Chapter 2 - 5E-02(pp 064-073) 1/17/06 1:24 PM Page 64...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5E-02(pp 064-073) 1/17/06 1:24 PM Page 64 CHAPTER 2 The idea of a limit is illustrated by secant lines approaching a tangent line. L imits and Rates of Change 5E-02(pp 064-073) 1/17/06 1:25 PM Page 65 In A Preview of Calculus (page 2) we saw how the idea of a limit underlies the various branches of calculus. It is therefore appropriate to begin our study of calculus by investigating limits and their properties. |||| 2.1 The Tangent and Velocity Problems In this section we see how limits arise when we attempt to find the tangent to a curve or the velocity of an object. The Tangent Problem The word tangent is derived from the Latin word tangens, which means “touching.” Thus, a tangent to a curve is a line that touches the curve. In other words, a tangent line should have the same direction as the curve at the point of contact. How can this idea be made precise? For a circle we could simply follow Euclid and say that a tangent is a line that intersects the circle once and only once as in Figure 1(a). For more complicated curves this definition is inadequate. Figure l(b) shows two lines l and t passing through a point P on a curve C. The line l intersects C only once, but it certainly does not look like what we think of as a tangent. The line t, on the other hand, looks like a tangent but it intersects C twice. Locate tangents interactively and explore them numerically. Resources / Module 1 / Tangents / What Is a Tangent? t P t C l FIGURE 1 (a) y Q { x, ≈} FIGURE 2 t P (1, 1) 0 To be specific, let’s look at the problem of trying to find a tangent line t to the parabola x 2 in the following example. EXAMPLE 1 Find an equation of the tangent line to the parabola y y y=≈ (b) x x 2 at the point P 1, 1 . SOLUTION We will be able to find an equation of the tangent line t as soon as we know its slope m. The difficulty is that we know only one point, P, on t, whereas we need two points to compute the slope. But observe that we can compute an approximation to m by choosing a nearby point Q x, x 2 on the parabola (as in Figure 2) and computing the slope mPQ of the secant line PQ. We choose x 1 so that Q P. Then mPQ x2 x 1 1 65 5E-02(pp 064-073) 66 ❙❙❙❙ 1/17/06 1:25 PM Page 66 CHAPTER 2 LIMITS AND RATES OF CHANGE For instance, for the point Q 1.5, 2.25 we have mPQ x 3 2.5 2.1 2.01 2.001 lim mPQ Q lP x 1.25 0.5 2.5 and m lim xl1 x2 x 1 1 2 Assuming that the slope of the tangent line is indeed 2, we use the point-slope form of the equation of a line (see Appendix B) to write the equation of the tangent line through 1, 1 as mPQ 0 0.5 0.9 0.99 0.999 1 1 The tables in the margin show the values of mPQ for several values of x close to 1. The closer Q is to P, the closer x is to 1 and, it appears from the tables, the closer mPQ is to 2. This suggests that the slope of the tangent line t should be m 2. We say that the slope of the tangent line is the limit of the slopes of the secant lines, and we express this symbolically by writing mPQ 2 1.5 1.1 1.01 1.001 2.25 1.5 1 1.5 1.9 1.99 1.999 y 1 2x 1 or y 2x 1 Figure 3 illustrates the limiting process that occurs in this example. As Q approaches P along the parabola, the corresponding secant lines rotate about P and approach the tangent line t. y y y Q t t t Q Q P P 0 P 0 x x 0 x Q approaches P from the right y y y t t Q P t P P Q 0 x Q 0 x 0 x Q approaches P from the left FIGURE 3 In Module 2.1 you can see how the process in Figure 3 works for five additional functions. Many functions that occur in science are not described by explicit equations; they are defined by experimental data. The next example shows how to estimate the slope of the tangent line to the graph of such a function. 5E-02(pp 064-073) 1/17/06 1:25 PM Page 67 SECTION 2.1 THE TANGENT AND VELOCITY PROBLEMS t 67 E XAMPLE 2 The flash unit on a camera operates by storing charge on a capacitor and releasing it suddenly when the flash is set off. The data at the left describe the charge Q remaining on the capacitor (measured in microcoulombs) at time t (measured in seconds after the flash goes off ). Use the data to draw the graph of this function and estimate the slope of the tangent line at the point where t 0.04. [ Note: The slope of the tangent line represents the electric current flowing from the capacitor to the flash bulb (measured in microamperes).] Q 0.00 0.02 0.04 0.06 0.08 0.10 ❙❙❙❙ 100.00 81.87 67.03 54.88 44.93 36.76 SOLUTION In Figure 4 we plot the given data and use them to sketch a curve that approximates the graph of the function. Q 100 90 80 A P 70 60 50 0 B C 0.02 0.04 0.06 0.08 0.1 t F IGURE 4 Given the points P 0.04, 67.03 and R 0.00, 100.00 on the graph, we find that the slope of the secant line PR is 100.00 0.00 mPR R (0.00, 100.00) (0.02, 81.87) (0.06, 54.88) (0.08, 44.93) (0.10, 36.76) mPR 824.25 742.00 607.50 552.50 504.50 67.03 0.04 824.25 The table at the left shows the results of similar calculations for the slopes of other secant lines. From this table we would expect the slope of the tangent line at t 0.04 to lie somewhere between 742 and 607.5. In fact, the average of the slopes of the two closest secant lines is 1 2 742 607.5 674.75 So, by this method, we estimate the slope of the tangent line to be 675. Another method is to draw an approximation to the tangent line at P and measure the sides of the triangle ABC, as in Figure 4. This gives an estimate of the slope of the tangent line as |||| The physical meaning of the answer in Example 2 is that the electric current flowing from the capacitor to the flash bulb after 0.04 second is about –670 microamperes. AB BC 80.4 0.06 53.6 0.02 670 The Velocity Problem If you watch the speedometer of a car as you travel in city traffic, you see that the needle doesn’t stay still for very long; that is, the velocity of the car is not constant. We assume from watching the speedometer that the car has a definite velocity at each moment, but how is the “instantaneous” velocity defined? Let’s investigate the example of a falling ball. 5E-02(pp 064-073) 68 ❙❙❙❙ 1/17/06 1:25 PM Page 68 CHAPTER 2 LIMITS AND RATES OF CHANGE EXAMPLE 3 Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 5 seconds. SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air resistance.) If the distance fallen after t seconds is denoted by s t and measured in meters, then Galileo’s law is expressed by the equation 4.9t 2 st The difficulty in finding the velocity after 5 s is that we are dealing with a single instant of time t 5 , so no time interval is involved. However, we can approximate the desired quantity by computing the average velocity over the brief time interval of a tenth of a second from t 5 to t 5.1: average velocity distance traveled time elapsed s 5.1 The CN Tower in Toronto is currently the tallest freestanding building in the world. s5 0.1 4.9 5.1 2 4.9 5 2 0.1 49.49 m s The following table shows the results of similar calculations of the average velocity over successively smaller time periods. Time interval 5 5 5 5 5 t t t t t Average velocity (m s) 6 5.1 5.05 5.01 5.001 53.9 49.49 49.245 49.049 49.0049 It appears that as we shorten the time period, the average velocity is becoming closer to 49 m s. The instantaneous velocity when t 5 is defined to be the limiting value of these average velocities over shorter and shorter time periods that start at t 5. Thus, the (instantaneous) velocity after 5 s is v 49 m s You may have the feeling that the calculations used in solving this problem are very similar to those used earlier in this section to find tangents. In fact, there is a close connection between the tangent problem and the problem of finding velocities. If we draw the graph of the distance function of the ball (as in Figure 5) and we consider the points P a, 4.9a 2 and Q a h, 4.9 a h 2 on the graph, then the slope of the secant line PQ is mPQ 4.9 a a h2 h 4.9a 2 a 5E-02(pp 064-073) 1/17/06 1:26 PM Page 69 S ECTION 2.1 THE TANGENT AND VELOCITY PROBLEMS ❙❙❙❙ 69 which is the same as the average velocity over the time interval a, a h . Therefore, the velocity at time t a (the limit of these average velocities as h approaches 0) must be equal to the slope of the tangent line at P (the limit of the slopes of the secant lines). s s s=4.9t@ s=4.9t@ Q slope of secant line average velocity a 0 slope of tangent instantaneous velocity P P a+h t 0 a t FIGURE 5 Examples 1 and 3 show that in order to solve tangent and velocity problems we must be able to find limits. After studying methods for computing limits in the next four sections, we will return to the problems of finding tangents and velocities in Section 2.6. |||| 2.1 Exercises 1. A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes. t (min) 5 10 15 20 25 30 V (gal) 694 444 250 111 28 0 (a) If P is the point 15, 250 on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph with t 5, 10, 20, 25, and 30. (b) Estimate the slope of the tangent line at P by averaging the slopes of two secant lines. (c) Use a graph of the function to estimate the slope of the tangent line at P. (This slope represents the rate at which the water is flowing from the tank after 15 minutes.) 2. A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after t minutes. When the data in the table are graphed, the slope of the tangent line represents the heart rate in beats per minute. t (min) Heartbeats 36 38 40 42 44 2530 2661 2806 2948 3080 The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient’s heart rate after 42 minutes using the secant line between the points with the given values of t. (a) t 36 and t 42 (b) t 38 and t 42 (c) t 40 and t 42 (d) t 42 and t 44 What are your conclusions? 3. The point P (1, 2 ) lies on the curve y 1 x 1 x. (a) If Q is the point x, x 1 x , use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x : (i) 0.5 (ii) 0.9 (iii) 0.99 (iv) 0.999 (v) 1.5 (vi) 1.1 (vii) 1.01 (viii) 1.001 (b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P (1, 1 ). 2 (c) Using the slope from part (b), find an equation of the tangent line to the curve at P (1, 1 ). 2 4. The point P 5, 3 lies on the curve y sx 4. (a) If Q is the point ( x, sx 4 ), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x : (i) 4.5 (ii) 4.9 (iii) 4.99 (iv) 4.999 (v) 5.5 (vi) 5.1 (vii) 5.01 (viii) 5.001 (b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P 5, 3 . 