Limits and Continuity

Chapter 1 Limits and Continuity 1.1 Introduction 1.1.1 What is Calculus? The origins of calculus can be traced back to ancient Greece. The ancient Greeks raised many questions about tangents, motion, area, the in nitely small, the in nitely large. The Greeks provided a few answers to their questions. But most of them remained unanswered until the invention of modern calculus. After the Greeks, progress was very slow. Algebra, founded by Arab scholars in the ninth century was not fully systemized until the sixteenth century. In the seventeenth century, Descartes established analytic geometry. With algebra and analytic geometry, the stage was set for calculus to evolve. The actual invention of modern calculus is credited to the Englishman Sir Isaac Newton (1642-1727) and the German mathematician Gottfried Wilhelm Leibniz (1646-1716). Newton started his work in 1665, Leibniz in 1673. Within 100 years, calculus reached pretty much the state under which it is known today, though some theoretical subtleties were not fully resolved until the twentieth century. To a Roman during the Roman Empire, a "calculus" was a pebble used in counting or gambling. The word then evolved to "calculare", which meant "to compute", to "reckon", "to gure out". To a mathematician or scientist of today, calculus is elementary mathematics (algebra, geometry, trigonometry), enhanced by the limit process. Calculus is also the mathematics of motion and change. In other subjects related to mathematics such as algebra, arithmetic, geometry, you have learned primarily to calculate with numbers and variables, to simplify algebraic expressions, to deal with points, lines and gures in the plane. But in each case, the quantities you dealt with were static. Calculus deals with quantities which change, quantities which approach other quantities. A technique often used in calculus when solving a di cult problem is to look at a simpler problem; one we know how to solve. We then change the simpler problem little by little. The 1

2 CHAPTER 1. LIMITS AND CONTINUITY changes satisfy two criteria. The rst is that as we apply the changes, the simpler problem looks more and more like the original, more di cult problem. The second criterion is that at each step, the changes should be such that we know how to solve the newly obtained problem. This technique illustrates the limit process mentioned above. We now look at some problems arising in Calculus which illustrate this technique. 1.1.2 Area of a Circle It is easy to nd the area of a circular region of radius r because we know the formula. It is r 2. How did people do it before the formula was known? One approach was to use inscribed regular polygons. For example, we could approximate the area of a circle by the area of the inscribed equilateral triangle as shown in gure 1.1. Let A 3 denote the area of the triangle. We have replaced the problem of nding the area of a circle by nding the area of a triangle. Which is much easier to do. Of course, you should argue that the two areas are not the same. So, we see that the simpler problem is indeed easier to solve, however, it does not give us the correct answer. We don t want to give up that quickly though. Let us increase the number of sides of the polygon inscribed in the circle. In general, let A n denote the area of the regular polygon with n sides inscribed in the circle. As gures 1.2, 1.3, 1.4 and 1.5 suggest, the larger n is, the closer A n is to the area of the circle.. We say that the area A of the circle is the limit of the areas of the inscribed polygons as n approaches in nity, and we write A = lim n!1 A n Of course, there is a slight problem. Even if we can derive a formula in terms of n for A n, how do we compute lim A n? Since 1 is not a number, we cannot n!1 plug it in the formula. Figure 1.1: Inscribed Triangle

1.1. INTRODUCTION 3 Figure 1.2: Inscribed square Figure 1.3: Inscribed Pentagon Figure 1.4: Inscribed Hexagon

4 CHAPTER 1. LIMITS AND CONTINUITY Figure 1.5: Inscribed dodecagon 1.1.3 The Area Problem The problem of nding the area of a circle can be generalized to that of nding the area of a region bounded by a curve. Consider the problem of nding the area S between the x-axis and the graph of y = f (x), between the vertical lines x = a and x = b. Such a region is shown in Figure 1.6. This region is too complex; there is not an easy formula which gives its area. We use the following technique: We subdivide the interval [a; b] into n subinterval of equal length, and draw rectangles as shown in Figures 1.7, 1.8 and 1.9. Let S n denote the sum of the areas of each rectangle. We approximate S by S n. Obviously, for small n, like the case n = 2 shown in Figure 1.7, our approximation is not very good. But, as n gets larger, S n is closer and closer to S. Once again, using the notation of the previous example, we have S = lim n!1 S n Of course, we have the same problem. Even if we can derive a formula in terms of n for S n, how do we compute lim S n? Since 1 is not a number, we cannot n!1 plug it in the formula. This problem is related to the de nite integral and will be studied when we study integration. You can experiment with the technique described here using the applet at http : ==science:kennesaw:edu=~plaval=applets=riemann:html Details of this procedure will be done in chapter 4. 1.1.4 The Tangent Problem This is a classical problem in Calculus. We are going to look at it in greater detail. Before we start, let us review some essential formulas. Proposition 1 The slope m of the line through two points of coordinates (x 1 ; y 1 ) and (x 2 ; y 2 ) is given by m = y 2 y 1 x 2 x 1 (1.1)

