MATH1013 Calculus I. Limits (part I) 1

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1 MATH1013 Calculus I Limits (part I) 1 Edmund Y. M. Chiang Department of Mathematics Hong Kong University of Science & Technology March 5, Based on Briggs, Cochran and Gillett: Calculus for Scientists and Engineers: Early Transcendentals, Pearson

2 Instantaneous Velocities Newton s paradox Limits Properties of Limits

3 The Dawn The need to investigate dynamical problems in the 18th century verses static problems in the past was strongly related to the cultural and economics developments at that time. It was Galilei Galileo ( ), called the father of sciences who headed the Scientific Revolution in the 17th century advocating beliefs should be built upon experiments and mathematics and that Philosophy is written in this grand book, the universe... It is written in the language of mathematics, and its characters are triangles, circles, and other geometric figures;... (Wiki) He showed that the velocity of a falling body only depends on its mass and has nothing to do with its shape and size He invented telescope and use it to discover the four largest satellites of the planet Jupiter, etc

4 Galilei Galileo ( ) Figure: (Portrait in crayon by Leoni (source Wiki))

5 Galilei Galileo ( ) Italian physicist Advocates heliocentricism instead of geocentrism He was investigated by the Roman Inquisition in 1615, that it could be supported as a possibility, but not as an established fact. Dialogue Concerning the Two Chief World Systems. It s content is about discussions amongst two philosophers and a layman. Under house arrest (1633). Wrote Two New Sciences where he show how mathematics should be used to deal with kinematics and strength of materials. Einstein (1954). Propositions arrived at by purely logical means are completely empty as regards reality. Because Galileo realised this, and particularly because he drummed it into the scientific world, he is the father of modern physics indeed, of modern science altogether.

6 Galileo s philosophy (a) Figure: (Kline Vol. I, p.330))

7 Galileo s philosophy (b) Figure: (Kline Vol. I, p.330))

8 Galileo s philosophy (c) Figure: (M. Kline, Mathematical Thoughts from Ancient to Modern Times, Vol. I, Oxford Univ. Press p.329))

9 Newton s role As M. Kline puts (1972) Great advances in mathematics and science are almost always built on the work of many men who contribute bit by bit over hundreds of years; eventually one man sharp enough to distinguish the valuable ideas of his predecessors from the welter (muddle) of suggestions and pronouncements, imaginative enough to fit the bits into a new account, and audacious (brave) enough to build a master plan takes the culminating and definitive step. In the case of Calculus, this was Newton. In fact, G. W. Leibniz ( ) also invented Calculus. Our calculus notation is also due to Leibniz. Newton s success is built upon Galileo s philosophy and on Kepler s experimental laws.

Sir Issac Newton (1642-1727) Figure: (1689

10 Sir Issac Newton ( ) Figure: (1689 by Sir Godfrey Kneller (Newton Institute))

11 G. W. Leibniz ( ) Figure: Gottfried Wilhelm Leibniz (Wikipedia)

12 Four major types of problems Given the formula of for the distance of an object as a function of time, to find the speed and acceleration of the object, and the converse problem, given the acceleration of the object, to recover the speed and distance. Finding tangents to a curve (e.g.reflections through a lens, direction of a moving body). In fact, even the very definition of a tangent is in question. Finding maximum and minimum values of a function. Finding lengths of curves, areas bounded by curves and volumes bounded by surfaces, gravitational attraction between planets. The Greeks use methods of exhaustion on relatively simple geometric areas and volumes. Their methods work on ad hoc base (with much ingenuity) that lacks generality.

13 Continuity against discreteness Even when a body with distance function S(t) is moving with non-uniform motion between two time S(t 1 ) and S(t 2 ) (t 2 > t 1 ), one can measure average velocity (the common usage speed has no direction): average velocity = distance travelled time taken = S(t 2) S(t 1 ) t 2 t 1 which means distance travelled per unit length. When a body starts from rest to reach certain velocity v 0 later, then its (average) velocity must change at different periods. How to describe an object at every instant? In fact, the very meaning of instantaneous velocity is in question. In fact, one should recognise that mathematics, like any other language, is a language that we use to represent our thoughts (understanding) about the nature. But mathematics (our thoughts) is not the nature itself so there is a limit to this language.

