MATH 114 Calculus Notes on Chapter 2 (Limits) (pages 60-? in Stewart)
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1 Still under construction. MATH 114 Calculus Notes on Chapter 2 (Limits) (pages 60-? in Stewart) As seen in A Preview of Calculus, the concept of it underlies the various branches of calculus. Hence we begin by studying its and their properties. 2.1 The Tangent and Velocity Problems In this section we see how its arise in trying to find the tangent to a curve or (as a special case of that) the (instantaneous) velocity of a falling object. A tangent to a curve is a line that touches the curve at some point and has the same slope as does the curve at that point. (See Figure 1 for pictures of a tangent to a circle and to a typical curve.) (Can make the analogy of a car driving on a curvey road; the direction in which the car is facing corresponds to the slope of the tangent at the point where the car is.) Example 1: Find the slope m of the tangent line t to the parabola y = x 2 at the point P (1, 1). Solution: If we choose a point Q on the parabola that is near to P, and then calculate the slope m P Q of the (secant) line P Q, we get an approximation to m. Taking Q to be (1.5, 2.25), for example, yields m P Q = = = 2.5 The table under Figure 2 on page 61 lists the results when we choose other points for Q. The table suggests (i.e., provides empirical evidence) that, as Q comes closer to P from the right, m P Q is decreasing towards 2, and as Q comes closer to P from the left, m P Q is increasing towards 2. We might write this as Q P m P Q = 2 See Figure 3 and its animation at garrett/qy/tracetangent.html. If we define the slope m of the tangent line to be this it, then we are led to believe (based upon only empirical evidence, mind you) that m = 2. To formalize what was done in building the table on page 61, define m(x) to be the slope of line P Q, where Q is the point (x, x 2 ) (and P is still (1, 1)). To identify a line by two of its points, those points must be distinct, of course. Hence, the definition of m(x) assumes that Q is distinct from P, which is to say that x 1. So the domain of m(x) includes all real numbers, except for 1, at which m is undefined. We have (taking ratio of the difference of the y-coordinates of P and Q with the difference of their x-coordinates) m(x) = x2 1 x 1 1
2 What the tables (on page 61) suggest is that the it of m(x), as x approaches 1 (corresponding to the idea of Q approaching P ), is 2. That is, x 2 1 x 1 x 1 = 2 Again, we have not proved this; we have simply gathered some empirical evidence suggesting it. (Of course, given that our notion of it is still only informal, it would be impossible to prove anything about it at this point!) Example 3: A problem about velocity Suppose that an object is moving, and its positions are measured at two points in time. Let t 1 and t 2 be the points in time and p 1 and p 2 be the positions at those times, respectively. Then the ratio distance between p 1 and p 2 time elpased between t 1 and t 2 represents the average velocity of the object over the time period from t 1 to t 2. (Note that the object s velocity could have been changing during the given interval of time.) For example, if you drive 100 miles in two hours, your average velocity during those two hours is 50 miles per hour. But what about instantaneous velocity? Your car s speedometer displays (an approximation to) this. But what does it really mean? If velocity is a ratio between distance traveled and elapsed time, what does it mean for a velocity to be instantaneous? (By definition of that word, it refers to a state that exists at one point in time, as opposed to over an interval.) To be able to make a sensible, rigorous definition of instantaneous velocity, we use the concept of it! Specifically, we define the instantaneous velocity of an object at a given point in time to be the it of its average velocities over shorter and shorter time periods ending at (or beginning at, or surrounding) that point in time. Example 3 asks us to find the (instantaneous) velocity of a ball that was dropped five seconds ago. (The assumption is that the ball was dropped from such a height that it will still be falling five seconds later, and also that the ball follows Galileo s equation characterizing the distance traveled by a falling object as a function of how long ago it was dropped: s(t) = 4.9t 2 The units here are seconds and meters. The table near top of page 64 shows the average velocities of the ball for several time intervals that begin at t = 5. As an example, the average velocity over the period [5, 5.1] is s(5.1) s(5) = (49)(5.1)2 (49)(5 2 ) 0.1 = 49.49m/s 2