1.1. INTRODUCTION 5 Figure 1.6: Area of a Region Figure 1.7: Two Rectangles

6 CHAPTER 1. LIMITS AND CONTINUITY Figure 1.8: Four Rectangles Figure 1.9: Eight Rectangles

1.1. INTRODUCTION 7 Proposition 2 The equation of a line through a point of coordinates (x 1 ; y 1 ) with slope m is y y 1 = m (x x 1 ) (1.2) If instead of being given the slope and a point we are given two points on the line, we can nd its equation by rst using formula 1.1 to derive the slope of the line. Then, using the slope just derived, one of the given points and formula 1.2, we can nd the equation of the line. Example 3 Find the equation of the line through the points (1; 1) and (2; 1). The slope of this line is The equation of the line is m = 1 ( 1) 2 1 = 2 1 = 2 y ( 1) = 2 (x 1) y + 1 = 2x 2 y = 2x 3 We used the point (1; 1) to derive the equation. We could have also used the second point, we would have obtained the same result. Now, we can look at the tangent problem. Suppose we are trying to nd the equation of the tangent line to the graph of y = f (x) at a point P on the graph. Let (a; b) be the coordinates of P (since P is on the graph, we have b = f (a)). For now, think of the tangent as the line which touches the graph at P. Such a line is shown in Figure 1.10.The tangent line is a line. To nd the equation of a line, we either need two points on the line, or the slope of the line and a point. Unfortunately, we do not have all the necessary information. We are only given one point. Instead, we consider an easier problem, one we can solve. The problem of nding the equation of a line given two points. We use the following approach: 1. Pick a second point Q on the graph, let (x; y) be its coordinates (note that since the point is on the graph, we have y = f (x)). 2. Find the slope of the secant line through P and Q, call it m P Q. Once we have the slope, we can nd the equation of that secant line. This secant line is shown in Figures 1.11 and 1.12. Of course, the secant line is not the same as the tangent line. From the equations above, we see that m P Q = f (x) x f (a) a (1.3)

8 CHAPTER 1. LIMITS AND CONTINUITY Figure 1.10: Tangent Line at P 3. Observe that by taking Q closer and closer to P, the secant line is getting closer and closer to the tangent line. In other words, the slope m of the tangent line through P can be though of as the limit of the slopes m P Q of the secant lines through P and Q, as Q approaches P. Using the same notation as above, we have m = lim Q!P m P Q To make Q closer to P is the same as making x closer to a. Thus, m = lim Q!P m P Q (1.4) = lim x!a m P Q (1.5) f (x) = lim x!a x f (a) a (1.6) Here again, we can see that this computation poses a problem because as x! a, the denominator of this fraction approaches 0, which is not allowed. This procedure is also illustrated by a Java applet which can be found at http : ==science:kennesaw:edu=~plaval=tools=dierentiation:html In fact, the graphs shown in Figures 1.10, 1.11 and 1.12 come from this applet. The above procedure allows us to express the slope of the tangent. Once we know how to compute limits, we will be able to nd the slope of the tangent.

