Continuity

Continuity In this section we will explore a very useful class of functions - continuous functions. Definition 0 .1.1 ( Continuity a t a P oint). A f unction f : X → R , X ⊂ R i s c ontinuous a t a point x0 ∈ X if for every ε > 0, there exists δ > 0 so that for all x ∈ X with |x − x0 | < δ we have |f (x) − f (x0 )| < ε. This definition is written in terms of ε and δ, just like the formal definition of a li mit. Essentially, this definition o f c ontinuity i s t he s ame a s a f unction h aving a l imit a t a g iven p oint, e xcept that here the limit value is the value of the function. However, this definition includes functions mapping X → R, in which these notions are not strictly equivalent1 . This issue is not really related to our goals at the moment, so let’s restrict ourselves to functions defined on intervals, where these notions are equivalent. Theorem 0.1.1 (Continuity at a Point). Suppose f : I → R, I an interval. Let x0 ∈ I. There are three situations: 1. x0 is an interior point of I. f is continuous at x0 iff lim f (x) = f (x0 ). x→x0 2. x0 is a left endpoint of I. f is continuous at x0 iff lim f (x) = f (x0 ). x&x0 3. x0 is a right endpoint of I. f is continuous at x0 iff lim f (x) = f (x0 ). x%x0 This theorem states that for functions defined on intervals, continuity is equivalent to having a limit that is equal to the value of the function, if we are careful to use one-sided limits for endpoints. The criterion given for continuity by this theorem actually contains three criteria. For a function f to be continuous at an interior point x0 , we need that: 1. f (x0 ) is defined (x0 is in the domain of f (x)) 2. limx→x0 f (x) exists 3. limx→x0 f (x) = f (x0 ) For endpoints we simply use the appropriate one-sided limits in the above criteria. Although continuity is a local property, defined in terms of points, we are really more interested in a global notion of continuity - functions for which every point in their domain they are continuous. Definition 0.1.2 (Continuity). A function f is continuous if it is continuous at every point in its domain. In essence, continuity is related to the notion of how fast a function can change. If a function is continuous at a point, it means that we can restrict nearby function values to be as close as we want, simply by restricting ourselves to near enough input values. This property most clearly manifests itself over intervals. If a function is continuous on an interval, then it cannot contain any holes or jumps within the interval; if it contained any holes or jumps in an interval, then there would be some points where we cannot restrict how much the function changes by looking at a small enough interval, because no matter how small the interval, the function would change by the 4 5 4 f (x) f (x) 3 2 1 3 2 1 0 0 0 0.5 1 x 1.5 (a) Discontinuity at a hole. 2 0 1 2 x 3 4 (b) Discontinuity at a jump. Figure 1: Example discontinuities. height of the jump or the distance between the removed point and the rest of the function. Stated in other words, a function is discontinuous at any point where it has a jump or hole (see figure 1). From the perspective of limits, it is clear to see that for a hole the limit value does not equal the function value. For a finite jump, the one-sided limits are different, so the limit does not exist, and thus the function cannot be continuous there. The fact that a function that is continuous on an interval cannot contain any holes or jumps has an immediate consequence, called the intermediate value property. Theorem 0.1.2 (The Intermediate Value Theorem). Let f (x) be continuous on the closed interval [a, b]. For any value y0 between f (a) and f (b), there exists some c ∈ [a, b] such that f (c) = y0 . We use the statement between f (a) and f (b) because it is not clear which one is larger than the other, if in fact f (a) 6= f (b). If we have f (a) = f (b) then the above statement tells us nothing, because there are no values between f (a) and f (b), and we already know our function takes on the value f (a) = f (b) at the endpoints of the interval. Why is this property true? Suppose that there was some value between f (a) and f (b) skipped by the function. How could f skip such a value? Either by having that value removed at a point of the function, or by having a jump. Either of these methods would require the function have a discontinuity, which it doesn’t, by hypothesis. The intermediate value property is particularly useful for finding roots of a function, insofar that if we have a function f (x), and we know that for some a we have f (a) < 0 and for some b we have f (b) > 0, then there is some c between a and b where f (c) = 0. Continuity also has implications for the invertibility of a function. Consider a continuous function f : I → R. Since f is a continuous function defined on an interval, it cannot have any holes or jumps. In order for f to be invertible then, it follows that f must be strictly monotone (always increasing or always decreasing). Clearly if f were constant for some period of time it would not be invertible. However, we can go further. If f is increasing over some subinterval (or decreasing for that matter) unless it continues to increase, it will begin to repeat output values, because it cannot jump over them. For instance, suppose f (x) increases from 1 to 5 over some subinterval of I, and then later on has a value of 2. The only way it could get back down to 2 is by moving through all of the values 5 to 2, by the intermediate value theorem. Such a function would have multiple inputs with the same output, and hence not be invertible. Thus, although strict monotonicity is normally sufficient (but not necessary) for invertibility, in the case of continuous functions (defined on intervals) it also becomes necessary. Theorem 0.1.3 (Invertibility of Continuous Functions). Let f : I → R be continuous. It follows f is invertible if and only if f is strictly monotone. Just as we had rules for combining limits, we can also combine continuous functions to create continuous functions. Theorem 0.1.4 (Properties of Continuous Functions). Suppose g(x) is continuous at x0 1. If f is continuous at x0 , then f + g is continuous at x0 . 