3 The entries in the table suggest that, as the intervals are getting shorter, the average velocity is approaching 49. Is this problem any different, in nature, from that in Example 1 (where we estimated the slope of a line tangent to a parabola)? No!. This becomes very clear (if it wasn t already) if we draw a graph of Galileo s falling body function s (which is nothing more than a parabola!) and recognize that what we really did was to attempt to determine the slope of the tangent to that function at the point (5, s(5)) by finding the slopes of secant lines through that point and other points closer and closer to it. To formalize (and generalize) this a little, consider that, if we let the point of interest P be (a, s(a)) and the varying point Q be (a + h, s(a + h)) (where h 0), then the slope of P Q is m P Q = (4.9)(a + h)2 4.9a 2 (a + h) a representing the average velocity of the ball over the interval [a, a + h] (or [a + h, a] in the case that h < 0). Then we define the instantaneous velocity at time a to be (4.9)(a + h) 2 4.9a 2 h 0 (a + h) a Of course, the denominator simplifies to h. More generally, the slope of the tangent line to f(x) at the point (a, f(a)) is 2.2 The Limit of a Function f(a + h) f(a) h 0 h Having seen in Section 2.1 how its arise in finding the tangent to a curve (or, as a special case, the instantaneous velocity of a falling object), now we turn to its in general and to numerical and graphical methods for computing them. Take the function f(x) = x 2 x + 2 for values of x near 2. (See table on page 66.) The table entries suggest that, when x is near 2 (either on the left or right ), f(x) has value near 4. (This is not at all surprising considering that f(2) is itself 4!) Indeed, it turns out that we can make f(x) as close to 4 as we want to by taking x to be sufficiently close to 2, which we denote by x 2 (x 2 x + 2) = 4. This is the essence of the notion of a it. Indeed, a preinary (and somewhat imprecise) definition of it is as follows: Definition: If f(x) can be made arbitrarily close to L (i.e., as close as we wish) by choosing x sufficiently close (but not equal) to a (on either side of a), then the it of f(x), as x approaches a, is L, which is usually written f(x) = L 3
4 Roughly speaking, this means that, as x approaches a, from both left and right, the value f(x) gets closer and closer to L. It is important to note that f(x) does not depend upon the value f(a), or even upon whether or not f is defined at a! See Figure 2, on page 67, which shows the graphs of three functions differing only at x = a; in the leftmost (which is smooth ), f(a) = f(x), in the middle one, f(a) f(x), and in the rightmost, f(a) is undefined. (The crux is that all three functions have the same it as x approaches a, even though their behaviors at x = a are different.) x 1 Example 1: Guess x 1 x 2 1. The tables on page 67 suggest that the it we seek is 1 2. (Considering that the function is identical to the function 1 x+1, except when x = 1, and that 1 x+1 gets close to 1 2 when x is close to 1, this is not at all surprising.) If we were to take g to be the function identical to the given one (call it f), except at x = 1, where g has value 2 (or any other value you d like), we d have f(x) = g(x) x 1 x 1 Why? Because the it of a function, as x approaches a, does not depend upon the function s value at x = a!! Example 2: Estimate t 0 t As the table on page 68 indicates, for values of t fairly close to 0, the function s value is near 1 6. But if we use an electronic calculator to get the value of the function when t is very close to 0 (e.g., within ), it yields 0. It turns out that the actual it is, indeed, 1 6. So why does the calculator tell us that the function s value becomes 0 when t is very close to 0? Because it s WRONG!! The error is a result of the fact that, when t is very close to 0, the numerator is found to be 0, even though it is really slightly greater. Why? Because when the calculator adds a very small number to 9, the result is 9, due to the the number of significant digits in the numbers it stores being bounded. Suppose, for example, that you were asked to add 9.0 and and to express the result using at most seven significant digits. The best answer you could give would be 9.0! sin x Example 3: Guess x 0 x. This example illustrates, as Example 1 already did, that a function s it as x approaches a does not depend upon the function s value at a or even upon whether is is defined at a. Example 4: Investigate x 0 sin π x. t 2 If we evaluate the given function at values of x of the form 1 n (n a positive or negative integer), we always get 0. (That s because sin π 1/n = sin nπ = 0 for all such n.) However, if we take x = 2 4n+1 (and note that 1 2n+1 < 2 4n+1 < 1 2n, we find that the function has value 1: π (4n + 1)π sin = sin = sin (2n + 1 2/4n )π = sin π 2 = 1 4