1.1. INTRODUCTION 9 Figure 1.11: Secant Line Through P and Q Figure 1.12: Secant Line Through P and Q

10 CHAPTER 1. LIMITS AND CONTINUITY Example 4 The purpose of this example is to illustrate the procedure outlined above. Consider the function f (x) = x 2 + 5. Suppose we are trying to nd the equation of the tangent at x = 3. To do so, we approximate it by nding the equation of the secant line through x = 3 and another point we will call x = a. Then, we let a get closer and closer to 3. This amounts to using values of a which are closer and closer to 3. Answer the questions below: 1. Find the equation of the secant line through the points corresponding to x = 3 and x = 4. Note that the two points in questions are P = (3; f (3)) and Q = (4; f (4)) that is (3; 14) and (4; 21). Using formula 1.3, we see that the slope of the secant line is f (4) f (3) m P Q = 4 3 21 14 = 1 = 7 Thus, the equation of the secant line (using the point (3; 14) is y 14 = 7 (x 3) y 14 = 7x 21 or y = 7x 7 2. Find the slope of the secant line through the points corresponding to x = 3and x = a, for a = 4, 3:5, 3:1, 3:01, 3:001. We proceed as above. The slope of the secant line is f (a) f (3) m P Q = a 3 = a2 + 5 14 a 3 = a2 9 a 3 (a 3) (a + 3) = a 3 = a + 3 as long as a 6= 3 The table below gives us the slope of the secant for the required values of a. a 4 3:5 3:1 3:01 3:01 m P Q 7 6:5 6:1 6:01 6:001

1.1. INTRODUCTION 11 3. Write a formula for the slope of the tangent to y = f (x) at x = 3. If we call m the slope of the tangent, using formula 1.6, we have 1.1.5 The Velocity Problem Velocity is de ned to be m = f (x) f (3) lim x!3 x 3 = x 2 + 5 14 lim x!3 x 3 = x 2 9 lim x!3 x 3 velocity = distance time For example if you drive 100 miles in 2 hours, then your velocity was 50 miles/hour. This is known as the average velocity, that is the velocity over a fairly large time interval. It does not mean your velocity was 50 miles/hour during the whole trip. Indeed, if you looked at the speedometer of your car, you probably notice that it was changing all the time. This is because the speedometer measures the velocity at every instant. It measures what we call the instantaneous velocity. We illustrate these two notions. Suppose that the position of an object is given by a function s (t). For example, if t is in seconds and s (t) in meters, then s (2) represents the distance in meters traveled by the object after 2 seconds, s (5) represents the distance traveled after 5 seconds, and so on. Therefore, s (b) s (a) represents the distance traveled between t = a and t = b. So, we have De nition 5 (Average Velocity) If the position of an object is given by the function s (t) then the average velocity of the object between t = a and t = b is given by average velocity = s (b) s (a) (1.7) b a Remark 6 Geometrically, the average velocity represents the slope of the secant to y = s (t) through the points corresponding to t = a and t = b. So, we see that the velocity problem is identical to the secant problem. To nd the instantaneous velocity, we proceed the same way we did for the tangent problem. Since an instantaneous velocity is an average velocity over a very small interval, we can approximate the instantaneous velocity at t = a by computing the average velocity between a and t, then we let t approach a, thus making the time interval smaller and smaller. In other words, we have the following: De nition 7 (Instantaneous Velocity) If the position of an object is given by the function s (t) then the instantaneous velocity of the object at time t = ais

12 CHAPTER 1. LIMITS AND CONTINUITY given by s (t) v = lim t!a t s (a) a (1.8) Remark 8 Geometrically, the instantaneous velocity at t = a is the slope of the tangent to the graph of y = s (t) at t = a. We illustrate this with an example. Example 9 When an object fall, the distance (in meters) it travels is given by s (t) = 4:9t 2 where t is in seconds. 1. Find the average velocity of the object between t = 1 and t = 2. From formula 1.7, we have s (2) s (1) average velocity = 2 1 19:6 4:9 = 1 = 14:7 m=s 2. Express the instantaneous velocity of the object at t = 1 as a formula. From formula 1.8, we have 1.1.6 Conclusion s (t) s (1) v = lim t!1 t 1 4:9t 2 4:9 = lim t!1 t 1 4:9 t 2 1 = lim t!1 t 1 When doing Calculus, like in many other disciplines, it is important to know the details. It is also important to know the general idea. We only presented a few problems here in which the same idea, the idea of nding a quantity as the limit of other quantities, is used. This idea is central to Calculus. We will look at all the examples presented above in more details. However, before we do this, we must explain in more details this idea of limits. 1.1.7 A Note on Functions It is assumed that the reader is already familiar with the notion of functions. The functions encountered in this class fall in two categories. They are either an elementary function or a combination of elementary functions. Given a function, understanding its pattern, that is how it is made, is essential to understanding the properties of this function and to knowing how to work with it. We review