5E-02(pp 064-073) 70 ❙❙❙❙ 1/17/06 1:26 PM Page 70 CHAPTER 2 LIMITS AND RATES OF CHANGE (c) Draw the graph of s as a function of t and draw the secant lines whose slopes are the average velocities found in part (a). (d) Draw the tangent line whose slope is the instantaneous velocity from part (b). (c) Using the slope from part (b), find an equation of the tangent line to the curve at P 5, 3 . (d) Sketch the curve, two of the secant lines, and the tangent line. 5. If a ball is thrown into the air with a velocity of 40 ft s, its height in feet after t seconds is given by y 40 t 16 t 2. (a) Find the average velocity for the time period beginning when t 2 and lasting (i) 0.5 second (ii) 0.1 second (iii) 0.05 second (iv) 0.01 second (b) Find the instantaneous velocity when t 2. 8. The position of a car is given by the values in the table. t (seconds) 1 2 3 4 5 s (feet) 0 10 32 70 119 178 (a) Find the average velocity for the time period beginning when t 2 and lasting (i) 3 seconds (ii) 2 seconds (iii) 1 second (b) Use the graph of s as a function of t to estimate the instantaneous velocity when t 2. 6. If an arrow is shot upward on the moon with a velocity of 58 m s, its height in meters after t seconds is given by h 58t 0.83 t 2. (a) Find the average velocity over the given time intervals: (i) [1, 2] (ii) [1, 1.5] (iii) [1, 1.1] (iv) [1, 1.01] (v) [1, 1.001] (b) Find the instantaneous velocity after one second. 9. The point P 1, 0 lies on the curve y 7. The displacement (in feet) of a certain particle moving in a straight line is given by s t 3 6, where t is measured in seconds. (a) Find the average velocity over the following time periods: (i) [1, 3] (ii) [1, 2] (iii) [1, 1.5] (iv) [1, 1.1] (b) Find the instantaneous velocity when t 1. |||| 2.2 0 ; sin 10 x . (a) If Q is the point x, sin 10 x , find the slope of the secant line PQ (correct to four decimal places) for x 2, 1.5, 1.4, 1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, and 0.9. Do the slopes appear to be approaching a limit? (b) Use a graph of the curve to explain why the slopes of the secant lines in part (a) are not close to the slope of the tangent line at P. (c) By choosing appropriate secant lines, estimate the slope of the tangent line at P. The Limit of a Function Having seen in the preceding section how limits arise when we want to find the tangent to a curve or the velocity of an object, we now turn our attention to limits in general and numerical and graphical methods for computing them. Let’s investigate the behavior of the function f defined by f x x 2 x 2 for values of x near 2. The following table gives values of f x for values of x close to 2, but not equal to 2. y x ƒ approaches 4. y=≈- x+2 4 0 2 As x approaches 2, FIGURE 1 fx x fx 1.0 1.5 1.8 1.9 1.95 1.99 1.995 1.999 2.000000 2.750000 3.440000 3.710000 3.852500 3.970100 3.985025 3.997001 3.0 2.5 2.2 2.1 2.05 2.01 2.005 2.001 8.000000 5.750000 4.640000 4.310000 4.152500 4.030100 4.015025 4.003001 x From the table and the graph of f (a parabola) shown in Figure 1 we see that when x is close to 2 (on either side of 2), f x is close to 4. In fact, it appears that we can make the 5E-02(pp 064-073) 1/17/06 1:26 PM Page 71 S ECTION 2.2 THE LIMIT OF A FUNCTION ❙❙❙❙ 71 values of f x as close as we like to 4 by taking x sufficiently close to 2. We express this by saying “the limit of the function f x x 2 x 2 as x approaches 2 is equal to 4.” The notation for this is lim x 2 x l2 x 2 4 In general, we use the following notation. 1 Definition We write lim f x xla and say L “the limit of f x , as x approaches a, equals L” if we can make the values of f x arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a (on either side of a) but not equal to a. Roughly speaking, this says that the values of f x get closer and closer to the number L as x gets closer and closer to the number a (from either side of a) but x a. A more precise definition will be given in Section 2.4. An alternative notation for lim f x xla is f x lL as L xla which is usually read “ f x approaches L as x approaches a.” Notice the phrase “but x a” in the definition of limit. This means that in finding the limit of f x as x approaches a, we never consider x a. In fact, f x need not even be defined when x a. The only thing that matters is how f is defined near a. Figure 2 shows the graphs of three functions. Note that in part (c), f a is not defined and in part (b), f a L. But in each case, regardless of what happens at a, it is true that lim x l a f x L. y y y L L L 0 a x (a) 0 a 0 x (b) a x (c) FIGURE 2 lim ƒ=L in all three cases xa EXAMPLE 1 Guess the value of lim x l1 x x2 1 . 1 x 1 x 2 1 is not defined when x 1, but that doesn’t matter because the definition of lim x l a f x says that we consider values SOLUTION Notice that the function f x 5E-02(pp 064-073) 72 ❙❙❙❙ x 1/17/06 1:27 PM Page 72 CHAPTER 2 LIMITS AND RATES OF CHANGE 1 0.5 0.9 0.99 0.999 0.9999 of x that are close to a but not equal to a. The tables at the left give values of f x (correct to six decimal places) for values of x that approach 1 (but are not equal to 1). On the basis of the values in the tables, we make the guess that fx 0.666667 0.526316 0.502513 0.500250 0.500025 lim x l1 x x2 1 1 0.5 Example 1 is illustrated by the graph of f in Figure 3. Now let’s change f slightly by giving it the value 2 when x 1 and calling the resulting function t : x 1 1.5 1.1 1.01 1.001 1.0001 fx 0.400000 0.476190 0.497512 0.499750 0.499975 x x2 t (x) 1 1 1 if x 2 if x 1 This new function t still has the same limit as x approaches 1 (see Figure 4). y y 2 y= x-1 ≈-1 y=© 0.5 0 0.5 1 x 0 FIGURE 3 1 x FIGURE 4 EXAMPLE 2 Estimate the value of lim st 2 tl0 9 t 3 2 . SOLUTION The table lists values of the function for several values of t near 0. t 1.0 0.5 0.1 0.05 0.01 t 0.0005 0.0001 0.00005 0.00001 st 2 9 t2 0.16800 0.20000 0.00000 0.00000 3 st 2 9 t 3 2 0.16228 0.16553 0.16662 0.16666 0.16667 As t approaches 0, the values of the function seem to approach 0.1666666 . . . and so we guess that 1 st 2 9 3 lim 2 tl0 t 6 In Example 2 what would have happened if we had taken even smaller values of t ? The table in the margin shows the results from one calculator; you can see that something strange seems to be happening. 5E-02(pp 064-073) 1/17/06 1:27 PM Page 73 SECTION 2.2 THE LIMIT OF A FUNCTION ❙❙❙❙ 73 If you try these calculations on your own calculator you might get different values, but eventually you will get the value 0 if you make t sufficiently small. Does this mean that the answer is really 0 instead of 1 ? No, the value of the limit is 1 , as we will show in the 6 6 | next section. The problem is that the calculator gave false values because st 2 9 is very close to 3 when t is small. (In fact, when t is sufficiently small, a calculator’s value for |||| For a further explanation of why calculators st 2 9 is 3.000 . . . to as many digits as the calculator is capable of carrying.) sometimes give false values, see the web site Something similar happens when we try to graph the function www.stewartcalculus.com Click on A dditional Topics and then on L ies My Calculator and Computer Told Me . In particular, see the section called The Perils of Subtraction. st 2 ft 9 t 3 2 of Example 2 on a graphing calculator or computer. Parts (a) and (b) of Figure 5 show quite accurate graphs of f , and when we use the trace mode (if available) we can estimate easily that the limit is about 1. But if we zoom in too far, as in parts (c) and (d), then we get 6 inaccurate graphs, again because of problems with subtraction. 0.2 0.2 0.1 0.1 (a) _5, 5 by _ 0.1, 0.3 (b) _ 0.1, 0.1 by _ 0.1, 0.3 (c) _10– ^, 10– ^ by _ 0.1, 0.3 (d) _10– &, 10– & by _ 0.1, 0.3 F IGURE 5 EXAMPLE 3 Guess the value of lim xl0 sin x . x SOLUTION The function f x sin x x is not defined when x 0. Using a calculator (and remembering that, if x , sin x means the sine of the angle whose radian measure is x), we construct the following table of values correct to eight decimal places. From the table and the graph in Figure 6 we guess that lim xl0 sin x x 1 This guess is in fact correct, as will be proved in Chapter 3 using a geometric argument. x 1.0 0.5 0.4 0.3 0.2 0.1 0.05 0.01 0.005 0.001 sin x x 0.84147098 0.95885108 0.97354586 0.98506736 0.99334665 0.99833417 0.99958339 0.99998333 0.99999583 0.99999983 y 1 _1 FIGURE 6 y= 0 1 sin x x x 5E-02(pp 074-085) 74 ❙❙❙❙ 1/17/06 1:16 PM Page 74 CHAPTER 2 LIMITS AND RATES OF CHANGE |||| COMPUTER ALGEBRA SYSTEMS Computer algebra systems (CAS) have commands that compute limits. In order to avoid the types of pitfalls demonstrated in Examples 2, 4, and 5, they don’t find limits by numerical experimentation. Instead, they use more sophisticated techniques such as computing infinite series. If you have access to a CAS, use the limit command to compute the limits in the examples of this section and to check your answers in the exercises of this chapter. EXAMPLE 4 Investigate lim sin xl0 x . SOLUTION Again the function f x sin for some small values of x, we get f1 sin f ( 1) 3 sin 3 f 0.1 x is undefined at 0. Evaluating the function f (1) 2 0 sin 10 Similarly, f 0.001 f 0.0001 tempted to guess that 0 sin 2 0 f (1) 4 0 sin 4 0 f 0.01 sin 100 0 0. On the basis of this information we might be lim sin xl0 0 x | but this time our guess is wrong. Note that although f 1 n sin n 0 for any integer 1 for infinitely many values of x that approach 0. [In fact, n, it is also true that f x sin x 1 when x and, solving for x, we get x 2 2 4n 2n 1 .] The graph of f is given in Figure 7. y y=sin(π/x) 1 Listen to the sound of this function trying to approach a limit. Resources / Module 2 / Basics of Limits / Sound of a Limit that Does Not Exist _1 1 x _1 F IGURE 7 Module 2.2 helps you explore limits at points where graphs exhibit unusual behavior. The dashed lines indicate that the values of sin x oscillate between 1 and 1 infinitely often as x approaches 0 (see Exercise 37). Since the values of f x do not approach a fixed number as x approaches 0, lim sin xl0 x x3 cos 5x 10,000 EXAMPLE 5 Find lim x 3 xl0 1 0.5 0.1 0.05 0.01 1.000028 0.124920 0.001088 0.000222 0.000101 x does not exist cos 5x . 