5 (The next-to-last step is due to the fact that sin has period 2π, which is to say that, for any integer k, sin (2k + c)π = sin cπ.) Hence, the function oscillates as x approaches 0 and therefore has no it there. Example 5: Skip. Example 6: Define the Heaviside function H(t) to be 0 for t < 0 and to be 1 for t 0. As t approaches 0 from the left, H(t) approaches 0; from the right, H(t) approaches 1. Hence, the function has no it at 1. But this gives rise to notion of one-sided its. We write f(x) = L and say that the left-hand it of f(x) as x approaches a (or the it of f(x) as x approaches a from the left) is L if f(x) can be made arbitrarily close to L (i.e., as close as we wish) by choosing x sufficiently close to, but less than, a. An analogous definition applies to the notion of a right-hand it. Note that f(x) = L is equivalent to the conjunction f(x) = L f(x) = L + In other words, for L to be the it of f as x approaches a requires that L be both the leftand right-hand its of f as x approaches a. (Hence, the term it really means two-sided it.) See Example 7 (and accompanying Figure 10) on page 71 to illustrate left-hand and right-hand its and their relationship to (two-sided) its. Infinite Limits If, as x approaches a, f(x) grows without bound, we say that the it is (infinity). (Analogously, if f(x) gets smaller without bound (meaning that it is negative and its absolute value grows without bound), we say that the it is.) (See formal, but still somewhat imprecise, definition on page 72.) The notation for saying this is f(x) = and it is spoken as the it of f(x), as x approaches a, is infinity, or f(x) increases without bound as x approaches a. (See example graph in Figure 12. An analogous graph in Figure 13 shows an example of a function that decreases without bound (i.e., the case).) This is not to say that is a number; it simply describes this particular way that the it fails to exist.) Example 8: Find x 0 1 x 2. (See Figure 11, page 72.) Due to the fact that 1 grows beyond any bound as x approaches x 2 0 (that is, for any (large) value M you choose, we could find a value of x near 0 such that f(x) > M) we say that the it is +. 5
6 For similar reasons, if the function had been 1 x 2, the it would have been. Similar definitions apply to one-sided its. See Figure 14 illustrating positive and negative one-sided its. Vertical Asymptote: See definition 6, page 73. Determining a curve s vertical asymptotes is very helpful in sketching its graph. Example 9, top page 74 (Figure 15). 2.3: Calculating Limits Using the Limit Laws Page 77. See Example 1, page 78 (Figure 1). 2.4: The Precise Definition of Limit Definition 2 (Limit) Let f be a function defined on some open interval that contains a, except possibly at a itself. Then f(x) = L means that, for every number ɛ > 0 (no matter how small) there exists a number δ > 0 such that if x lies in the interval (a δ, a + δ) (but excluding a itself) then f(x) lies in the interval (L ɛ, L + ɛ). Definition 6 (Infinite Limit) Let f be defined on some open interval that contains the number a, except possibly at a itself. Then f(x) = + means that for every positive number M there exists a positive number δ such that if 0 < x a < δ then f(x) > M. In other words, no matter how big M is, you can find an open interval around a, namely (a δ, a + δ), such that for all x in that interval (except possibly a itself), f(x) > M. Example 5: : Continuity You will recall the Direct Substitution Property in Section 2.3, which says that if f is a polynomial function or a rational function with a in its domain, then f has the property that its it, as x approaches a, is simply f(a). Some other functions have this property, too. Any function having this property is said to be continuous at a: Definition 1: A function f is continuous at a if f(x) = f(a) Any of three conditions makes f discontinuous (i.e., not continuous) at a: 1. f(a) is undefined, 6