1.1. INTRODUCTION 13 both the elementary functions and the ways to combine them. In what follows, C will denote a constant, n will denote a positive integer constant, a will denote a positive real number constant not equal to 1, f and g will denote two functions and x will denote the independent variable. 1. The elementary functions the reader will encounter in this class are: (a) C (constant function) (b) x n (power function) (c) np x (n th root function) (d) e x and a x (exponential functions) (e) ln x and log a x (logarithmic functions) (f) sin x, cos x, tan x, cot x, sec x, csc x (trigonometric functions) (g) sin 1 x or arcsin x, cos 1 x or arccos x, tan 1 x or arctan x (inverse of the trigonometric functions) 2. Functions can be combined as follows: (a) f + g where (f + g) (x) = f (x) + g (x) (addition) (b) f g where (f g) (x) = f (x) g (x) (subtraction) (c) Cf where (Cf) (x) = Cf (x) (multiplication by a constant) (d) fg where (fg) (x) = f (x) g (x) (multiplication of two functions) (e) f f g where (x) = f (x) (division of two functions) g g (x) (f) f g where (f g) (x) = f (g (x)) (composition of two functions) When trying to decide if a given function has a certain property, this will depend on what the function looks like. If the function is one of the elementary functions, then it is simply a matter of knowing whether that elementary function has the given property. However, if the given function is a combination of elementary functions, then one needs to see if each elementary function has the property as well as whether that property is preserved by the way the elementary functions are combined. We illustrate this with a few examples. Example 10 What is the domain of f (x) = (x 1) ln x? The property in question here is "being de ned". We are asking where f (x) is de ned. First, we note that f is a product of two functions ln x and x 1. We know that a product is always de ned as long as each member of the product is de ned. Since x 1 is always de ned and ln x is de ned only when x > 0, we conclude that f (x) is de ned when x > 0 that is on (0; 1).

14 CHAPTER 1. LIMITS AND CONTINUITY Example 11 What is the domain of g (x) = ln x (x 1)? The property in question here is "being de ned". We are asking where g (x) is de ned. First, we note that g is a division of two functions ln x and x 1. For the division to be de ned, each function has to be de ned. In addition, the denominator cannot be 0. We see that in this case, the division adds an extra condition. From the previos example, we know that x 1 is always de ned. ln x is de ned when x > 0. Since the denominator cannot be 0, it follows that x 1 6= 0 or x 6= 1. In conclusion, g (x) is de ned when x > 0 and x 6= 1. Remark 12 The last example shows that a property depends not only on each elementary functions making up a given function, but also on how the elementary functions are combined. The technique used in nding the domain in the two examples above will be used throughout Calculus when deciding if a function has a certain property or when applying a certain rule to a function. It will always involve understanding how a given function is made up. Which elementary functions it contains and how they are combined. 1.1.8 Sample Problems The rst three problems are related to the material of this section. Problem 4 is a review problem. In the next few sections, the concept of the domain of a function will play an important role. These problems are designed to check that students have mastered this concept. If it is not the case, students are strongly encouraged to review this concept. The remaining problems are designed to prepare students for the material to come. 1. Consider the function f (x) = x 2 3 (a) Find the equation of the secant line through the points corresponding to x = 1 and x = 2. (answer y = 3x 5) (x 1) (x + 1) (b) Write a formula for the slope of the tangent at x = 1. answer lim x!1 x 1 2. Consider the function f (x) = sin x (a) Find the equation of the secant line through the points corresponding to x = 0 and x = 2. answer y = 2 x (b) Write a formula for the slope of the tangent at x = 2. answer lim x7! 2 3. The position of an object in feet is given by s (t) = 40t 16t 2. (a) Find the average velocity of the object between t = 1 and t = 2. (answer 8 ft= sec) sin x 1 x 2

1.1. INTRODUCTION 15 (b) Express the instantaneous velocity of the object at time t = 1 as a 8 (t 1) (2t 3) formula. answer lim t7!1 t 1 4. Find the domain of the functions below (a) ln (x 2). (answer (2; 1)) (b) p (x 2). (answer [2; 1)) (c) ln (5 x). (answer ( 1; 5)) ln x (d). (answer (0; 5) [ (5; 1)) x 5 (e) x 5. (answer (0; 1) [ (1; 1)) ln x 5. Consider a function y = f (x). What can you say about the output values (y values) corresponding to input values which would be very close to one another? You may look at speci c examples to help you answer this question. 6. Can you think of a function y = f (x) for which f (a) and f (b) would be far apart, even though a and b are very close? First, think what the graph of that function would look like. Then, try to nd a formula for it.