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, Ay/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 = Xo. The conditions for continuity are (1) x = x() must be in the domain of the function j[ (2)y must have a limit as x xo, and (3) the said limit must be equal to /(x0). When these are satisfied, we can write lim /(x) = /(xo) [continuity condition] (6.13)

In contrast, when the "limit" concept is applied to the difference quotienl Ay/Ax as Ax —► 0, we deal instead with the question of whether the function /is differentiahle at x — xo, i.e., whether the derivative dv/dx exists at x = xo, or whether fix o) exists. The term differentiahle is used here because the process of obtaining the derivative dy/dx is known as differentiation (also called derivation). Since /'(xy) exists if and only if the limit of Ay /Ax exists at x = xu as Ax 0. the symbolic expression of the differentiabililv of /is f(x0) = lim

= lim —[differentiability condition] (6.14)

These two properties, continuity and differentiability, are very intimately related to each other the continuity off is a necessary condition for its differentiability (although, as we shall see later, this condition is not sufficient). What this means is that, to be diffcrentiable atx = jtfl, the function must first pass the test of being continuous at* = xo. To prove this, we shall demonstrate that, given a functiony = /(x), its continuity atx = x0 follows from its differentiability at x ~ x^; 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 A'and (2) replacing (xo + 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 uAx 011 becomes "x Nwhich is analogous to the expression v -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 - /V is a nonzero number, it is permissible to write the following identity;

FIGURE 6.4

fix)

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

Right side = lim ^—- lirn Cjc — N) [product limit theorem]

= /'{/V)(lim .v - lim JV) [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 expres$ion) in (6,15) would not possess a limit. Ii f(N) does exist, however, the two sides will have limits as shown in the previous equations. 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 differentiate 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 y = f(x) = |jc - 2| + 1 (6.16)

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, / {2) is also found to be 1. Thus all three requirements of continuity are met. To show that the function J is not differentiable at

FIGURE 6.5

FIGURE 6.5

x = 2, we must show that the limit of the difference quotient r fW-fW v I*-21 + 1-1 r I*

does not exist This involves the demonstration of a disparity between the left-side and the right-side limits. Since, in considering the right-side limit, x must exceed 2, according to the definition of absolute value in (6.8) we have \x - 2\ = x — 2. Thus the right-side limit is r I*"2' ,■ v i i lim -= lim — = lim 1 = 1

On the other hand, in considering the left-side limit, x must be less than 2; thus, according to (6,8), |x - 2| = —(x - 2), Consequently, the left-side limit is

which is different from the right-side limit. This shows that continuity does not guarantee differentiability. In sum, all differentiable functions arc continuous, but not all continuous functions are differentiable,

Jn Fig. 6.5, the nondifferentiability of the function at x = 2 is manifest in the f act that the point (2, 1) has no tangent line defined, and hcnce no definite slope can be assigned to the point. Specifically, to the left of that point, the curve has a slope of -1, but to the right it has a slope of+1, and the slopes on the two sides display no tendency to approach a common magnitude at x = 2. The point (2, i ) is, of course, a special point; il is the only sharp point on the curve. At other points on the curve, the derivative is defined and the function is differentiable* More specifically, the function in (6.16) can be divided into two linear functions as follows:

The left part is differentiable in the interval (-co, 2) , and the right part is differentiable m the interval (2, oc) in the domain.

In general, differentiability is a more restrictive condition than continuity, because it requires something beyond continuity. Continuity at a point only rules out the presence of a gap, whereas differentiability rules out "sharpness" as well. Therefore, differentiability calls for "smoothness" of the function (curve) as well as its continuity. Most of the specific functions employed in economics have the property that they are differentiable everywhere. When general functions are used, moreover, they arc often assumed to be everywhere differentiable, as we shall in the subsequent discussion.

EXERCISE 6,7

1. A function y - f(x) is discontinuous at x = when any of the three requirements for continuity is violated at x — xq. Construct three graphs to illustrate the violation of each of those requirements.

2. Taking the set of all finite real numbers as the domain of the function q = g(v) = v2 -5v — 2:

(a) Find the limit of q as v tends to N (a finite real number).

(b) Check whether this limit is equal to g(N).

(c) Check whether the function is continuous at N and continuous in its domain.

(a) Use the limit theorems to find lim q, N being a finite real number (ib) Check whether this limit is equal to g(N).

(c) Check the continuity of the function g(v) at N and in its domain (-cc, oo).

(a) Is it possible to apply the quotient limit theorem to find the limit of this function as x 4?

(ib) Is this function continuous at x - 4? Why?

(c) find a function which, for x ^ 4, is equivalent to the given function, and obtain from the equivalent function the limit of y as x 4.

5. In the rational function in Example 2,; the numerator is evenly divisible by the denominator, and the quotient is v -f 1 > Can we for that reason replace that function outright by q = v+1 ? Why or why not?

6. Or* the basis of the graphs of the six functions in Fig. 2.8, would you conclude that each such function ¡s differentiate at every point in its domain? Explain.

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