## Continuity of a Function

When a function q = g(v) possesses a limit as v tends to the point N in the domain, and when this limit is also equal to g(N)—that is, equal to the value of the function at v = N—the function is said to be continuous at N. As stated above, the term continuity involves no less than three requirements: (1) the point N must be in the domain of the function; i.e., g(N) is defined; (2) the function must have a limit as v -* N; i.e., lim g(v) exists; and (3) that limit must be equal

It is important to note that while—in discussing the limit of the curve in Fig. 6.3—the point (N, L) was excluded from consideration, we are no longer excluding it in the present context. Rather, as the third requirement specifically states, the point (N, L) must be on the graph of the function before the function can be considered as continuous at point N.

Let us check whether the functions shown in Fig. 6.2 are continuous. In diagram a, all three requirements are met at point N. Point N is in the domain; q has the limit L as v -* N; and the limit L happens also to be the value of the function at N. Thus, the function represented by that curve is continuous at N. The same is true of the function depicted in Fig. 62b, since L is the limit of the function as v approaches the value N in the domain, and since L is also the value of the function at N. This last graphic example should suffice to establish that the continuity of a function at point N does not necessarily imply that the graph of the function is "smooth" at t> = N, for the point (N, L) in Fig. 6.2b is actually a "sharp" point and yet the function is continuous at that value of v.

When a function q = g(v) is continuous at all values of v in the interval (a, b), it is said to be continuous in that interval. If the function is continuous at all points in a subset S of the domain (where the subset S may be the union of several disjoint intervals), it is said to be continuous in S. And, finally, if the function is continuous at all points in its domain, we say that it is continuous in its domain. Even in this latter case, however, the graph of the function may nevertheless show a discontinuity (a gap) at some value of v, say, at v = 5, if that value of v is not in its domain.

Again referring to Fig. 6.2, we see that in diagram c the function is discontinuous at N because a limit does not exist at that point, in violation of the second requirement of continuity. Nevertheless, the function does satisfy the requirements of continuity in the interval (0, N) of the domain, as well as in the interval [iV, oo). Diagram d obviously is also discontinuous at v = N. This time, discontinuity emanates from the fact that N is excluded from the domain, in violation of the first requirement of continuity.

On the basis of the graphs in Fig. 6.2, it appears that sharp points are consistent with continuity, as in diagram b, but that gaps are taboo, as in diagrams c and d. This is indeed the case. Roughly speaking, therefore, a function that is continuous in a particular interval is one whose graph can be drawn for the said interval without lifting the pencil or pen from the paper—a feat which is possible even if there are sharp points, but impossible when gaps occur.

### Polynomial and Rational Functions

Let us now consider the continuity of certain frequently encountered functions. For any polynomial function, such as q = g(v) in (6.11), we have found from (6.12) that lim q exists and is equal to the value of the function at N. Since TV is a v — N

point (any point) in the domain of the function, we can conclude that any polynomial function is continuous in its domain. This is a very useful piece of information, because polynomial functions will be encountered very often.

What about rational functions? Regarding continuity, there exists an interesting theorem (the continuity theorem) which states that the sum, difference, product, and quotient of any finite number of functions that are continuous in the domain are, respectively, also continuous in the domain. As a result, any rational function (a quotient of two polynomial functions) must also be continuous in its domain.

Example 1 The rational function

is defined for all finite real numbers; thus its domain consists of the interval (—oo, oo). For any number N in the domain, the limit of q is (by the quotient limit theorem)

Hm(4u2) 4n2

which is equal to g(N). Thus the three requirements of continuity are all met at N. Moreover, we note that N can represent any point in the domain of this function; consequently, this function is continuous in its domain.

Example 2 The rational function j.d ( ' - v ,f v3+ v2-4v-4"rjzll^il -7—T, ©

is not defined at v = 2 and at v = — 2. Since those two values of v are not in the domain, the function is discontinuous at v = — 2 and v = 2, despite the fact that a limit of q exists as v -* — 2 or 2. Graphically, this function will display a gap at each of these two values of v. But for other values of v (those which are in the domain), this function is continuous.