10,000 SOLUTION As before, we construct a table of values. From the table in the margin it appears that lim x 3 xl0 cos 5x 10,000 0 5E-02(pp 074-085) 1/17/06 1:16 PM Page 75 S ECTION 2.2 THE LIMIT OF A FUNCTION x 0.005 0.001 x3 lim x 3 xl0 cos 5x 10,000 Later we will see that lim x l 0 cos 5x | 0.000100 1 10,000 1; then it follows that the limit is 0.0001. Examples 4 and 5 illustrate some of the pitfalls in guessing the value of a limit. It is easy to guess the wrong value if we use inappropriate values of x, but it is difficult to know when to stop calculating values. And, as the discussion after Example 2 shows, sometimes calculators and computers give the wrong values. In the next two sections, however, we will develop foolproof methods for calculating limits. EXAMPLE 6 The Heaviside function H is defined by y 1 FIGURE 8 75 But if we persevere with smaller values of x, the second table suggests that cos 5x 10,000 0.00010009 0.00010000 0 ❙❙❙❙ 0 1 Ht t if t if t 0 0 [This function is named after the electrical engineer Oliver Heaviside (1850–1925) and can be used to describe an electric current that is switched on at time t 0.] Its graph is shown in Figure 8. As t approaches 0 from the left, H t approaches 0. As t approaches 0 from the right, H t approaches 1. There is no single number that H t approaches as t approaches 0. Therefore, lim t l 0 H t does not exist. One-Sided Limits We noticed in Example 6 that H t approaches 0 as t approaches 0 from the left and H t approaches 1 as t approaches 0 from the right. We indicate this situation symbolically by writing lim H t tl0 0 and lim H t tl0 1 The symbol “t l 0 ” indicates that we consider only values of t that are less than 0. Likewise, “t l 0 ” indicates that we consider only values of t that are greater than 0. 2 Definition We write lim f x xla L and say the left-hand limit of f x as x approaches a [or the limit of f x as x approaches a from the left] is equal to L if we can make the values of f x arbitrarily close to L by taking x to be sufficiently close to a and x less than a. Notice that Definition 2 differs from Definition 1 only in that we require x to be less than a. Similarly, if we require that x be greater than a, we get “the right-hand limit of f x as x approaches a is equal to L” and we write lim f x xla L 5E-02(pp 074-085) 76 ❙❙❙❙ 1/17/06 1:17 PM Page 76 CHAPTER 2 LIMITS AND RATES OF CHANGE Thus, the symbol “x l a ” means that we consider only x trated in Figure 9. y a. These definitions are illus- y L ƒ 0 FIGURE 9 x ƒ L 0 x a a x x (b) lim ƒ=L (a) lim ƒ=L x a+ x a_ By comparing Definition l with the definitions of one-sided limits, we see that the following is true. lim f x 3 y L xla if and only if lim f x xla and lim f x xla L EXAMPLE 7 The graph of a function t is shown in Figure 10. Use it to state the values (if they exist) of the following: 4 3 (a) lim t x (b) lim t x (c) lim t x (d) lim t x y=© (e) lim t x (f) lim t x xl2 xl2 xl5 1 0 L 1 2 3 4 5 x xl2 xl5 xl5 SOLUTION From the graph we see that the values of t x approach 3 as x approaches 2 from the left, but they approach 1 as x approaches 2 from the right. Therefore (a) lim t x F IGURE 10 xl2 and 3 (b) lim t x xl2 1 (c) Since the left and right limits are different, we conclude from (3) that lim x l 2 t x does not exist. The graph also shows that (d) lim t x xl5 and 2 (e) lim t x xl5 2 (f) This time the left and right limits are the same and so, by (3), we have lim t x xl5 Despite this fact, notice that t 5 2 2. I nfinite Limits EXAMPLE 8 Find lim xl0 1 if it exists. x2 SOLUTION As x becomes close to 0, x 2 also becomes close to 0, and 1 x 2 becomes very large. (See the table on the next page.) In fact, it appears from the graph of the function fx 1 x 2 shown in Figure 11 that the values of f x can be made arbitrarily large 5E-02(pp 074-085) 1/17/06 1:17 PM Page 77 SECTION 2.2 THE LIMIT OF A FUNCTION 1 0.5 0.2 0.1 0.05 0.01 0.001 77 by taking x close enough to 0. Thus, the values of f x do not approach a number, so lim x l 0 1 x 2 does not exist. 1 x2 x ❙❙❙❙ y 1 4 25 100 400 10,000 1,000,000 y= 1 ≈ x 0 F IGURE 11 To indicate the kind of behavior exhibited in Example 8, we use the notation lim xl0 1 x2 | This does not mean that we are regarding as a number. Nor does it mean that the limit exists. It simply expresses the particular way in which the limit does not exist: 1 x 2 can be made as large as we like by taking x close enough to 0. In general, we write symbolically lim f x xla to indicate that the values of f x become larger and larger (or “increase without bound”) as x becomes closer and closer to a. Explore infinite limits interactively. Resources / Module 2 / Limits that Are Infinite / Examples A and B 4 Definition Let f be a function defined on both sides of a, except possibly at a itself. Then lim f x xla means that the values of f x can be made arbitrarily large (as large as we please) by taking x sufficiently close to a, but not equal to a. Another notation for lim x l a f x is y fxl y=ƒ Again the symbol 0 a FIGURE 12 lim ƒ=` xa xla is not a number, but the expression lim x l a f x “the limit of f x , as x approaches a, is infinity” x x=a as or “ f x becomes infinite as x approaches a” or “ f x increases without bound as x approaches a ” This definition is illustrated graphically in Figure 12. is often read as 5E-02(pp 074-085) 78 ❙❙❙❙ 1/17/06 1:17 PM Page 78 CHAPTER 2 LIMITS AND RATES OF CHANGE y A similar sort of limit, for functions that become large negative as x gets close to a, is defined in Definition 5 and is illustrated in Figure 13. x=a 5 a 0 Definition Let f be defined on both sides of a, except possibly at a itself. Then x lim f x y=ƒ xla means that the values of f x can be made arbitrarily large negative by taking x sufficiently close to a, but not equal to a. FIGURE 13 lim ƒ=_` xa The symbol lim x l a f x can be read as “the limit of f x , as x approaches a, is negative infinity” or “ f x decreases without bound as x approaches a.” As an example we have 1 lim xl0 x2 Similar definitions can be given for the one-sided infinite limits lim f x lim f x x la x la lim f x lim f x x la x la remembering that “x l a ” means that we consider only values of x that are less than a, and similarly “x l a ” means that we consider only x a. Illustrations of these four cases are given in Figure 14. y y a 0 (a) lim ƒ=` x a_ x y a 0 x (b) lim ƒ=` x y a 0 (c) lim ƒ=_` a+ x a 0 x x (d) lim ƒ=_` a_ x a+ FIGURE 14 6 Definition The line x a is called a vertical asymptote of the curve y if at least one of the following statements is true: lim f x xla lim f x xla lim f x x la lim f x x la fx lim f x x la lim f x x la For instance, the y-axis is a vertical asymptote of the curve y 1 x 2 because lim x l 0 1 x 2 . In Figure 14 the line x a is a vertical asymptote in each of the four cases shown. In general, knowledge of vertical asymptotes is very useful in sketching graphs. 5E-02(pp 074-085) 1/17/06 1:18 PM Page 79 SECTION 2.2 THE LIMIT OF A FUNCTION E XAMPLE 9 Find lim x l3 2x x 3 2x and lim x x l3 3 ❙❙❙❙ 79 . 3 is a small posiSOLUTION If x is close to 3 but larger than 3, then the denominator x tive number and 2 x is close to 6. So the quotient 2 x x 3 is a large positive number. Thus, intuitively we see that 2x lim x l3 x 3 y 2x y= x-3 5 Likewise, if x is close to 3 but smaller than 3, then x 3 is a small negative number but 2 x is still a positive number (close to 6). So 2 x x 3 is a numerically large negative number. Thus 2x lim x l3 x 3 x 0 x=3 The graph of the curve y cal asymptote. FIGURE 15 2x x 3 is given in Figure 15. The line x EXAMPLE 10 Find the vertical asymptotes of f x y 1 0 π 2 π 3π 2 sin x cos x there are potential vertical asymptotes where cos x 0. In fact, since cos x l 0 as xl 2 and cos x l 0 as x l 2 , whereas sin x is positive when x is near 2, we have lim tan x lim tan x and x xl xl 2 2 2 is a vertical asymptote. Similar reasoning shows This shows that the line x 2n 1 2, where n is an integer, are all vertical asymptotes of that the lines x fx tan x. The graph in Figure 16 confirms this. FIGURE 16 y=tan x |||| 2.2 tan x. SOLUTION Because tan x 3π π _ 2 _π _ 2 3 is a verti- Exercises 1. Explain in your own words what is meant by the equation 4. For the function f whose graph is given, state the value of the lim f x given quantity, if it exists. If it does not exist, explain why. (a) lim f x (b) lim f x xl2 5 xl0 Is it possible for this statement to be true and yet f 2 Explain. 3? 2. Explain what it means to say that lim f x xl1 3 and xl3 (c) lim f x (d) lim f x xl3 xl3 (e) f 3 y lim f x xl1 7 In this situation is it possible that lim x l 1 f x exists? Explain. 4 2 3. Explain the meaning of each of the following. (a) lim f x xl 3 (b) lim f x xl4 0 2 4 x 5E-02(pp 074-085) 80 ❙❙❙❙ 1/17/06 1:19 PM Page 80 CHAPTER 2 LIMITS AND RATES OF CHANGE 5. Use the given graph of f to state the value of each quantity, if it exists. If it does not exist, explain why. (a) lim f x (b) lim f x (c) lim f x xl1 xl1 (d) lim f x (a) lim R x (b) lim R x (c) lim R x (d) x l2 xl1 xl5 xl 3 (e) f 5 xl5 8. For the function R whose graph is shown, state the following. lim R x xl 3 (e) The equations of the vertical asymptotes. y y 4 2 0 2 2 x 5 x 4 6. For the function t whose graph is given, state the value of each xl 2 (d) lim f x (f) lim t x (g) lim t x (h) t 2 (k) t 0 (c) lim f x (e) lim f x xl 3 (i) lim t x ( j) lim t x (b) lim f x xl 7 xl 2 (e) lim t x 9. For the function f whose graph is shown, state the following. (a) lim f x quantity, if it exists. If it does not exist, explain why. (a) lim t x (b) lim t x (c) lim t x xl 2 0 _3 xl0 (l) lim t x (d) t 2 xl2 x l2 xl4 xl6 xl6 (f) The equations of the vertical asymptotes. xl2 y xl4 xl0 y _7 0 _3 6 x 2 1 10. A patient receives a 150-mg injection of a drug every 4 hours. _3 _2 _1 0 1 2 3 4 x _1 The graph shows the amount f t of the drug in the bloodstream after t hours. (Later we will be able to compute the dosage and time interval to ensure that the concentration of the drug does not reach a harmful level.) Find and lim f t t l 12 7. For the function t whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why. (a) lim t t (b) lim t t (c) lim t t tl0 tl0 (d) lim t t (e) lim t t (g) t 2 and explain the significance of these one-sided limits. (h) lim t t tl2 lim f t t l 12 tl0 f(t) ( f ) lim t t tl2 tl2 300 tl4 y 150 4 0 2 2 4 t 4 8 12 16 t 1 1 2 1 x to state the ; 11. Use the graph of the function f x value of each limit, if it exists. If it does not exist, explain why. (a) lim f x xl0 (b) lim f x xl0 (c) lim f x xl0 5E-02(pp 074-085) 1/17/06 1:21 PM Page 81 ❙❙❙❙ S ECTION 2.2 THE LIMIT OF A FUNCTION 12. Sketch the graph of the following function and use it to deter- 29. mine the values of a for which lim x l a f x exists: ■ 2 x x fx x 1 if x 1 if 1 x if x 1 2 1 4, 3, f3 lim f x f 2 14. lim f x 1, lim f x 1, xl0 xl2 ■ ■ ■ 2, xl3 lim f x ; 2, xl 2 1, xl0 f2 ■ 1, ■ lim f x xl2 ■ ■ ; ■ ■ ■ ; 2x x2 2x x2 0.999, , x 0, 0.5, 0.9, 0.95, 2 1.5, 1.1, 1.01, 1.001 x 2, 1 0.99, x2 ■ 1, x 0.5, 0.2, 0.1, 0.05, 0.01 36. (a) Evaluate h x xl1 ■ ■ 2 x 1x to five 1 x 1x . ■ ■ ■ ■ ■ ■ 4 x 2 20. lim xl0 1 1 ■ 22. lim ■ ■ ■ ■ ■ ■ ; 5x ■ 0.04, 0.02, 0.01, 0.005, 0.003, and x x 3 for x tan x 1, 0.5, 0.1, 0.05, tan x x . x3 (c) Evaluate h x for successively smaller values of x until you finally reach 0 values for h x . Are you still confident that your guess in part (b) is correct? Explain why you eventually obtained 0 values. (In Section 7.7 a method for evaluating the limit will be explained.) (d) Graph the function h in the viewing rectangle 1, 1 by 0, 1 . Then zoom in toward the point where the graph crosses the y-axis to estimate the limit of h x as x approaches 0. Continue to zoom in until you observe distortions in the graph of h. Compare with the results of part (c). xl0 ■ x xl0 ■ 2x 1000 (b) Guess the value of lim tan 3 x tan 5 x 9x 1 x. 0.01, and 0.005. |||| Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically. x6 x10 x x decimal places. (b) Illustrate part (a) by graphing the function y (b) Evaluate f x for x 0.001. Guess again. 