14 Newton s infinitesimals 1. For convenience sake, we can say that Newton invented virtual distance and virtual time to measure virtual velocity. That is, virtual velocity = virtual distance virtual time or just instantaneous velocity which is not a clearly defined term (in any sense). Newton calls these infinitesimal. 2. Let S(t) be the distance function and let dt represents an infinitesimal change in t. So ds(t) = S(t + dt) S(t) is an infinitesimal change in S(t) over the time interval dt 3. So 4. Instantaneous velocity = ds dt = S(t + dt) S(t) dt is essentially what Newton interpreted as the instantaneous velocity although the notation ds, dt was due to Leibniz.

15 So do we have dt = 0? If so, then one would have ds dt = 0 0. That was the question that Newton could not answer satisfactorily during his life time. Newton s trouble Suppose an object moves according to the rule S(t) = t 2 where S measures the distance of the object from the initial position t seconds later. We now compute instantaneous velocity of the object at time t: let dt and ds be the virtual time and virtual distance respectively. Then the change of virtual distance is given by ds = S(t + dt) S(t). So the virtual velocity is ds dt = S(t + dt) S(t) dt = 4(t + dt)2 4t 2 dt Newton then delete the last dt: ds dt = 8 t + 4 /// dt = 8 t. = 8 t + 4 dt.

16 Exercises Find, with Newton s technique to calculate the instantaneous velocity of a body moving according to the distance function s(t) = 16t t, when t = 1, 2, 3 s(t) = t t, when t = 1, 2, 3 We do not need to restrict ourselves to problems only related to distance and time. After all, velocity is rate of change, so that instantaneous velocity is a special case of instantaneous rate of change. Find the (instantaneous) rate of change of f (x) = 1 x Find the (instantaneous) rate of change of f (x) = ax 2 + bx + c where a, b, c are fixed constants.

17 Newton s thought So he simply considers that is a virtual distance ds traveled by the object in a virtual time dt. He considers both to be infinitesimal small quantities. So do we have dt = 0? If so, then one would have ds dt = 0 0. That was the question that Newton could not answer satisfactorily during his life time. To put the question differently, is an infinitesimal quantity equal to zero? If dt is infinitely small then it would have to be less than any positive quantity, and we conclude it must be equal to zero. For suppose dt 0 then dt > 0. Hence dt = r > 0 is an actual positive quantity. But then we could find r/2 < dt, contradicting the fact that dt is smaller then any positive quantity. Hence dt = 0. Newton was actually attacked by many people, and among them was the Bishop Berkeley. But he method of calculation of instantaneous velocity has been used by other since then.

18 Newton s Principles of Natural Philosophy (1729) Figure: (M. Kline, Mathematical Thoughts from Ancient to Modern Times, vol. 1)

19 A dynamical problem (p. 54) A rock is launched vertically upward from the ground with a speed of 96 ft/s. Neglecting air resistance, a well-known formula from physics states that the position of the rock after t seconds is given by s(t) = 16t t. The position s is measured in feet with s = 0 corresponding to the ground. Find the average velocity of the rock between 1. t = 1 and t = 3, 2. t = 1 and t = 2.

20 A dynamical problem (figure 2.1) Figure: 2.1 Publisher

21 Figure: 2.2a Publisher A dynamical problem (figure 2.2a)

22 A dynamical problem (figure 2.2b) Figure: 2.2b Publisher

23 A dynamical problem (figure 2.3) Figure: 2.3 Publisher

24 A dynamical problem (table 2.1) Figure: 2.1 Publisher

25 A dynamical problem (figure 2.4) Figure: 2.4 Publisher

26 A dynamical problem (figure 2.5) Figure: 2.5 Publisher

27 Revisit: S(t) = t 2 Let s get close but not having dt = 0. Find the average velocity of the object between t = 2 and t = 2.5 ( S(2.5) S(2) ) (2.5) 2 ( (2) 2) = = 18; t = 2 and t = 2.1 ( S(2.1) S(2) ) (2.1) 2 ( (2) 2) = = t = 2 and t = 2.01 ( S(2.01) S(2) ) (2.01) 2 ( (2) 2) = = t = 2 and t = ( S(2.001) S(2) ) (2.001) 2 ( (2) 2) = = t = 2 and t = S(2.0001) S(2) ( ) (2.0001) 2 ( (2) 2) = =

