seen in Example 5.4.2.
Example 5.1.2 Use Definition 5.1.l to prove that the function f(x) = 3x^2 -
2x - 1 is continuous at x 0 = 2.
Solution. Let f(x) = 3x^2 - 2x - 1. Note that f(2) = 7.
(a) Scratchwork: Let c > 0.
We must find 8 > 0 3 Ix - 21 < 8 ==;. j(3x^2 - 2x - 1) - 71 < c. i.e.,
j3x^2 - 2x - 81 < c, or l3x + 4llx - 21 < c.