19–22 21. lim ■ (b) Confirm your answer to part (a) by graphing the function. lim x 2 sx 4 , x 17, 16.5, 16.1, 16.05, 16.01, x 16 15, 15.5, 15.9, 15.95, 15.99 sx ■ x2 2 x 1000 for x 1, 0.8, 0.6, 0.4, 0.2, 0.1, and 0.05, and guess the value of x l 16 xl0 ■ 35. (a) Evaluate the function f x 18. lim 19. lim ■ 1 1 and lim 3 x l1 x x3 1 1 (a) by evaluating f x 1 x 3 1 for values of x that approach 1 from the left and from the right, (b) by reasoning as in Example 9, and (c) from a graph of f . xl0 sin x 17. lim , xl0 x tan x ■ ■ function y 2 x at the point 0, 1 is lim x l 0 2 x Estimate the slope to three decimal places. , x 2.5, 2.1, 2.05, 2.01, 2.005, 2.001, x2 x 2 1.9, 1.95, 1.99, 1.995, 1.999 ■ ■ 34. The slope of the tangent line to the graph of the exponential x2 x l2 xl ■ x1 x sin x 33. (a) Estimate the value of the limit lim x l 0 1 ■ |||| Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places). 16. lim ■ y 15–18 15. lim ■ 0 f 0 is undefined ■ ■ xl1 32. (a) Find the vertical asymptotes of the function 1 lim f x ■ 30. lim sec x 2 x l1 |||| Sketch the graph of an example of a function f that satisfies all of the given conditions. xl3 lim 31. Determine lim 13–14 13. lim f x xl 81 ■ ■ ■ ; 37. Graph the function f x 23–30 |||| 23. lim 6 xl5 x 25. lim 2 x 24. lim 5 x 12 x l1 27. lim xl 2 x of Example 4 in the viewsin ing rectangle 1, 1 by 1, 1 . Then zoom in toward the origin several times. Comment on the behavior of this function. Determine the infinite limit. x x2 x xl5 26. lim xl0 1 2 6 x 5 x x2 x 28. lim csc x xl 1 2 38. In the theory of relativity, the mass of a particle with velocity v is m m0 s1 v2 c2 where m 0 is the rest mass of the particle and c is the speed of light. What happens as v l c ? 5E-02(pp 074-085) 82 ❙❙❙❙ 1/17/06 1:21 PM Page 82 CHAPTER 2 LIMITS AND RATES OF CHANGE ; 39. Use a graph to estimate the equations of all the vertical asymp- ; 40. (a) Use numerical and graphical evidence to guess the value of the limit totes of the curve lim y tan 2 sin x xl1 x 1 1 (b) How close to 1 does x have to be to ensure that the function in part (a) is within a distance 0.5 of its limit? Then find the exact equations of these asymptotes. |||| 2.3 x3 sx Calculating Limits Using the Limit Laws In Section 2.2 we used calculators and graphs to guess the values of limits, but we saw that such methods don’t always lead to the correct answer. In this section we use the following properties of limits, called the Limit Laws, to calculate limits. Limit Laws Suppose that c is a constant and the limits lim f x and xla lim t x xla exist. Then 1. lim f x tx 2. lim f x tx xla xla 3. lim c f x xla lim f x xla lim f x xla xla lim f x xla xla fx tx lim t x xla c lim f x 4. lim f x t x 5. lim lim t x xla xla lim t x xla lim f x xla lim t x if lim t x xla 0 xla These five laws can be stated verbally as follows: Sum Law 1. The limit of a sum is the sum of the limits. Difference Law 2. The limit of a difference is the difference of the limits. Constant Multiple Law 3. The limit of a constant times a function is the constant times the limit of the function. Product Law 4. The limit of a product is the product of the limits. Quotient Law 5. The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0). It is easy to believe that these properties are true. For instance, if f x is close to L and t x is close to M, it is reasonable to conclude that f x t x is close to L M. This gives us an intuitive basis for believing that Law 1 is true. In Section 2.4 we give a precise definition of a limit and use it to prove this law. The proofs of the remaining laws are given in Appendix F. 5E-02(pp 074-085) 1/17/06 1:22 PM Page 83 S ECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS y following limits, if they exist. 1 (a) lim 1 83 EXAMPLE 1 Use the Limit Laws and the graphs of f and t in Figure 1 to evaluate the f 0 ❙❙❙❙ x xl 2 fx 5t x (b) lim f x t x (c) lim xl1 xl2 fx tx SOLUTION g (a) From the graphs of f and t we see that lim f x 1 xl 2 FIGURE 1 and lim t x 1 xl 2 Therefore, we have lim xl 2 fx 5t x lim f x lim 5t x lim f x (b) We see that lim x l 1 f x right limits are different: 5 (by Law 3) xl 2 xl 2 1 (by Law 1) 5 lim t x xl 2 xl 2 1 4 2. But lim x l 1 t x does not exist because the left and lim t x 2 xl1 lim t x xl1 1 So we can’t use Law 4. The given limit does not exist, since the left limit is not equal to the right limit. (c) The graphs show that lim f x xl2 1.4 and lim t x xl2 0 Because the limit of the denominator is 0, we can’t use Law 5. The given limit does not exist because the denominator approaches 0 while the numerator approaches a nonzero number. If we use the Product Law repeatedly with t x Power Law n 6. lim f x x la n [ lim f x ] x la f x , we obtain the following law. where n is a positive integer In applying these six limit laws, we need to use two special limits: 7. lim c xla 8. lim x c xla a These limits are obvious from an intuitive point of view (state them in words or draw graphs of y c and y x), but proofs based on the precise definition are requested in the exercises for Section 2.4. If we now put f x x in Law 6 and use Law 8, we get another useful special limit. 9. lim x n xla an where n is a positive integer 5E-02(pp 074-085) 84 ❙❙❙❙ 1/17/06 1:23 PM Page 84 CHAPTER 2 LIMITS AND RATES OF CHANGE A similar limit holds for roots as follows. (For square roots the proof is outlined in Exercise 37 in Section 2.4.) n 10. lim sx n sa xla where n is a positive integer (If n is even, we assume that a 0.) More generally, we have the following law, which is proved as a consequence of Law 10 in Section 2.5. n 11. lim sf x) Root Law x la n s lim f x la x) where n is a positive integer [If n is even, we assume that lim f x 0. x la Explore limits like these interactively. Resources / Module 2 / The Essential Examples / Examples D and E EXAMPLE 2 Evaluate the following limits and justify each step. (a) lim 2 x 2 x l5 3x 4 (b) lim x3 xl 2 2x2 1 5 3x SOLUTION lim 2 x 2 (a) x l5 3x 4 lim 2 x 2 lim 3x x l5 2 lim x 2 3 lim x x l5 2 52 lim 4 x l5 lim 4 x l5 35 (by Laws 2 and 1) x l5 (by 3) x l5 4 (by 9, 8, and 7) 39 (b) We start by using Law 5, but its use is fully justified only at the final stage when we see that the limits of the numerator and denominator exist and the limit of the denominator is not 0. lim xl 2 x3 2x 2 1 5 3x lim x 3 2x2 xl 2 lim 5 2 lim x 2 xl 2 lim 5 3 5 lim 1 xl 2 xl 2 2 (by Law 5) 3x xl 2 lim x 3 1 xl 2 3 lim x 2 22 32 (by 1, 2, and 3) xl 2 1 (by 9, 8, and 7) 1 11 If we let f x 2 x 2 3x 4, then f 5 39. In other words, we would have gotten the correct answer in Example 2(a) by substituting 5 for x. Similarly, direct substitution provides the correct answer in part (b). The functions in Example 2 are a polynomial and a rational function, respectively, and similar use of the Limit Laws proves that NOTE ■ 5E-02(pp 074-085) 1/17/06 1:23 PM Page 85 S ECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS |||| NEWTON AND LIMITS Isaac Newton was born on Christmas Day in 1642, the year of Galileo’s death. When he entered Cambridge University in 1661 Newton didn’t know much mathematics, but he learned quickly by reading Euclid and Descartes and by attending the lectures of Isaac Barrow. Cambridge was closed because of the plague in 1665 and 1666, and Newton returned home to reflect on what he had learned. Those two years were amazingly productive for at that time he made four of his major discoveries: (1) his representation of functions as sums of infinite series, including the binomial theorem; (2) his work on differential and integral calculus; (3) his laws of motion and law of universal gravitation; and (4) his prism experiments on the nature of light and color. Because of a fear of controversy and criticism, he was reluctant to publish his discoveries and it wasn’t until 1687, at the urging of the astronomer Halley, that Newton published Principia Mathematica. In this work, the greatest scientific treatise ever written, Newton set forth his version of calculus and used it to investigate mechanics, fluid dynamics, and wave motion, and to explain the motion of planets and comets. The beginnings of calculus are found in the calculations of areas and volumes by ancient Greek scholars such as Eudoxus and Archimedes. Although aspects of the idea of a limit are implicit in their “method of exhaustion,” Eudoxus and Archimedes never explicitly formulated the concept of a limit. Likewise, mathematicians such as Cavalieri, Fermat, and Barrow, the immediate precursors of Newton in the development of calculus, did not actually use limits. It was Isaac Newton who was the first to talk explicitly about limits. He explained that the main idea behind limits is that quantities “approach nearer than by any given difference.” Newton stated that the limit was the basic concept in calculus, but it was left to later mathematicians like Cauchy to clarify his ideas about limits. ❙❙❙❙ 85 direct substitution always works for such functions (see Exercises 53 and 54). We state this fact as follows. Direct Substitution Property If f is a polynomial or a rational function and a is in the domain of f , then lim f x fa xla Functions with the Direct Substitution Property are called continuous at a and will be studied in Section 2.5. However, not all limits can be evaluated by direct substitution, as the following examples show. EXAMPLE 3 Find lim xl1 x2 x 1 . 