28 Re-assessing the problem Let us begin with the above example about the movement of the object P. Since we are interested to know the magnitude of the average velocity of P near 2, so let us rewrite the expression in the following form: g(x) = S(2 + x) S(2). x This is a function g depends on the variable x, which can be made as close to 16 as we wish by choosing t close to 2. That is, g(x) approaches the value 16 as x approaches 0. On the other hand, we cannot put x = 0 in the function g(x), since both the numerator S(2 + x) S(x) and the denominator x would be zero. We say that the function g has limit equal to 16 as x approaches 0 abbreviated as lim g(x) = 16. x 0

29 Limit definition Note that the above statement is merely an abbreviation for the statement: The function g can get as close to 16 as possible if we let x approach 0 as close as we wish. It is important to note that we are not allowed to put x = 0 above Definition Let a and l be two real numbers. If the value of the function f (x) approaches l as close as we wish as x approaches a, then we say the limit of f is equal to l as x tends to a. The statement is denoted by lim f (x) = l. x a Alternatively, we may also write f (x) l as x a.

30 Examples x 3 8 Find lim x 2 x 2. Note that we can not substitute x = 2 in the expression. For then both the numerator and denominator will be zero. Consider x 3 8 lim x 2 x 2 = lim (x 2)(x 2 + 2x + 4) x 2 x 2 = lim (x 2 + 2x + 4) = 12. x 2 The above is an abbreviation of the expression: x 3 8 x 2 = (x 2)(x 2 + 2x + 4) = x 2 + 2x + 4 x 2 tends to the value 12 as x tends to 2. or more briefly x 3 8 x 2 = x 2 + 2x , as x 2.

31 Illegal step If f (x) = x 3 8 x 2, then it is absolutely forbidden to write lim f (x) = x 3 8 x 2 x 2 = f (2) since the function f is simply undefined at x = 2.

32 A discontinuity Figure: 2.7 (Publisher)

33 Continuity point Figure: 2.8 (Publisher)

34 Towards discontinuity Figure: 2.9 (Publisher)

35 Towards continuity Figure: 2.10 (Publisher)

36 Examples x 3/2 8 Exercises Find lim x 4 x 1/2 2 x 2 4 lim x 2 lim x 4 (12) x 2, (4) x 2 4 x, ( 1/4) (2 + h) 4 16 lim, (32) h 0 h The above examples could be misleading. There could be situations that no easy simplification when finding limit as in the above examples. We will show in the next chapter that e x 1 x 1, x 0.

37 Artificial examples Remark We remark that the above definition does not mention whether we could substitute x = a in f (x). In fact, f (a) may or may not be meaningful. This is slightly different from the physical problem about the object P where we were not allowed to put x = 0 in g(x). The following examples do not have the kind of physical context about having 0/0 problem that we encountered earlier. They are simply created to illustrate what one should interpret the limit definition properly, even though they seems to be trivial: Example Let f (x) = 4x Then 1. lim x 1 (4x ) = 24, 2. lim x 1 (4x ) = 24, 3. lim x 3 (4x ) = 56.