7 2. f(x) does not exist, or 3. f(a) f(x) (where the first is defined and the second exists) See the curve in Figure 2 (page 97) for examples of each of these three conditions, at x = 1, x = 3, and x = 5, respectively. Example 2 (page 98) defines four functions and asks reader to find the points at which they are discontinuous. (See graphs in Figure 3, page 98.) The function described in (a) is just y = x + 1, except at x = 2, where it is undefined, so it has a removable discontinuity there. (It is removable in the sense that if we simply defined the function to have value 2 at x = 1, there would be no discontinuity.) The function described in (c) is the same as in (a), except that it is defined to be 1 at x = 1. It, too, has a removable discontinuity at x = 1 that can be remedied by changing its definition to be 2 at x = 1. The function described in (d), y = x, has a jump discontinuity at x = k, for each integer k. The function described in (b) (y = 1 x 2 except at x = 0) has an infinite discontinuity at x = 0 In analogy with left- and right-its, we can define (see Definition 2, page 99) the concept of continuous from the right (respectively, left) at a by + f(x) = f(a) and f(x) = f(a), respectively. Example 3 points out that, for every integer k, f(x) = x is continuous from the right at x = k, but discontinuous from the left. Definition 3: f is continuous on an interval if it is continuous at every number in the interval. (For the left endpoint of the interval, however, it suffices to show that f is continuous from the right; similarly, for the right endpoint, it suffices to show that f is continuous from the left. Intuitively, if a function is continuous on an interval, graphing the corresponding curve on that interval can be done without lifting the pencil. (Any place where the pencil would have to be lifted would correspond to a point of discontinuity.) Example 4 (see graph in Figure 4, page 99) shows that f(x) = 1 1 x 2 is continuous on the interval [ 1, 1]. For points in the interior of that interval, the Limit Laws suffice to show continuity. For the two endpoints ( 1 and 1), similar calculations show that the function is continuous from the right and from the left, respectively. Theorem 4 (page 100) tells us that continuity at a point is preserved by common operations that combine functions to form new ones: If f and g are continuous at a, then so are the functions f + g, f g, cg (where c is a constant), fg, and f/g (provided that g(a) 0). One can easily prove these (as Stewart does for f + g on page 100) using the Limit Laws and definition of continuity at a. 7
8 Theorem 4 and Definition 3 lead to the conclusion that, if both f and g are continuous on an interval, so are f + g, f g, cf (where c is a constant), fg, and (if g is non-zero throughout the interval) f/g. Using the notion of continuity on an interval, we can restate the Direct Substitution Property from Section 2.3 as Theorem 5 (page 100): A polynomial is continuous everywhere (i.e., on the interval (, + )) and a rational function is continuous at every number in its domain. Sketch of Proof: Let a be an arbitrary number, and let p(x) be a polynomial, which is to say that, for some n 0 and some constants c 0, c 1,..., c n, p(x) = c 0 x 0 + c 1 x 1 + c 2 x c n x n For i satisfying 0 i n, let p i (x) = c i x i. Hence, we have p(x) = p 0 (x) + p 1 (x) + + p n (x) Theorem 4 tells us that, to show that p is continuous at a, it suffices to show that each of the p i s is continuous at a. (After all, p is just p 0 + p p n and (repeated application of) part 1 of Theorem 4 says that the sum of functions is continuous at a if each of the functions is.) So consider p i = c i x i, for any i satisfying 0 i n. Taking c = c i and t(x) = x i, we have p i (x) = ct(x). Theorem 4 tells us that, to show that p i is continuous at a, it suffices to show that t is continuous at a. (After all, part 3 of Theorem 4 tells us that the product of a constant and a function is continuous at a if the function is.) So consider t(x) = x i. By Limit Law (9), x i = a i, which is to say that t(x) = t(a), which is to say that t is continuous at a. This completes the proof that p is continuous at a. Because a was chosen arbitrarily, we have proved that p is continuous everywhere! As for rational functions, let r(x) = p(x)/q(x), where p and q are polynomials. From what we just proved, we know that both p and q are continuous everywhere. From part 5 of Theorem 4, we have that r is continuous at every point where q is non-zero. But this is precisely the domain of r. End of Proof of Theorem 5 Theorem 7 (page 102) extends Theorem 5 to include root functions (i.e., ones of the form f(x) = k x, where k is a positive integer) and trigonometric functions (e.g., sin, cos). (Actually, Limit Law 10 from Section 2.3 told us that root functions are continuous everywhere.) Indeed, probably most functions familiar to you are continuous at every number in their domains. Example 6 (page 102) has three functions and asks the reader to identify the (maximal) intervals on which they are continuous. As (a) is a polynomial, it is continuous on (, + ). As (b) is a rational function defined everywhere except x = 1, 1, it is continuous on (, 1) 8