### Differentiability of a Function

The previous discussion has provided us with the tools for ascertaining whether any function has a limit as its independent variable approaches some specific value. Thus we can try to take the limit of any function y = f(x) as x approaches some chosen value, say, x0. However, we can also apply the "limit" concept at a different level and take the limit of the difference quotient of that function, Aj>/Ax, as Ax approaches zero. The outcomes of limit-taking at these two different levels relate to two different, though related, properties of the function /.

Taking the limit of the function y = f(x) itself, we can, in line with the discussion of the preceding subsection, examine whether the function / is continuous at x = x0. The conditions for continuity are (1) x = x0 must be in the domain of the function /, (2) y must have a limit as x -> x0, and (3) the said limit must be equal to /(x0). When these are satisfied, we can write

When the "limit" concept is applied to the difference quotient A^/Ax as Ax -» 0, on the other hand, we deal instead with the question of whether the function / is differentiable at x = x0, i.e., whether the derivative dy/dx exists at x = x0, or whether f'(x0) exists. The term "diiferentiable" is used here because the process of obtaining the derivative dy/dx is known as differentiation (also called derivation). Since f'(x0) exists if and only if the limit of Ay/Ax exists at x = xQ as Ax -> 0, the symbolic expression of the differentiability of / is

These two properties, continuity and differentiability, are very intimately related to each other—the continuity of / is a necessary condition for_ilS differentiability (although, as we shall see later, this condition is not sufficient). What this means is that, to be differentiable at x = x0, the function must first pass the test of being continuous at x = x0. To prove this, we shall demonstrate that, given a function^ = f(x), its continuity at x = x0 follows from its differentiability at x = x0, i.e., condition (6.13) follows from condition (6.14). Before doing this, however, let us simplify the notation somewhat by (1) replacing x0 with the symbol N and (2) replacing (xf) + Ax) with the symbol x. The latter is justifiable because the postchange value of x can be any number (depending on the magnitude of the change) and hence is a variable denotable by x. The equivalence of the two notation systems is shown in Fig. 6.4, where the old notations appear (in brackets) alongside the new. Note that, with the notational change, Ax now becomes (x - N), so that the expression "Ax -> 0" becomes lim

[differentiability condition]

Figure 6.4

"x —> N," which is analogous to the expression v —> N used before in connection with the function q = g(v). Accordingly, (6.13) and (6.14) can now be rewritten, respectively, as

What we want to show is, therefore, that the continuity condition (6.13') follows from the differentiability condition (6.14'). First, since the notation x -» N implies that x + N, so that x - N is a nonzero number, it is permissible to write the following identity:

Taking the limit of each side of (6.15) as x -» N yields the following results: Left side = lim f(x) — lim f(N) [difference limit theorem]

Right side = lim -—~ ^m (x - N) [product limit theorem.]

= f'(N)( lim x - lim iv) [by (6.14') and difference

limit theorem]

Note that we could not have written these results, if condition (6.14') had not been granted, for if f'(N) did not exist, then the right-side expression (and hence also the left-side expression) in (6.15) would not possess a limit. If f'(N) does exist, however, the two sides will have limits as shown above. Moreover, when the left-side result and the right-side result are equated, we get lim /(x) - f(N) = 0, which is identical with (6.13'). Thus we have proved that continuity, as shown in (6.13'), follows from differentiability, as shown in (6.14'). In general, if a function is differentiable at every point in its domain, we may conclude that it must be continuous in its domain.

Although differentiability implies continuity, the converse is not true. That is, continuity is a necessary, but not a sufficient, condition for differentiability. To demonstrate this, we merely have to produce a counterexample. Let us consider the function

which is graphed in Fig. 6.5. As can be readily shown, this function is not differentiable, though continuous, when x = 2. That the function is continuous at x = 2 is easy to establish. First, x = 2 is in the domain of the function. Second, the limit of y exists as x tends to 2; to be specific, lim y = lim y = 1. Third, x->2+ x->2~

/(2) is also found to be 1. Thus all three requirements of continuity are met. To show that the function /is not differentiable at x = 2, we must show that the limit of the difference quotient

0 0