38 Simple exercises Example Let f (x) = x Then 1. lim x 1 x = = 4 = 2; 2. lim x 1 x = ( 1) = 4 = 2. Let g(x) = 1 x 2. Then 1 lim x 3 x 2 = = 1; 1 lim x 1 x 2 = = 1/3, Exercises lim x 2 5 = (5) lim x 1(x 3 1) = (0) lim x 1(x 3 1) = ( 2) lim x 1(ax 3 1) = ( a 1) 4x+2 lim x 0 = (1) 2 ( lim 1 x 1 + ) 1 x x+1 = (3/2) lim x 2(2 + x) 5 1 = (4 5 1) lim x 3(x ( 2 3x + 2) = (2) lim 1 ) x 1 2x 5 = ( 1/7)

39 Example Consider More examples f (x) = { 2 if x 1, 1 if x = 1. We see that lim x a f (x) = 2 whenever a 1. This is different from the value of f at 1. So lim x 1 f (x) = 2 1 = f (1). Example Consider the function x + 1 if x 1, 2; f (x) = 3 if x = 1; 1 if x = 2. Thus x = a and other than a = 1, 2, then f (x) approaches the value a + 1 as x approaches a. In fact f (a) = a + 1. Although when a = 1, 2, we still have lim x a f (x) = a + 1, it is not equal to the values of f (1) = 3 and f (2) = 1. Thus there are two jumps on the graph of f.

40 More examples Example Suppose x 3 if x < 2; f (x) = 1 x 1 if x 2. For any a < 2, f (x) approaches a 2 3 as x approaches a, and for any b > 2, f (x) approaches 1/(b 1) as x approaches b. When x = 2, x 2 3 approaches 1 as x approaches 2 on the left, and 1/(x 1) approaches 1 as x approaches 2 on the right. Hence we conclude that f approaches 1 as x approaches 2, i.e., The limit lim x 2 f (x) = 1 exists.

41 Right limit Let a and l be two real numbers. If the values of the function f (x) approaches l as close as we wish as x approaches a from the right then we say the right limit of f is equal to l as x tends to a from above. The statement is denoted by We may also write lim f (x) = l. x a+ f (x) l as x a +. We have a similar definition for left limit, denoted by lim x a f (x) = l or f (x) l as x a. We note again that both definitions do not say anything about f at the point x = a.

42 Left and Right limits It is not difficult to see that lim x a f (x) = l exists if and only if both lim f (x) = l = lim f (x). x a+ x a The previous example shown three slides before clearly illustrates this statement Example Show x has limit at all points on the real line. Example (p. 68) Let f (x) = x x, x 0. Does lim x a f (x) exist, where a = 0 or a 0? Sketch a graph of f (x).

43 Let f (x) = Example { 1 if x < 1 x + 1 if x 1. Since f remains at 1 for all x < 1, f approaches 1 when x tends to 1 on the left. So lim f (x) = 1. x 1 Note that f (1) = 2 lim x 1 f (x). On the other hand, lim f (x) = lim x + 1 = 2. x 1+ x 1+ And we have f (1) = 2 = lim x 1+ f (x).

44 Exercises Let 3x 1, if x < 0; f (x) = 0, if x = 0; 2x + 5, if x > 0. Evaluate 1. lim x 2 f (x), 2. lim x 3 f (x), 3. lim x 0+ f (x), 4. lim x 0 f (x), 5. lim x 0 f (x). 6. ( Answers (1) 9, (2) 10, (3) 5, (4) 1, (5) does not exist )

45 An example that has no limit { 2 if x 1, Recall that the earlier example f (x) = has no 1 if x = 1. limit at x = 1 which is a discontinuity of f. But we could still correct f to be continuous at x = 1 again by re-defining f (1) = 2. Consider the example on page 64: f (x) = cos 1 x on the interval (0, 1]. It is not defined at x = 0. We see that even a small change in x near zero would result in a large change of 1 x.so there would be an unlimited number of oscillations between the values {±1} throughout (0, a]. Hence no correction of value of f (x) would make f (x) continuous at x = 0 again.

46 Figure 2.14 Figure: 2.14 (Publisher)

47 How to avoid the infinitesimal? Here is the real difficulty: Our thinking process and/or language usage generally does not allow us to describe infinitesimal quantities clearly Mathematicians have found a way to get around describing infinitesimal directly. We say that the function can get as close to a number (limit l) as possible. But we need to pay a heavy price if we want to do so precisely. Here it is. The abbreviation lim x a f (x) = l really means: Given an arbitrary ε > 0, one can find a δ > 0 such that f (x) l < ε, whenever 0 < x a < δ. Both ε and δ represent positive real numbers. Given each/any ε > 0 one can (always) find a δ > 0 such that... holds we refer to this kind of statement as ε δ language interpretation.