1 x 2 1 x 1 . We can’t find the limit by substituting x 1 because f 1 isn’t defined. Nor can we apply the Quotient Law because the limit of the denominator is 0. Instead, we need to do some preliminary algebra. We factor the numerator as a difference of squares: SOLUTION Let f x x2 x 1 1 x 1x x1 1 The numerator and denominator have a common factor of x 1. When we take the limit as x approaches 1, we have x 1 and so x 1 0. Therefore, we can cancel the common factor and compute the limit as follows: lim xl1 x2 x 1 1 lim x xl1 1x x1 lim x 1 1 1 2 xl1 1 The limit in this example arose in Section 2.1 when we were trying to find the tangent to the parabola y x 2 at the point 1, 1 . NOTE In Example 3 we were able to compute the limit by replacing the given function f x x 2 1 x 1 by a simpler function, t x x 1, with the same limit. This is valid because f x t x except when x 1, and in computing a limit as x approaches 1 we don’t consider what happens when x is actually equal to 1. In general, if f x t x when x a, then ■ lim f x xla lim t x xla EXAMPLE 4 Find lim t x where x l1 tx x 1 if x if x 1 1 S OLUTION Here t is defined at x 1 and t 1 , but the value of a limit as x approaches 1 does not depend on the value of the function at 1. Since t x x 1 for 5E-02(pp 086-097) 86 ❙❙❙❙ 1/17/06 12:48 PM Page 86 CHAPTER 2 LIMITS AND RATES OF CHANGE x 1, we have lim t x lim x xl1 x 1 xl1 2 Note that the values of the functions in Examples 3 and 4 are identical except when 1 (see Figure 2) and so they have the same limit as x approaches 1. y y y=ƒ 3 2 The graphs of the functions f (from Example 3) and g (from Example 4) 2 1 FIGURE 2 y=© 3 1 0 1 EXAMPLE 5 Evaluate lim 2 h2 h 3 hl0 0 x 3 9 1 2 x 3 . SOLUTION If we define h2 h 3 Fh 9 then, as in Example 3, we can’t compute lim h l 0 F h by letting h undefined. But if we simplify F h algebraically, we find that Fh 9 lim st 2 tl0 h2 6h 9 t2 6 h h 0 when letting h approach 0.) Thus h2 h 3 hl0 EXAMPLE 6 Find lim 9 h (Recall that we consider only h Explore a limit like this one interactively. Resources / Module 2 / The Essential Examples / Example C h2 6h 0 since F 0 is 3 9 lim 6 h hl0 6 . SOLUTION We can’t apply the Quotient Law immediately, since the limit of the denominator is 0. Here the preliminary algebra consists of rationalizing the numerator: lim tl0 st 2 9 t 2 3 lim st 2 t tl0 lim tl0 lim tl0 9 3 2 t2 9 t (st 2 9 9 3) 2 st 2 1 9 st 2 st 2 9 9 3 3 t2 lim t (st 2 tl0 2 9 3) 1 3 s lim tl0 t 2 1 9 3 3 3 This calculation confirms the guess that we made in Example 2 in Section 2.2. 1 6 5E-02(pp 086-097) 1/17/06 12:48 PM Page 87 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ❙❙❙❙ 87 Some limits are best calculated by first finding the left- and right-hand limits. The following theorem is a reminder of what we discovered in Section 2.2. It says that a two-sided limit exists if and only if both of the one-sided limits exist and are equal. 1 Theorem lim f x L xla if and only if lim f x xla L lim f x xla When computing one-sided limits, we use the fact that the Limit Laws also hold for one-sided limits. EXAMPLE 7 Show that lim x 0. xl0 SOLUTION Recall that x x Since x |||| The result of Example 7 looks plausible from Figure 3. x for x if x x if x lim x 0 0, we have lim x y xl0 y=| x| For x 0 we have x xl0 x and so lim x lim xl0 0 x x xl0 lim x 0 xl0 E XAMPLE 8 Prove that lim xl0 x does not exist. x lim x x xl0 lim SOLUTION y 0 Therefore, by Theorem 1, FIGURE 3 |x| y= x 0 0 x x xl0 xl0 lim x x lim 1 1 lim 1 xl0 1 0 x xl0 lim x x xl0 1 _1 FIGURE 4 Since the right- and left-hand limits are different, it follows from Theorem 1 that lim x l 0 x x does not exist. The graph of the function f x x x is shown in Figure 4 and supports the one-sided limits that we found. EXAMPLE 9 If if x if x sx 4 8 2x fx 4 4 determine whether lim x l 4 f x exists. |||| It is shown in Example 3 in Section 2.4 that lim x l 0 sx 0. SOLUTION Since f x sx 4 for x lim f x xl4 4, we have lim s x xl4 4 s4 4 0 5E-02(pp 086-097) ❙❙❙❙ 88 1/17/06 12:48 PM Page 88 CHAPTER 2 LIMITS AND RATES OF CHANGE Since f x 8 2 x for x y 4, we have lim f x lim 8 xl4 2x xl4 8 24 0 The right- and left-hand limits are equal. Thus, the limit exists and 0 x 4 lim f x 0 xl4 The graph of f is shown in Figure 5. FIGURE 5 EXAMPLE 10 The greatest integer function is defined by x that is less than or equal to x. (For instance, 4 4, 4.8 1 1.) Show that lim x l 3 x does not exist. 2 |||| Other notations for x are x and ⎣ x⎦ . SOLUTION The graph of the greatest integer function is shown in Figure 6. Since x for 3 y x lim x x l3 3 y=[ x] Since x 3 4, we have 4 2 the largest integer 4, 3, s2 1, 2 for 2 x lim 3 3 lim 2 2 x l3 3, we have 1 0 1 2 3 4 5 lim x x x l3 x l3 Because these one-sided limits are not equal, lim x l 3 x does not exist by Theorem 1. FIGURE 6 The next two theorems give two additional properties of limits. Their proofs can be found in Appendix F. Greatest integer function Theorem If f x t x when x is near a (except possibly at a) and the limits of f and t both exist as x approaches a, then 2 lim f x lim t x xla 3 The Squeeze Theorem If f x tx xla h x when x is near a (except possibly at a) and lim f x y xla h g L f 0 FIGURE 7 a x then lim h x xla lim t x xla L L The Squeeze Theorem, which is sometimes called the Sandwich Theorem or the Pinching Theorem, is illustrated by Figure 7. It says that if t x is squeezed between f x and h x near a, and if f and h have the same limit L at a, then t is forced to have the same limit L at a. 5E-02(pp 086-097) 1/17/06 12:49 PM Page 89 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS E XAMPLE 11 Show that lim x 2 sin xl0 1 x ❙❙❙❙ 0. SOLUTION First note that we cannot use lim x 2 sin xl0 1 x lim x 2 lim sin xl0 xl0 1 x because lim x l 0 sin 1 x does not exist (see Example 4 in Section 2.2). However, since 1 sin 1 x 1 we have, as illustrated by Figure 8, x2 x 2 sin 1 x x2 y y=≈ 1 y=≈ sin x Watch an animation of a similar limit. Resources / Module 2 / Basics of Limits / Sound of a Limit that Exists x 0 y=_≈ FIGURE 8 We know that lim x 2 xl0 x 2, t x Taking f x obtain 0 and x 2 sin 1 x , and h x lim x 2 sin xl0 |||| 2.3 1 x x2 0 x 2 in the Squeeze Theorem, we 0 Exercises 1. Given that lim f x x la 3 (c) lim sh x 3 lim t x x la 0 lim h x x la 8 find the limits that exist. If the limit does not exist, explain why. (a) lim f x x la lim xl0 hx (b) lim f x x la 2 xla f h f (g) lim x la t (e) lim x la x x x x 1 fx tx (f ) lim x la f x 2f x (h) lim x la h x fx (d) lim x la 89 5E-02(pp 086-097) 90 ❙❙❙❙ 1/17/06 12:50 PM Page 90 CHAPTER 2 LIMITS AND RATES OF CHANGE 2. The graphs of f and t are given. Use them to evaluate each 21. lim limit, if it exists. If the limit does not exist, explain why. y tl9 y 23. lim y=ƒ x 1 (a) lim f x 1 0 x l0 xl 1 29. lim tl0 (f ) lim s3 x l2 3–9 fx x l1 7 1 t s1 ■ 2x2 xl 2 1 4. lim 5x 1 4 x3 7. lim 1 3x 4 x 2 3x 4 1 x l1 x l2 x2 6x 1 ul 2 ■ ■ ■ ■ ■ ■ ■ ■ 3 t 3 t ■ ■ ■ ■ ■ 10. (a) What is wrong with the following equation? x2 x x 6 2 x ■ ■ x2 x l2 x x 6 2 5 s1 x 3x 11. lim x 2 13. lim s3 ■ lim x tl 3 x x xl 4 6 2 x xl4 3 16. lim 5x 3x x2 14. lim 9 7t x x2 12. lim 2 t 2t 2 2 x xl 1 x2 4 4 4x 3x 4 2 2 17. lim 4 hl0 19. lim hl0 1 h h 16 h4 h 1 3 18. lim x l1 20. lim h l0 s3 x x x 2 sin 4x 3x 4 x x2 1 1 2 h3 h 0 x Illustrate by graphing the functions f, t, and h (in the notation of the Squeeze Theorem) on the same screen. 35. If 1 x2 fx 36. If 3x fx 2x x3 37. Prove that lim x 4 cos x l0 38. Prove that lim sx 1 x l0 2 ■ 1 ; 34. Use the Squeeze Theorem to show that 3 x l2 x l0 6 ■ to estimate the value of lim x l 0 f x to two decimal places. (b) Use a table of values of f x to estimate the limit to four decimal places. (c) Use the Limit Laws to find the exact value of the limit. lim sx 3 2 x2 x l2 15. lim x x x l2 ■ lim x l 0 x 2 cos 20 x 0. Illustrate by graphing the functions fx x 2, t x x 2 cos 20 x, and h x x 2 on the same screen. Evaluate the limit, if it exists. |||| ■ ; 32. (a) Use a graph of is correct. 11–30 ■ ; 33. Use the Squeeze Theorem to show that 3 (b) In view of part (a), explain why the equation lim 1 3 by graphing the function f x x (s1 3x 1). (b) Make a table of values of f x for x close to 0 and guess the value of the limit. (c) Use the Limit Laws to prove that your guess is correct. 6 ■ 1 h sx x 2 1 sx x2 xl4 3 t h x l1 fx 9. lim s16 1 30. lim 1 t x l0 3 8. lim su 4 t2 hl0 lim 4 3u tl 1 1 ; 31. (a) Estimate the value of 1 6. lim t 2 5. lim x 2 xl3 2x x 1 t 28. lim Evaluate the limit and justify each step by indicating the appropriate Limit Law(s). 3. lim 3x 4 16 2 tl0 |||| 2 x4 x x l2 81 3 ■ (e) lim x 3f x 3 x2 sx x l9 fx tx (d) lim 24. lim 26. lim 2 x h h h l0 1 x x 27. lim tx x l1 (c) lim f x t x sx s1 22. lim 1 4 25. lim xl 4 4 x 1 (b) lim f x tx x l2 t st x l7 y=© 1 9 3 2 for all x, find lim x l 2 for 0 2 x 1 f x. 2, evaluate lim x l 1 f x . x 0. sin2 2 x 0. 39–44 |||| Find the limit, if it exists. If the limit does not exist, explain why. 8 39. lim x xl 4 4 40. lim xl 4 x x 4 4 5E-02(pp 086-097) 1/17/06 12:51 PM Page 91 ❙❙❙❙ S ECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS x x 41. lim x l2 43. lim x l0 ■ 2 2 1 x ■ 42. lim x l 1.5 1 x ■ 2x 2 2x x l0 ■ ■ ■ ■ ■ (b) If n is an integer, evaluate (i) lim f x (ii) lim f x 3x 3 1 x 44. lim x ln 51. If f x ■ ■ ■ 45. The signum (or sign) function, denoted by sgn, is defined by 1 0 1 sgn x if x if x if x xl0 xl0 x , show that lim x l 2 f x exists but is not 52. In the theory of relativity, the Lorentz contraction formula xl0 x2 1 fx x2 x 54. If r is a rational function, use Exercise 53 to show that if x if x r a for every number a in the domain of r. x l1 x2 0 fx 2 2 prove that lim x l 0 f x 1 . 1 57. Show by means of an example that limx l a f x t x may exist even though neither lim x l a f x nor limx l a t x exists. 58. Evaluate lim (ii) lim F x x l2 s6 s3 x x 2 . 1 x l1 59. Is there a number a such that lim xl 2 48. Let x x2 8 if x if 0 x if x 0 x 2 xl0 (v) lim h x xl2 xl2 x l1 (vi) lim h x x l2 y 49. (a) If the symbol denotes the greatest integer function defined in Example 10, evaluate (i) lim x (ii) lim x (iii) lim x xl 2 (b) If n is an integer, evaluate (i) lim x (ii) lim x x ln xln (c) For what values of a does lim x l a x exist? x x. (a) Sketch the graph of f. ax a 3 x2 x 1 2 y 2 1 and a shrinking circle C2 with radius r and center the origin. P is the point 0, r , Q is the upper point of intersection of the two circles, and R is the point of intersection of the line PQ and the x-axis. What happens to R as C2 shrinks, that is, as r l 0 ? (b) Sketch the graph of h. xl 2 x2 60. The figure shows a fixed circle C1 with equation 2 xl0 (iv) lim h x 3x 2 exists? If so, find the value of a and the value of the limit. (a) Evaluate each of the following limits, if it exists. (i) lim h x (ii) lim h x (iii) lim h x 50. Let f x 0. t x may exist even though neither limx l a f x nor limx l a t x exists. (b) Does lim x l 1 F x exist? (c) Sketch the graph of F . hx if x is rational if x is irrational 56. Show by means of an example that lim x l a f x (a) Find (i) lim F x pa. 55. If (a) Find lim x l 2 f x and lim x l 2 f x . (b) Does lim x l 2 f x exist? (c) Sketch the graph of f . 