9 ( 1, 1) (1, + ). (Another way to express this set is R { 1, 1}, where R denotes the set of real numbers.) The most interesting function in Example 6 is h (part (c)), which is the sum/difference of a root function and two rational functions. More precisely, h(x) = F (x) + G(x) + H(x), where F (x) = x G(x) = x + 1 x 1 H(x) = x + 1 x Appealing to (parts 1 and 2 of) Theorem 4 (page 100), we reason that h is continuous at any number at which each of F, G, and H is continuous. (In other words, the set of numbers at which h is continuous is the intersection of the sets of numbers at which F, G, and H, respectively, are continuous.) By Theorem 7, each of F, G, and H is continuous on its domain. Hence, obtaining the answer boils down to finding the intersection of the domains of F, G, and H. By inspecting their definitions, we find that F s domain is [0, + ), G s domain is R {1}, and H s domain is R. The intersection is all nonnegative real numbers, excluding 1, which can be written [0, 1) (1, + ). Continuity and Function Composition Theorem 8: (page 103) If f is continuous at b and g(x) = b, then f(g(x)) = f( g(x)) Intuitively, this theorem sounds right, because if x is close to a, then g(x) is close to b, which means (due to f s being continuous at b) that f(g(x)) is close to f(b). Theorem 8 serves as a stepping stone to: Theorem 9: (page 103) If g is continuous at a and f is continuous at g(a), then f g is continuous at a. In other words, if g(x) = g(a) and f(x) = f(g(a)) x g(a) then (f g)(x) = (f g)(a) Proof: We have (f g)(x)) = f(g(x)) by definition of = f( g(x)) by Theorem 8 (and using our two assumptions) = f(g(a)) because g is continuous at a = (f g)(a) by definition of End of proof. 9
10 Example 8(a): Where is h(x) = sin x 2 continuous? Answer: h can be expressed as the composition f g, where g(x) = x 2 and f(x) = sin x. Each of f and g is continuous everywhere; hence (by Theorem 9) so is h = f g. Example 8(b): Where is F (x) = 1 x continuous? Answer: F can be expressed as the composition f g h k, where k(x) = x 2 + 7, h(x) = x, g(x) = x 4, and f(x) = 1 x. By Theorem 7, all of these are continuous on their domains, so by Theorem 9, F is continuous on its domain, which includes all reals except for 3 and 3. All this sets the stage for an important theorem that, although it may seem very obvious, can help us to prove some interesting things. Theorem 10 The Intermediate Value Theorem (IMVT): Let f be continuous on the closed interval [a, b], where f(a) f(b) and let N be any number between f(a) and f(b). (That is, either f(a) < N < f(b) or f(a) > N > f(b).) Then there exists c (a, b) such that f(c) = N. See Figure 7. Intuitively, this says that, as a function s value varies from one number to another (over an interval on which the function is continuous), all intermediate values are assumed by the function along the way! Example 9: Show that the equation 4x 3 6x 2 + 3x 2 = 0 has a root between 1 and 2. The expression on the left-hand side describes a polynomial f(x), which (by Theorem 5) is continuous everywhere, including on the interval [1, 2]. Evaluating in the usual way, we get f(1) = 1 and f(2) = 12. The IMVT tells us that f(x) = 0 for some x between 1 and 2. Another example: Draw a circle around the world. Then, on this circle, there must exist two points, directly opposite from one another, having the same temperature. Why? Draw a line segment L from the center of the circle to a point on the circle. Let t(θ) be the temperature at the point on the circle on which lies the line segment obtained from L by rotating it through θ radians (keeping its one endpoint on the center of the circle). Let f(θ) = t(θ) t(θ +π); i.e., f is the function describing the differences between the temperatures on directly opposite sides of the circle. (Recall that π radians corresponds to an angle of 180 degrees, which takes you exactly half way around a circle.) To solve the problem, it suffices to show that f has a root. (Why? Because if, for some θ 1, f(θ 1 ) = t(θ 1 ) t(θ 1 + π) = 0, then it must be that t(θ 1 ) = t(θ 1 + π), which is to say that there are two points, directly opposite from one another, having the same temperature.) Note that, for any θ, f(θ + π) = t(θ + π) t(θ + 2π) by definition of f = t(θ + π) t(θ) thas period 2π = (t(θ) t(θ + π)) algebra = f(θ) by definition of f 10
11 from which we conclude that f(θ) and f(θ+π) have opposite signs (or both are zero). Assuming that t is continuous (which is what we usually do for functions describing natural phenomena such as temperature), it follows that f is, too (from Theorem 4). Applying the Intermediate Value Theorem, we conclude that there must be some θ 1 satisfying θ θ 1 θ + π such that f(θ 1 ) = 0. That is, f has a root, which is exactly what we were trying to show. 11
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