48 A linear function example How do we use δ ε to describe lim x 3 = 5? Figure: 2.56 (Publisher)

49 ε = 1 How do we use δ ε to describe lim x 3 = 5? Figure: 2.57a (Publisher)

50 Figure: 2.57b (Publisher) δ = 2 The corresponding δ = 2. That is, 0 < x 3 < 2 guarantees f (x) 5 < 1.

51 ε = 1/2 Figure: 2.58a (Publisher)

52 Figure: 2.58b (Publisher) δ = 1 The corresponding δ = 2. That is, 0 < x 3 < 1 guarantees f (x) 5 < 1/2.

53 ε = 1/8, δ = 1/4 Figure: 2.59 (Publisher)

54 General ε δ That is, 0 < x 3 < δ guarantees f (x) 5 < ε. Figure: 2.60 (Publisher)

Example p. 124 f (x) = x 3 6x 2 + 12x 5.

55 Example p. 124 f (x) = x 3 6x x 5. In order to show lim x 2 f (x) = 3, given ε = 1, find the corresponding δ. Figure: 2.61 (Publisher)

56 Example p. 115 (cont.) f (x) = x 3 6x x 5. In order to show lim x 2 f (x) = 3, given ε = 1, find the corresponding δ. Figure: 2.62 (Publisher)

57 ε δ definition example Let f (x) = 2x. Show lim x 0 2x = 0 by the ε δ argument of limit. Given ε > 0, we want to find a δ > 0 such that 2x 0 < ε, whenever 0 < x 0 < δ. Notice that 2x 0 = 2x = 2 x. So if we impose that 0 < δ < ε/2 and that x < δ < ε/2. Hence under this restriction of δ and x, we have 2x 0 = 2x = 2 x < 2 δ < 2 ε 2 = ε Thus given the ε > 0, we have found δ > 0 (namely δ < ε/2). Since this argument works for every ε > 0. We conclude that lim x 0 2x = 0.

58 ε δ definition example Let f (x) = 3x + 1. Show lim x 1 3x + 1 = 4 by the ε δ argument of limit. Given ε > 0, we want to find a δ > 0 such that (3x + 1) 4 < ε, whenever 0 < x 1 < δ. Notice that (3x + 1) 4 = 3x 3 = 3 x 1. So if we impose that 0 < δ < ε/3 and that x 1 < δ < ε/3. Hence under this restriction of δ and x, we have (3x + 1) 4 = 3x 3 = 3 x 1 < 3 δ < 3 ε 3 = ε Thus given the ε > 0, we have found δ > 0 (namely δ < ε/3). Since this argument works for every ε > 0. We conclude that lim x 1 3x + 1 = 4.

59 ε δ definition example Let f (x) = x 2. Show lim x 2 x 2 = 4 by the ε δ argument of limit. Given ε > 0, we want to find a δ > 0 such that x 2 4 < ε, whenever 0 < x 2 < δ. Notice that x 2 4 = (x 2)(x + 2). We can control the factor x 2, but the other factor x + 2 depends on x which is unlike those of previous examples. Since we are close to 2 anyway, so WLOG, we may impose x 2 < 1. So x 2 x 2 < 1. So x < 3 and x + 2 x + 3 < 5. We impose x < 3 and x 2 < δ < ε/5, and whichever is smaller. i.e., δ < min(1, ε 5 ). Then we have x 2 4 = (x 2) (x + 2) < 5 x 2 < 5 δ < 5 ε 5 = ε Thus given any ε > 0, we have found a δ > 0. We conclude that lim x 2 x 2 = 4.