47. Let F x v2 c2 53. If p is a polynomial, show that lim xl a p x 46. Let 4 x L 0 s1 expresses the length L of an object as a function of its velocity v with respect to an observer, where L 0 is the length of the object at rest and c is the speed of light. Find lim v l c L and interpret the result. Why is a left-hand limit necessary? lim x l a r x (iv) lim sgn x xl0 x equal to f 2 . L 0 0 0 (a) Sketch the graph of this function. (b) Find each of the following limits or explain why it does not exist. (i) lim sgn x (ii) lim sgn x (iii) lim sgn x x ln (c) For what values of a does lim x l a f x exist? 1 x ■ 91 x l 2.4 P Q C™ 0 R C¡ x 5E-02(pp 086-097) 92 ❙❙❙❙ 1/17/06 12:52 PM Page 92 CHAPTER 2 LIMITS AND RATES OF CHANGE |||| 2.4 The Precise Definition of a Limit The intuitive definition of a limit given in Section 2.2 is inadequate for some purposes because such phrases as “x is close to 2” and “ f x gets closer and closer to L” are vague. In order to be able to prove conclusively that cos 5 x 10,000 lim x 3 xl0 0.0001 or lim xl0 sin x x 1 we must make the definition of a limit precise. To motivate the precise definition of a limit, let’s consider the function 2x 6 fx 1 if x if x 3 3 Intuitively, it is clear that when x is close to 3 but x 3, then f x is close to 5, and so lim x l 3 f x 5. To obtain more detailed information about how f x varies when x is close to 3, we ask the following question: How close to 3 does x have to be so that f x differs from 5 by less than 0.l? |||| It is traditional to use the Greek letter (delta) in this situation. The distance from x to 3 is x 3 and the distance from f x to 5 is f x problem is to find a number such that fx 5 If x 3 0, then x ber such that 0.1 x that is, 5 3 fx x 3 but x 3 3, so an equivalent formulation of our problem is to find a numfx Notice that if 0 if 5 , so our 0.1 0.1 2 5 2x fx 5 1 0.1 if 0 x 3 0.05, then 5 if 2x 6 2x 0 x 3 3 0.1 0.05 Thus, an answer to the problem is given by 0.05; that is, if x is within a distance of 0.05 from 3, then f x will be within a distance of 0.1 from 5. If we change the number 0.l in our problem to the smaller number 0.01, then by using the same method we find that f x will differ from 5 by less than 0.01 provided that x differs from 3 by less than (0.01) 2 0.005: fx 5 0.01 if 0 x 3 0.005 0.001 if 0 x 3 0.0005 Similarly, fx 5 The numbers 0.1, 0.01, and 0.001 that we have considered are error tolerances that we might allow. For 5 to be the precise limit of f x as x approaches 3, we must not only be able to bring the difference between f x and 5 below each of these three numbers; we 5E-02(pp 086-097) 1/17/06 12:52 PM Page 93 S ECTION 2.4 THE PRECISE DEFINITION OF A LIMIT y ƒ is in here 93 must be able to bring it below any positive number. And, by the same reasoning, we can! If we write (the Greek letter epsilon) for an arbitrary positive number, then we find as before that 5+∑ 5 5-∑ fx 1 0 x 3 3-∂ 3+∂ when x is in here (x≠3) FIGURE 1 ❙❙❙❙ 5 if 0 x 3 2 This is a precise way of saying that f x is close to 5 when x is close to 3 because (1) says that we can make the values of f x within an arbitrary distance from 5 by taking the values of x within a distance 2 from 3 (but x 3). Note that (1) can be rewritten as 5 fx 5 whenever 3 x 3 x 3 and this is illustrated in Figure 1. By taking the values of x ( 3) to lie in the interval 3 ,3 we can make the values of f x lie in the interval 5 ,5 . Using (1) as a model, we give a precise definition of a limit. 2 Definition Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then we say that the limit of f x as x approaches a is L, and we write lim f x xla if for every number 0 there is a number fx L whenever L 0 such that 0 x a Another way of writing the last line of this definition is if 0 x a then fx L Since x a is the distance from x to a and f x L is the distance from f x to L, and since can be arbitrarily small, the definition of a limit can be expressed in words as follows: L means that the distance between f x and L can be made arbitrarily small lim x l a f x by taking the distance from x to a sufficiently small (but not 0). Alternatively, L means that the values of f x can be made as close as we please to L lim x l a f x by taking x close enough to a (but not equal to a). We can also reformulate Definition 2 in terms of intervals by observing that the inequality x a is equivalent to xa , which in turn can be written as a xa . Also 0 x a is true if and only if x a 0, that is, x a. Similarly, the inequality f x L is equivalent to the pair of inequalities L fx L . Therefore, in terms of intervals, Definition 2 can be stated as follows: L means that for every lim x l a f x 0 (no matter how small and x 0 such that if x lies in the open interval a ,a the open interval L ,L . is) we can find a, then f x lies in 5E-02(pp 086-097) ❙❙❙❙ 94 1/17/06 12:53 PM Page 94 CHAPTER 2 LIMITS AND RATES OF CHANGE We interpret this statement geometrically by representing a function by an arrow diagram as in Figure 2, where f maps a subset of onto another subset of . f FIGURE 2 x a f(a) ƒ The definition of limit says that if any small interval L ,L is given around L, then we can find an interval a ,a around a such that f maps all the points in a ,a (except possibly a) into the interval L ,L . (See Figure 3.) f x FIGURE 3 ƒ a a-∂ a+∂ L-∑ L L+∑ Another geometric interpretation of limits can be given in terms of the graph of a function. If 0 is given, then we draw the horizontal lines y L and y L and the graph of f (see Figure 4). If lim x l a f x L, then we can find a number 0 such that if we restrict x to lie in the interval a and take x a, then the curve ,a lies between the lines y L and y L . (See Figure 5.) You can see that y fx if such a has been found, then any smaller will also work. It is important to realize that the process illustrated in Figures 4 and 5 must work for every positive number no matter how small it is chosen. Figure 6 shows that if a smaller is chosen, then a smaller may be required. y y y y=ƒ L+∑ y=L+∑ L ƒ is in here ∑ ∑ y=L-∑ y=L+∑ L y=L+∑ ∑ ∑ y=L-∑ y=L-∑ L-∑ 0 a x 0 a-∂ 0 x a a-∂ a+∂ x a a+∂ when x is in here (x≠ a) F IGURE 4 FIGURE 5 FIGURE 6 EXAMPLE 1 Use a graph to find a number x3 5x 6 2 0.2 such that whenever In other words, find a number that corresponds to the function f x x 3 5x 6 with a 1 and L x 1 0.2 in the definition of a limit for 2. 5E-02(pp 086-097) 1/17/06 12:53 PM Page 95 S ECTION 2.4 THE PRECISE DEFINITION OF A LIMIT 15 ❙❙❙❙ 95 SOLUTION A graph of f is shown in Figure 7; we are interested in the region near the point 1, 2 . Notice that we can rewrite the inequality x3 _3 3 6 2 1.8 as 5x x3 5x 6 0.2 2.2 So we need to determine the values of x for which the curve y x 3 5x 6 lies between the horizontal lines y 1.8 and y 2.2. Therefore, we graph the curves y x 3 5x 6, y 1.8, and y 2.2 near the point 1, 2 in Figure 8. Then we use the cursor to estimate that the x-coordinate of the point of intersection of the line y 2.2 and the curve y x 3 5x 6 is about 0.911. Similarly, y x 3 5x 6 intersects the line y 1.8 when x 1.124. So, rounding to be safe, we can say that _5 FIGURE 7 2.3 y=2.2 y=˛-5x+6 y=1.8 0.8 1.7 FIGURE 8 x3 1.8 (1, 2) 1.2 5x 6 2.2 whenever 0.92 x 1.12 This interval 0.92, 1.12 is not symmetric about x 1. The distance from x 1 to the left endpoint is 1 0.92 0.08 and the distance to the right endpoint is 0.12. We can choose to be the smaller of these numbers, that is, 0.08. Then we can rewrite our inequalities in terms of distances as follows: x3 5x 6 2 0.2 whenever x 1 0.08 This just says that by keeping x within 0.08 of 1, we are able to keep f x within 0.2 of 2. Although we chose 0.08, any smaller positive value of would also have worked. The graphical procedure in Example 1 gives an illustration of the definition for 0.2, but it does not prove that the limit is equal to 2. A proof has to provide a for every . In proving limit statements it may be helpful to think of the definition of limit as a challenge. First it challenges you with a number . Then you must be able to produce a suitable . You have to be able to do this for every 0, not just a particular . Imagine a contest between two people, A and B, and imagine yourself to be B. Person A stipulates that the fixed number L should be approximated by the values of f x to within a degree of accuracy (say, 0.01). Person B then responds by finding a number such that fx L xa whenever 0 . Then A may become more exacting and challenge B with a smaller value of (say, 0.0001). Again B has to respond by finding a corresponding . Usually the smaller the value of , the smaller the corresponding value L. of must be. If B always wins, no matter how small A makes , then lim x l a f x E XAMPLE 2 Prove that lim 4 x x l3 5 7. SOLUTION 1. Preliminary analysis of the problem ( guessing a value for positive number. We want to find a number 4x But 4x 5 7 5 7 4x 4x 3 whenever 12 4x ). Let be a given such that 3 whenever 0 4x 0 x 3 3 . Therefore, we want x 3 5E-02(pp 086-097) 96 ❙❙❙❙ 1/17/06 12:54 PM Page 96 CHAPTER 2 LIMITS AND RATES OF CHANGE that is, 3 whenever 4 0 This suggests that we should choose 4. 2. Proof (showing that this works ). Given , then 0 x3 y y=4x-5 7+∑ x x 3 0, choose 4. If 7 4x 7-∑ 5 7 4x 12 4x 3 4 4 4 Thus 4x 5 7 whenever 0 x 3 Therefore, by the definition of a limit, 0 3 3-∂ FIGURE 9 x lim 4 x 3+∂ x l3 5 7 This example is illustrated by Figure 9. Note that in the solution of Example 2 there were two stages—guessing and proving. We made a preliminary analysis that enabled us to guess a value for . But then in the second stage we had to go back and prove in a careful, logical fashion that we had made a correct guess. This procedure is typical of much of mathematics. Sometimes it is necessary to first make an intelligent guess about the answer to a problem and then prove that the guess is correct. The intuitive definitions of one-sided limits that were given in Section 2.2 can be precisely reformulated as follows. 3 Definition of Left-Hand Limit lim f x xla if for every number fx 4 L 0 there is a number L 0 such that a whenever x a Definition of Right-Hand Limit lim f x xla if for every number fx 0 there is a number L whenever L 0 such that a x a Notice that Definition 3 is the same as Definition 2 except that x is restricted to lie in the left half a . In Definition 4, x is restricted to lie , a of the interval a ,a in the right half a, a of the interval a ,a . E XAMPLE 3 Use Definition 4 to prove that lim s x xl0 0. 