2. If f is continuous at x0 , then f · g is continuous at x0 . 3. If f is continuous at x0 , g(x0 ) 6= 0, then f /g is continuous at x0 . 4. If f is continuous at g(x0 ), then f ◦ g is continuous at x0 . As before, these rules will only be useful if we have some knowledge about basic continuous functions. We will be able to use these functions as building blocks to verify the continuity of more complex functions. We already know that for any polynomial function, the limit is the same as the value of the polynomial at every point, so it follows that all polynomial functions are continuous on R. Sine, cosine, and the exponential function are also continuous on all of R. In fact, most functions that we encounter are continuous, as long as we restrict them to the proper domain. For instance, ln(x) is continuous for x > 0, which is the entire domain over which it is defined. It follows from the continuity of these basic functions, and the above properties, that much more complicated functions such as x sin(x) f (x) = 2 x +2 are continuous (when combining functions in this way one must be careful to avoid division by 0, which we do here because (x2 + 2) > 0 for all x). At this point let us turn our attention to discontinuities, the points where functions fail to be continuous. In this discussion, unless otherwise stated we will be looking at functions defined on R. There are two classes of discontinuities - removable and nonremovable. If we have a removable discontinuity, it means that we can get remove the discontinuity just by redefining the function at the point of discontinuity. This corresponds to a function which is not continuous at a given point, but that ought to be. Think about a hole removed from a function and put in a different place. All we need to do is fill up that hole, and the discontinuity will be removed (see figure 2). Thus, holes in a function are an example of removable discontinuities. In contrast, nonremovable discontinuities are places where a function is broken so badly that there is no simple way to fix the function. Think about the function |x|/x. It is discontinuous at x = 0, because it is not defined. Nevertheless, no matter what value we try to assign to the function at that point, it will still be discontinuous from one side or the other. In this case there is a jump in the function that cannot be removed simply by redefining a single point. In order to remove this discontinuity from the function, we would need to redefine it over an entire interval, significantly changing the characteristics of the function. Any discontinuity we cannot remove by redefining the function at a single point is called a nonremovable discontinuity. In general, if a function is discontinuous because the function value is not equal to the limit, we simply redefine 4 3 3 f (x) f (x) 4 2 1 2 1 0 0 0 0.5 1 x 1.5 2 (a) Removable discontinuity. 0 0.5 1 x 1.5 2 (b) Discontinuity removed. Figure 2: Removing a removable discontinuity. the function value to the limit value and remove the discontinuity. If there is a place in the function where a limit does not exist, then simply redefining a single value will not make the limit exist, so the discontinuity is nonremovable. Example 1. Is sin(1/x) continuous over the real numbers? If not, does it have any discontinuities which can be redefined to make it continuous over the real numbers? Solution Since there is a division by 0 at x = 0, this function is not defined at x = 0, so it is discontinuous over a domain of real numbers. As x → 0 the function oscillates faster and faster between 1 and −1, so it does not approach any single value. Thus, there is no way that we could define f (0) in order to make the function continuous over the domain of real numbers. However, if the domain is restricted to the real numbers minus 0 this function is continuous. Example 2. Is x·sin(1/x) continuous over the real numbers? If not, does it have any discontinuities which can be redefined to make it continuous over the real numbers? Solution Since this function is undefined at x = 0, it is discontinuous over a domain of real numbers. However, as x → 0 despite the fact that the sine function oscillates faster and faster between 1 and −1, it is multiplied by x, so it approaches 0. Thus, if we define f (0) = 0, then it will be a continuous function over the domain of reals. Example 3. Is sin(x)/x continuous over the real numbers? If not, does it have any discontinuities which can be redefined to make it continuous over the real numbers? Solution Once again this function is undefined at x = 0, and thus is discontinuous there. However, we have previously shown that sin(x) = 1, x→0 x lim so if we define f (0) = 1, then we have a continuous function over the real numbers. x2 + x − 6 discontinuous? Are these discontinuities removable? x2 − 4 Solution We have a quotient of two continuous functions, so this rational function is continuous wherever we do not have division by 0. Since there is division by 0 when x = ±2, we have two discontinuities. Are either of them removable? In order to find out we need to look at the limits as x → 2 and x → −2. If we look as x → 2 we find Example 4. Where is x2 + x − 6 (x + 3)(x − 2) (x + 3) 5 = lim = lim = . 2 x→2 x→2 (x + 2)(x − 2) x→2 (x + 2) x −4 4 lim By redefining f (2) = 5/4, we remove this discontinuity. Looking as x → −2 we find x2 + x − 6 = ∞, x&−2 x2 − 4 lim so we find that this discontinuity is nonremovable (no matter how we redefine the function this limit will not exist).