60 ε δ limit exercises Employ ε δ arguments to prove the following limits: lim x 1 2x 1 = 1; lim x 1 2x 1 = 3; lim x 1 ax + b = a + b; lim x 1 x 2 = 1; lim x 1 x 2 = 1; lim x 1 1 x = 1; lim x 1 1 x 2 = 1. lim x a [f (x) + g(x)] = lim x a f (x) + lim x a g(x)

61 Limit laws Suppose lim x a f (x) = l, lim x a g(x) = m both exist. Let c be a constant, then the following hold: ( ) lim f (x) + g(x) = lim f (x) + lim g(x) = l + m x a x a x a ( ) lim c f (x) = c lim f (x) = cl x a x a ( ) lim f (x)g(x) = lim f (x) lim g(x) = lm x a x a x a f (x) lim x a g(x) = lim x a f (x) lim x a g(x) = l m provided m 0.

62 One-sided Limit laws These properties of limit have counterparts in the left and right limits formulations. Since the formulations are exactly the same as the above results except that the number a is replaced by either a or a+, so we omit the details here.

63 The real difficulty again So for lim x a [f (x) + g(x)] = lim x a f (x) + lim x a g(x), one needs to show, assuming that lim x a f (x) = l and lim x a g(x) = s Given an arbitrary ε > 0, one can find a δ > 0 such that [f (x) + g(x)] (l + s) < ε, whenever 0 < x a < δ. with the given assumption. This is slightly not easy. Some other laws are more difficult to verify using this language. So this explains why one needs to state these seemingly simple laws as separate entities.

64 Examples (p. 71) Given that lim x 2 f (x) = 4, lim x 2 g(x) = 5, lim x 2 h(x) = 8. lim x 2 [6f (x)g(x) + h(x)] = 6 lim [f (x)g(x)] + lim h(x)] x 2 x 2 = 6 lim f (x) lim g(x) + lim h(x) x 2 x 2 x 2 = 6 (4 5) + 8 = 128. f (x) g(x) lim = lim x 2[f (x) g(x)] x 2 h(x) lim x 2 h(x) = lim x 2 f (x) lim x 2 g(x) lim x 2 h(x) = 4 5 =

65 Examples By the above properties, lim (4x ) = lim 4x 2 + lim 20 x 2 x 2 x 2 = 4 lim x 2 + lim x 2 = 4(4) + 20 = 36. x 2 20 We note that since both lim x 2 x 2 and lim x 2 20 exist, so we can apply the above properties.

66 Examples 3x 2 1 lim x 3 1 6x Applying the above properties give 3x 2 1 lim x 3 1 6x = lim x 3(3x 2 1) lim x 3 (1 6x) = We again note both lim x 3 (3x 2 1) and lim x 3 (1 6x) exist. Hence we can apply the above result. lim x 3 (x 1) 2 (x + 1) So lim (x x 3 1)2 (x + 1) = lim (x 1) 2 lim (x + 1) x 3 x 3 = (3 1) 2 (3 + 1) = 16 We could apply some of the above limit laws, this is because that both lim x 3 (x 1) 2 and lim x 3 (x + 1) exist.

67 Examples (p. 72) lim x 2 2x x 1 4x + 1 = lim x 2( 2x x 1) lim x 2 4x + 1 limx 2 (2x = 3 + 9) + lim x 2 (3x 1) lim x 2 4x = (3 2 1) = =

68 Squeezed limits (p. 76) Theorem Assume that a functions f, g, f satisfy f (x) g(x) h(x) for all x near a, except possibly at a. If lim x a f (x) = L = lim x a h(x), then lim x a g(x) = L. (p. 76) E.g. It is clear from the graph that x sin x x, 0 1 cos x x hold on [ π/2, π/2]. Since lim x 0 x = 0, so the Squeeze theorem implies that lim x 0 sin x = 0. Similarly, lim x 0 cos x = 1. (p. 77) E.g. Show lim x 0 x 2 sin 1 x = 0.

MATH1014 Calculus II. A historical review on Calculus

MATH1014 Calculus II. A historical review on Calculus MATH1014 Calculus II A historical review on Calculus Edmund Y. M. Chiang Department of Mathematics Hong Kong University of Science & Technology September 4, 2015 Instantaneous Velocities Newton s paradox