5E-02(pp 086-097) 1/17/06 12:54 PM Page 97 S ECTION 2.4 THE PRECISE DEFINITION OF A LIMIT |||| CAUCHY AND LIMITS After the invention of calculus in the 17th century, there followed a period of free development of the subject in the 18th century. Mathematicians like the Bernoulli brothers and Euler were eager to exploit the power of calculus and boldly explored the consequences of this new and wonderful mathematical theory without worrying too much about whether their proofs were completely correct. The 19th century, by contrast, was the Age of Rigor in mathematics. There was a movement to go back to the foundations of the subject—to provide careful definitions and rigorous proofs. At the forefront of this movement was the French mathematician Augustin-Louis Cauchy (1789–1857), who started out as a military engineer before becoming a mathematics professor in Paris. Cauchy took Newton’s idea of a limit, which was kept alive in the 18th century by the French mathematician Jean d’Alembert, and made it more precise. His definition of a limit reads as follows: “When the successive values attributed to a variable approach indefinitely a fixed value so as to end by differing from it by as little as one wishes, this last is called the limit of all the others.” But when Cauchy used this definition in examples and proofs, he often employed delta-epsilon inequalities similar to the ones in this section. A typical Cauchy proof starts with: “Designate by and two very small numbers; . . .” He used because of the correspondence between epsilon and the French word erreur. Later, the German mathematician Karl Weierstrass (1815–1897) stated the definition of a limit exactly as in our Definition 2. SOLUTION 1. Guessing a value for . Let be a given positive number. Here a such that so we want to find a number sx 0 0 whenever 0 0 and L 97 0, x sx that is, whenever ❙❙❙❙ x or, squaring both sides of the inequality s x 2 x , we get whenever . 2 0, let s s so x 2 This suggests that we should choose 2. Showing that this works. Given sx 0 sx x , then 2 0 According to Definition 4, this shows that lim x l 0 sx EXAMPLE 4 Prove that lim x 2 . If 0 0. 9. xl3 SOLUTION 1. Guessing a value for . Let 0 be given. We have to find a number 0 such that x2 To connect x 2 want 9 whenever 3 we write x 2 9 with x x 3 x 0 3 x 3 x 3x 9 whenever 0 x Notice that if we can find a positive constant C such that x x 3 x 3 Cx 3 . Then we 3 3 C, then 3 and we can make C x 3 by taking x 3 C . We can find such a number C if we restrict x to lie in some interval centered at 3. In fact, since we are interested only in values of x that are close to 3, it is reasonable to assume that x is within a distance l from 3, that is, x 3 1. Then 2 x 4, so 5 x 3 7. Thus, we have x 3 7, and so C 7 is a suitable choice for the constant. But now there are two restrictions on x 3 , namely x 3 1 and x 3 C 7 To make sure that both of these inequalities are satisfied, we take to be the smaller of the two numbers 1 and 7. The notation for this is min 1, 7 . 2. Showing that this works. Given 0, let min 1, 7 . If 0 x3 , then x 3 1?2 x 4? x 3 7 (as in part l). We also have x3 7, so x2 9 This shows that lim x l 3 x 2 9. x 3 x 3 7 7 5E-02(pp 098-109) 98 ❙❙❙❙ 1/19/06 4:19 PM Page 98 CHAPTER 2 LIMITS AND RATES OF CHANGE As Example 4 shows, it is not always easy to prove that limit statements are true using the , definition. In fact, if we had been given a more complicated function such as f x 6 x 2 8 x 9 2 x 2 1 , a proof would require a great deal of ingenuity. Fortunately this is unnecessary because the Limit Laws stated in Section 2.3 can be proved using Definition 2, and then the limits of complicated functions can be found rigorously from the Limit Laws without resorting to the definition directly. For instance, we prove the Sum Law: If lim x l a f x L and lim x l a t x M both exist, then lim f x tx xla L M The remaining laws are proved in the exercises and in Appendix F. Proof of the Sum Law Let fx b tx L M 0 such that whenever 0 x a Using the Triangle Inequality we can write |||| Triangle Inequality: a 0 be given. We must find a b fx 5 tx L M fx (See Appendix A.) L fx We make f x tx LM and t x M less than 2. Since 2 0 and lim x l a f x fx L Let min if 0 , 1 x 2 L whenever 0 whenever 2 M by making each of the terms f x a 1 0 such that 2 0 x a 1 and 0 tx M 2 . Notice that a and so then 0 fx L x a and 2 x a 2 Therefore, by (5), fx tx L M fx 2 L tx M 2 To summarize, fx L 0 such that 1 x M , there exists a number M M tx L, there exists a number 2 Similarly, since lim x l a t x tx less than tx tx L M whenever 0 Thus, by the definition of a limit, lim f x xla tx L M x a 2 5E-02(pp 098-109) 1/19/06 4:19 PM Page 99 SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ❙❙❙❙ 99 I nfinite Limits Infinite limits can also be defined in a precise way. The following is a precise version of Definition 4 in Section 2.2. 6 Definition Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then lim f x xla y means that for every positive number M there is a positive number y=M M 0 x a a-∂ FIGURE 10 a+∂ fx M 0 whenever x such that a This says that the values of f x can be made arbitrarily large (larger than any given number M ) by taking x close enough to a (within a distance , where depends on M , but with x a). A geometric illustration is shown in Figure 10. 0 such that if we restrict Given any horizontal line y M , we can find a number x to lie in the interval a ,a but x a, then the curve y f x lies above the line y M . You can see that if a larger M is chosen, then a smaller may be required. EXAMPLE 5 Use Definition 6 to prove that lim xl0 SOLUTION 1. Guessing a value for . Given M 1 x2 . 0, we want to find 1 x2 M whenever 0 x that is, x2 1 M whenever 0 x or x 1 sM whenever 0 0 such that x This suggests that we should take 2. Showing that this works. If M 1 sM. 0 is given, let 0 1 sM . If 0 then Thus 1 x2 M ? x2 2 ? x 1 x2 1 whenever Therefore, by Definition 6, lim xl0 1 x2 M 2 0 x 0 x 0 , 5E-02(pp 098-109) ❙❙❙❙ 100 1/19/06 4:19 PM Page 100 CHAPTER 2 LIMITS AND RATES OF CHANGE y a-∂ Similarly, the following is a precise version of Definition 5 in Section 2.2. It is illustrated by Figure 11. a+∂ a 0 7 Definition Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then x lim f x y=N N xla means that for every negative number N there is a positive number such that FIGURE 11 fx |||| 2.4 N whenever 0 x a Exercises 1. How close to 2 do we have to take x so that 5 x 3 is within a y 1 is within a distance of (a) 0.01, (b) 0.001, and (c) 0.0001 from 29? 2.4 2 1.6 y=œx „ distance of (a) 0.1 and (b) 0.01 from 13? 2. How close to 5 do we have to take x so that 6 x 3. Use the given graph of f x 1 x to find a number such that 0 1 x 0.5 0.2 x whenever ? x ? 4 2 x 2 to find a number 6. Use the given graph of f x such that y x2 1 y= x 1 1 2 1 whenever x 1 y 0.7 0.5 y=≈ 1.5 0.3 1 0 10 7 2 4. Use the given graph of f to find a number fx 3 0.6 x 10 3 whenever 0.5 such that x 0 0 5 ? s4x 3.6 3 2.4 1 0.5 3 whenever x 2 ; 8. Use a graph to find a number such that | sin x 0 4 | 0.1 5. Use the given graph of f x 0.4 sx to find a number whenever x 4 whenever x 6 ; 9. For the limit x 5 5.7 1 2 lim 4 2 x ? ; 7. Use a graph to find a number such that y sx 1 such that xl1 x 3x 3 2 illustrate Definition 2 by finding values of 1 and 0.1. that correspond to 5E-02(pp 098-109) 1/19/06 4:20 PM Page 101 S ECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ; 10. For the limit 21. lim 4x lim x l 2 3x xl 5 1 4 4.5 23. lim x 25. lim x 2 that correspond to x 1x 100 2 1 0 whenever x 4x 1 x l2 xl 2 ; 12. For the limit lim cot 2x ■ ■ 0 xl0 29. lim x 2 1 4 28. lim s9 x xl9 ■ 5 30. lim x 2 1 ■ 32. lim x 3 ■ ■ ■ 4 8 8 x l2 ■ 0 x x l3 3 7 3 26. lim x 3 0 12 101 c xla 31. lim x 2 x x x 24. lim c 27. lim x xl0 x2 x l3 0 xl0 ; 11. Use a graph to find a number such that 22. lim 7 a xla illustrate Definition 2 by finding values of 0.5 and 0.1. 2 3x 5 4 ❙❙❙❙ ■ ■ ■ ■ xl0 illustrate Definition 6 by finding values of (a) M 100 and (b) M 1000. 33. Verify that another possible choice of that correspond to lim x l3 x 2 34. Verify, by a geometric argument, that the largest possible 13. A machinist is required to manufacture a circular metal disk with area 1000 cm2. (a) What radius produces such a disk? (b) If the machinist is allowed an error tolerance of 5 cm2 in the area of the disk, how close to the ideal radius in part (a) must the machinist control the radius? (c) In terms of the , definition of lim x l a f x L, what is x ? What is f x ? What is a? What is L ? What value of is given? What is the corresponding value of ? ; 14. A crystal growth furnace is used in research to determine how best to manufacture crystals used in electronic components for the space shuttle. For proper growth of the crystal, the temperature must be controlled accurately by adjusting the input power. Suppose the relationship is given by choice of CAS for showing that lim x l3 x 2 35. (a) For the limit lim x l 1 x 0.1w Tw 2.155w 36. Prove that lim x l2 1 x Prove the statement using the , illustrate with a diagram like Figure 9. sa if a xla | 15. lim 2 x 3 17. lim 1 4x xl1 xl 3 ■ ■ 19–32 ■ |||| 19. lim x l3 x 5 16. lim 5 xl 2 ■ ■ ■ ■ x sx | a . sa tion 2.2, prove, using Definition 2, that lim t l 0 H t does not exist. [Hint : Use an indirect proof as follows. Suppose that the 1 limit is L. Take 2 in the definition of a limit and try to arrive at a contradiction.] 39. If the function f is defined by 0 1 fx if x is rational if x is irrational prove that lim x l 0 f x does not exist. 40. By comparing Definitions 2, 3, and 4, prove Theorem 1 in xl6 2 3x ■ 20. lim 3) x xl4 Prove the statement using the , 3 5 ( 18. lim 7 13 sa 0. Section 2.3. 1 2 41. How close to 5 ■ ■ ■ definition of limit. x 4 3 9 2 3. 38. If H is the Heaviside function defined in Example 6 in Sec- definition of limit and |||| s9 1 . 2 37. Prove that lim s x 20 where T is the temperature in degrees Celsius and w is the power input in watts. (a) How much power is needed to maintain the temperature at 200 C ? (b) If the temperature is allowed to vary from 200 C by up to 1 C , what range of wattage is allowed for the input power? (c) In terms of the , definition of lim x l a f x L, what is x ? What is f x ? What is a? What is L ? What value of is given? What is the corresponding value of ? 