Notes: 1. Regard as the maximal output error and as the corresponding maximal input error

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1 Limits and Continuity One of the major tasks in analysis is to classify a function by how nice it is Of course, nice often depends upon what you wish to do or model with the function Below is a list of some of the function classifications we will encounter in this course: 1 continuity 2 uniform continuity 3 Lipschitz 4 differentiable 5 monotonic & bounded variation 6 (Riemann-Stieltjes) integrable We begin with an investigation of continuous functions The mathematical meaning of continuous is consistent with the standard English usage of the word Continuous: Uninterrupted in time, sequence, substance, or extent (Mindscape Complete Reference Library CD-ROM) Definition 25 Suppose that a function f is defined on an open interval I containing the point a We say that the function f is continuous at a iff for each there exists a such that whenever Notes: 1 Regard as the maximal output error and as the corresponding maximal input error 2 depends on the output error, the function, and the point a; that is, In general, continuity is difficult to establish via Definition 25 KEJ 34

2 6 The above definition of continuity is essentially due to Bernard Bolzano ( ) and appeared in an essay he wrote in 1817 although it was largely overlooked at the time Augustin-Louis Cauchy ( ) is often given credit for the first reasonable characterization of the notion of continuity for his work in Negation of the above definition: A function is not continuous at if for some and every there exists such that and Example 13 1 Show that is continuous at x = 4 2 Show that is continuous at x = a KEJ 35

3 3 Show that is continuous at x = 3 4 Show that is continuous at x = 9 5 Show that is continuous at x = a (a 0) Example 14 Define Determine the points, if any, where f is continuous Answer: x = 0, 1 Illustration 9 (1) Find m so that is continuous at (2) Find b so that Answers: is continuous at KEJ 36

4 Example 15 Suppose that is defined as follows: where is a fraction in lowest terms Then The graph of over [0,1] is shown below: The function is continuous at each rational x and discontinuous at each rational x We prove this fact later Illustration 10 (1) Prove that is continuous at KEJ 37

5 Solution Let be given Choose Then for we have that since the function for (do the calculus to verify this critical fact!) We conclude that is continuous at (2) Prove that the rational function is continuous at Solution Let be given Choose Then for we have that KEJ 38

6 since for (again, do the calculus to verify this fact!) We conclude that is continuous for Illustration 11 (1) Explain why the function fails to be continuous at (2) Explain why the function is everywhere continuous including at (3) Explain why the function fails to be continuous at no matter what value we assign to KEJ 39

7 Illustration 12 Prove that is continuous at Solution (sketch) Let be given Choose Let x be such that We observe that for all such that (graph this difference quotient on the interval (18,22) to verify the stated bound!) Consider We conclude that is continuous at The next result gives a characterization of continuity in terms of sequences; it states that continuous functions carry convergent sequences to convergent sequences Theorem 26 - Sequential Criterion for Continuity Let f be defined on an open interval I = (b,c) containing the point a (that is, ) Then the following are equivalent: 1 f is continuous at a; KEJ 40

8 2 if I with, then Proof ( ) Suppose f is continuous at x = a Let be given and let I with Because f is continuous at x = a, there exists a so that whenever Because there exists N so that if, It follows that for we have Thus, ( ) In this case, we assume that condition 1 is false and we show that condition 2 is also false Because f is discontinuous at x = a, there exists at least one so that for all possible there exists x with and By choosing (n = 1, 2, 3, ) we construct a sequence I with and Thus, converges to a while does not converge to The previous result is probably most useful in showing that a particular function is not continuous at a particular point x = a Example 16 We define the Dirichlet function as follows Show that the Dirichlet function is nowhere continuous KEJ 41

9 Solution Let a (Case 1) Suppose that a Q Define a sequence as follows (Note that is rational when n is odd and is irrational when n is even) Then but oscillates between 0 and 1 and, hence, it fails to converge to (Case 2) Exercise Example 17 Suppose that the function f is continuous at each point in the interval with for every rational number r Prove that for all x in Solution Let x be any number in Find a sequence of rational numbers in such that Since f is continuous at x, Theorem 26 implies that We conclude We now extend the pointwise notion of continuity to a setwise formulation Definition 27 Suppose The function f is continuous at a with respect to D iff 1 a D; 2 for each there exists such that KEJ 42

10 whenever and A function is said to be continuous on D if it is continuous with respect to D at every point of D A special case of the above is when D is a closed bounded interval [a,b] In this case the definition automatically takes of care of the one-sided continuity at the endpoints (Explain why?) Example 18 1 Polynomial functions are continuous on 2 Rational functions are continuous on their domain of definition 3 Power functions such as are continuous on their domain of definition Example 19 Consider the function which consists of the straight line segments connecting the point with the two points and where and The graph of is shown below: The above is clearly continuous away from zero given that it is piecewise linear To see that it is continuous at zero we note that and so continuity there follows by choosing KEJ 43

11 Point: The function is continuous on [-1,1] yet there is no way to sketch the graph of without lifting the pencil from the page (The problem is to connect (0,0) to (1,1) or (-1,-1) to (0,0)) Definition 28 Suppose that f is defined on for some Then the limit of f(x) as x approaches a is L, written, iff for each there exists a so that whenever According to Definition 28 the function f(x) under consideration need not be defined at x = a Further, if it is defined at x = a, its behavior at x = a is of no consequence with respect to the limit process Example 20 Let a, c 1 Show that (Any ä works!) 2 Show that (Choose ä = å) 3 Show that (Choose ) Example 21 Show that if KEJ 44

12 Solution Let be given Choose We consider two cases: (i) x > 4 and (ii) x < 4 (i) Suppose x < 4 with Then Consider (ii) Suppose that x > 4 with Then Consider So, if, then and so Note that in the above the behavior of the function at x = 4 was of no interest to us and its value at x = 4 played no role in the solution of the problem KEJ 45

13 of limits Below we state as a theorem the standard Calculus I characterization of continuity in terms Theorem 29 Suppose that f is defined on an open interval I containing the point x Then f is continuous at x = a if and only if One useful consequence of Theorem 29 is that it is typically the case that for each theorem we prove concerning limits there is a corresponding result for continuity Example 22 (See Example 15) Define f : by a Explain why for all a b Use part a and Theorem 29 to conclude that f is continuous precisely on the irrational numbers Theorem 30 Suppose that f is defined on for some r > 0 If and, then L = M KEJ 46

14 Example 23 Suppose that g(x) is bounded on (some r > 0) If, then show that Solution Let å > 0 be given Suppose that for x in Since, there exists ä > 0 so that whenever Let x be a number such that Then We conclude that It is not required in the above that exist; the function g need only be defined and bounded on The next example is quite useful! KEJ 47

15 Example 24 1 If and f is continuous at L, show that Solution Let å > 0 be given Since f is continuous at L, (Theorem 29) So, there exists such that whenever Because, there exists ä > 0 so that whenever Therefore, for x with we have Hence, In view of the above, we say that a continuous function commutes with limits 2 Evaluate Solution From Calculus I we recall that Set Then f is continuous on and, in particular, f is continuous at 1 By part 1, we have KEJ 48

16 Theorem 31 - Algebra of Limits Suppose that and 1 ; 2 ; 3 provided M 0; 4 provided M 0 Proof 1 Let å > 0 be given Since, there exists a so that whenever Similarly, since, there exists a so that KEJ 49

17 whenever Choose For x such that we have Therefore, (The subtraction case is similar and is omitted) 2 Let å > 0 be given Since, there exists a so that whenever Thus, f is bounded on Because and, it follows that and Therefore, by Example 17, Hence, KEJ 50

18 3 To obtain the desired results set and apply Example 18 4 Follows from parts 2 and 3 By a standard induction argument, the results of Theorem 31 extend to finite sums and products Example 25 If and both exist, does it follow that exists? Solution No, consider and The next result is an immediate consequence of the Algebra of Limits Theorem and the definition of a continuous function Corollary 32 Suppose the functions f and g are continuous at x = a 1 is continuous at x = a 2 is continuous at x = a 3 is continuous at x = a provided 4 is continuous at x = a provided that We note that Example 16 and Corollary 32 now establishes Example 15 Combining Example 20 with Corollary 32 we have the following result on composite functions KEJ 51

19 Corollary 33 If g is continuous at a and f is continuous at g(a), then is continuous at x = a We now state and prove a result that goes by a variety of names including the Sandwiching Theorem, Pinching Theorem, and Squeeze Theorem Theorem 34 - Sandwiching Theorem Suppose that f, g, and h are all defined on the set 1 for all x S and 2, (some r > 0) If then Proof - Optional Let å > 0 be given Since there and so that whenever and whenever Choose Then for real numbers x with we have Thus, whenever KEJ 52

20 Hence, Example 26 Suppose that for x Show that Solution Since and, Because, the Sandwiching Theorem implies that We now consider one-sided limits Definition 35 Suppose that f is defined on (a-r,a) (some r > 0) Then the limit of f(x) as x approaches a from the left is L, written iff for each å > 0 there exists a ä > 0 so that whenever There is an analogous definition for right-hand limits The Algebra of Limits Theorem holds for both left-hand and right-hand limits Further, there are versions of the Sandwiching Theorem for both left-hand and right-hand limits The following notations are convenient: KEJ 53

21 provided the limits exist The next result characterizes two-sided limits in terms of one-sided limits The result is most often used to show that a particular function does not have a two-sided limit at a specific point Theorem 36 Suppose that f is defined on (some r > 0) Then iff Proof (Sketch) ( ) Given å > 0, the same value of ä works for all limits ( ) Let å > 0 be given Choose In view of Theorems 29 and 36, a function f is continuous at x = a iff Definition 37 1 Suppose that f is defined on (0, ) and we set Then iff 2 Suppose that f is defined on (-,0) and we set Then iff We say that the horizontal line y = L is an asymptote for the graph of iff either or KEJ 54

22 Example 27 1 (k ) since 2 Show that Solution Set Then Because for all x,, by the Sandwich Theorem for right-hand limits, it follows that The desired results concerning f is established In Example 25, we note that since for x > 0, the Sandwiching Theorem for limits at infinity also provides the desired result Theorem 38 Suppose that f is continuous on the interval If and have opposite signs, then there exists such that Proof WLOG, assume Define KEJ 55

23 Since, and S is bounded above by b By the Least Upper Bound Property of, S has a least upper bound, say By Theorem 11, for each there exists such that Clearly, Because f is continuous at c, we have that Because each, it follows that Given that, we see that Construct a sequence where By the continuity of f at c, Because and, we conclude that and so It must be the case that Example 28 1 Show that has a root / zero / x-intercept on 2 ` Show that has a root / zero / x-intercept on Theorem 39 - Intermediate Value Theorem Suppose f is continuous on the interval If k is any number between and, then there exists a such that Proof Apply Theorem 38 to the function The next two results follow from the Bolzano-Weierstrass Theorem Theorem 40 Suppose f is continuous on the interval Then f is bounded there KEJ 56

24 Theorem 41 - Extreme-Value Theorem Suppose f is continuous on the interval Then there are points such that and KEJ 57