15–18 9 is 3 x1 3, use a graph to find a value of that corresponds to 0.4. (b) By using a computer algebra system to solve the cubic equation x 3 x 1 3 , find the largest possible value of that works for any given 0. (c) Put 0.4 in your answer to part (b) and compare with your answer to part (a). Hint: Use s x 2 for showing that min 2, 8 . 9 in Example 4 is 3 do we have to take x so that 1 10,000 x 34 ■ 42. Prove, using Definition 6, that lim xl 3 43. Prove that lim xl 1 5 x 1 3 . 1 x 34 . 5E-02(pp 098-109) 102 ❙❙❙❙ 1/19/06 4:20 PM Page 102 CHAPTER 2 LIMITS AND RATES OF CHANGE 44. Suppose that lim x l a f x and lim x l a t x a real number. Prove each statement. (a) lim f x tx xla |||| 2.5 c, where c is (b) lim f x t x if c xla 0 if c (c) lim f x t x xla 0 Continuity We noticed in Section 2.3 that the limit of a function as x approaches a can often be found simply by calculating the value of the function at a. Functions with this property are called continuous at a. We will see that the mathematical definition of continuity corresponds closely with the meaning of the word continuity in everyday language. (A continuous process is one that takes place gradually, without interruption or abrupt change.) Explore continuous functions interactively. Resources / Module 2 / Continuity / Start of Continuity 1 Definition A function f is continuous at a number a if lim f x x la |||| As illustrated in Figure 1, if f is continuous, then the points x, f x on the graph of f approach the point a, f a on the graph. So there is no gap in the curve. fa Notice that Definition l implicitly requires three things if f is continuous at a: 1. f a is defined (that is, a is in the domain of f ) 2. lim f x exists x la y ƒ approaches f(a). 3. lim f x y=ƒ x la f(a) 0 x a As x approaches a, FIGURE 1 y fa The definition says that f is continuous at a if f x approaches f a as x approaches a. Thus, a continuous function f has the property that a small change in x produces only a small change in f x . In fact, the change in f x can be kept as small as we please by keeping the change in x sufficiently small. If f is defined near a (in other words, f is defined on an open interval containing a, except perhaps at a), we say that f is discontinuous at a, or f has a discontinuity at a, if f is not continuous at a. Physical phenomena are usually continuous. For instance, the displacement or velocity of a vehicle varies continuously with time, as does a person’s height. But discontinuities do occur in such situations as electric currents. [See Example 6 in Section 2.2, where the Heaviside function is discontinuous at 0 because lim t l 0 H t does not exist.] Geometrically, you can think of a function that is continuous at every number in an interval as a function whose graph has no break in it. The graph can be drawn without removing your pen from the paper. EXAMPLE 1 Figure 2 shows the graph of a function f. At which numbers is f discontinu- ous? Why? 0 1 FIGURE 2 2 3 4 5 x SOLUTION It looks as if there is a discontinuity when a 1 because the graph has a break there. The official reason that f is discontinuous at 1 is that f 1 is not defined. The graph also has a break when a 3, but the reason for the discontinuity is different. Here, f 3 is defined, but lim x l 3 f x does not exist (because the left and right limits are different). So f is discontinuous at 3. What about a 5? Here, f 5 is defined and lim x l 5 f x exists (because the left and right limits are the same). But lim f x f5 xl5 So f is discontinuous at 5. 5E-02(pp 098-109) 1/19/06 4:20 PM Page 103 S ECTION 2.5 CONTINUITY ❙❙❙❙ 103 Now let’s see how to detect discontinuities when a function is defined by a formula. EXAMPLE 2 Where are each of the following functions discontinuous? x2 (a) f x Resources / Module 2 / Continuity / Problems and Tests x 2 x 2 x2 x (c) f x 2 2 1 if x 2 if x x 1 x2 1 (b) f x if x 0 if x 0 2 (d) f x x SOLUTION (a) Notice that f 2 is not defined, so f is discontinuous at 2. Later we’ll see why f is continuous at all other numbers. (b) Here f 0 1 is defined but lim f x lim xl0 xl0 1 x2 does not exist. (See Example 8 in Section 2.2.) So f is discontinuous at 0. (c) Here f 2 1 is defined and lim f x x l2 lim x l2 x2 x x 2 lim 2 x 2x x2 x l2 1 lim x x l2 1 3 exists. But lim f x f2 x l2 so f is not continuous at 2. (d) The greatest integer function f x x has discontinuities at all of the integers because lim x l n x does not exist if n is an integer. (See Example 10 and Exercise 49 in Section 2.3.) Figure 3 shows the graphs of the functions in Example 2. In each case the graph can’t be drawn without lifting the pen from the paper because a hole or break or jump occurs in the graph. The kind of discontinuity illustrated in parts (a) and (c) is called removable because we could remove the discontinuity by redefining f at just the single number 2. [The function t x x 1 is continuous.] The discontinuity in part (b) is called an infinite discontinuity. The discontinuities in part (d) are called jump discontinuities because the function “jumps” from one value to another. y y y y 1 1 1 1 0 (a) ƒ= FIGURE 3 1 2 ≈-x-2 x-2 0 x (b) ƒ= Graphs of the functions in Example 2 0 x 1/≈ if x≠0 1 if x=0 (c) ƒ= 1 2 x ≈-x-2 if x≠2 x-2 1 if x=2 0 1 2 (d) ƒ=[ x ] 3 x 5E-02(pp 098-109) 104 ❙❙❙❙ 1/19/06 4:20 PM Page 104 CHAPTER 2 LIMITS AND RATES OF CHANGE 2 Definition A function f is continuous from the right at a number a if lim f x fa xla and f is continuous from the left at a if lim f x fa xla EXAMPLE 3 At each integer n, the function f x x [see Figure 3(d)] is continuous from the right but discontinuous from the left because lim f x lim x x ln but n x ln lim f x lim x x ln fn n x ln 1 fn 3 Definition A function f is continuous on an interval if it is continuous at every number in the interval. (If f is defined only on one side of an endpoint of the interval, we understand continuous at the endpoint to mean continuous from the right or continuous from the left.) EXAMPLE 4 Show that the function f x interval S OLUTION If 1 x 2 is continuous on the s1 1, 1 . 1 a 1, then using the Limit Laws, we have lim (1 lim f x xla x2) s1 xla lim s1 x2 1 s xlim la 1 x2 1 s1 a2 1 xla (by Laws 2 and 7) (by 11) (by 2, 7, and 9) fa y Thus, by Definition l, f is continuous at a if ƒ=1 -œ„„„„„ 1 -≈ 1 0 FIGURE 4 lim f x xl 1 1 x 1 f 1 1 a and 1. Similar calculations show that lim f x x l1 1 f1 so f is continuous from the right at 1 and continuous from the left at 1. Therefore, according to Definition 3, f is continuous on 1, 1 . The graph of f is sketched in Figure 4. It is the lower half of the circle x2 y 1 2 1 Instead of always using Definitions 1, 2, and 3 to verify the continuity of a function as we did in Example 4, it is often convenient to use the next theorem, which shows how to build up complicated continuous functions from simple ones. 5E-02(pp 098-109) 1/19/06 4:20 PM Page 105 SECTION 2.5 CONTINUITY ❙❙❙❙ 105 4 Theorem If f and t are continuous at a and c is a constant, then the following functions are also continuous at a : 1. f t 2. f t 3. cf f if t a 0 4. f t 5. t Proof Each of the five parts of this theorem follows from the corresponding Limit Law in Section 2.3. For instance, we give the proof of part 1. Since f and t are continuous at a, we have lim f x fa xla and lim t x xla ta Therefore lim f xla tx lim f x tx xla lim f x lim t x xla fa ta f This shows that f (by Law 1) xla ta t is continuous at a. It follows from Theorem 4 and Definition 3 that if f and t are continuous on an interval, then so are the functions f t, f t, c f, f t, and (if t is never 0) f t. The following theorem was stated in Section 2.3 as the Direct Substitution Property. 5 Theorem (a) Any polynomial is continuous everywhere; that is, it is continuous on ,. (b) Any rational function is continuous wherever it is defined; that is, it is continuous on its domain. Proof (a) A polynomial is a function of the form Px cn x n cn 1 x n 1 c1 x c0 where c0 , c1, . . . , cn are constants. We know that lim c0 xla and lim x m xla am c0 m (by Law 7) 1, 2, . . . , n (by 9) This equation is precisely the statement that the function f x x m is a continuous m function. Thus, by part 3 of Theorem 4, the function t x c x is continuous. Since P is a sum of functions of this form and a constant function, it follows from part 1 of Theorem 4 that P is continuous. 5E-02(pp 098-109) 106 ❙❙❙❙ 1/19/06 4:20 PM Page 106 CHAPTER 2 LIMITS AND RATES OF CHANGE (b) A rational function is a function of the form Px Qx fx where P and Q are polynomials. The domain of f is D x Qx 0 . We know from part (a) that P and Q are continuous everywhere. Thus, by part 5 of Theorem 4, f is continuous at every number in D. As an illustration of Theorem 5, observe that the volume of a sphere varies continuously 4 3 with its radius because the formula V r 3 r shows that V is a polynomial function of r. Likewise, if a ball is thrown vertically into the air with a velocity of 50 ft s, then the height of the ball in feet after t seconds is given by the formula h 50t 16t 2. Again this is a polynomial function, so the height is a continuous function of the elapsed time. Knowledge of which functions are continuous enables us to evaluate some limits very quickly, as the following example shows. Compare it with Example 2(b) in Section 2.3. EXAMPLE 5 Find lim x3 xl 2 2x2 1 . 5 3x SOLUTION The function x3 fx 2x 2 1 5 3x is rational, so by Theorem 5 it is continuous on its domain, which is {x x Therefore lim xl 2 x3 2x2 1 5 3x lim f x f xl 2 2 3 5 2 22 32 5 3 }. 2 1 1 11 y P(Ł ¨, à ¨) 1 ¨ 0 (1, 0) x FIGURE 5 |||| Another way to establish the limits in (6) is to use the Squeeze Theorem with the inequality sin 0), which is proved in Sec(for tion 3.5. It turns out that most of the familiar functions are continuous at every number in their domains. For instance, Limit Law 10 (page 84) implies that root functions are continuous. [Example 3 in Section 2.4 shows that f x sx is continuous from the right at 0.] From the appearance of the graphs of the sine and cosine functions (Figure 18 in Section 1.2), we would certainly guess that they are continuous. We know from the definitions of sin and cos that the coordinates of the point P in Figure 5 are cos , sin . As l 0, we see that P approaches the point 1, 0 and so cos l 1 and sin l 0. Thus 6 lim cos l0 1 lim sin l0 0 Since cos 0 1 and sin 0 0, the equations in (6) assert that the cosine and sine functions are continuous at 0. The addition formulas for cosine and sine can then be used to deduce that these functions are continuous everywhere (see Exercises 54 and 55). It follows from part 5 of Theorem 4 that tan x sin x cos x 5E-02(pp 098-109) 1/19/06 4:21 PM Page 107 S ECTION 2.5 CONTINUITY y ❙❙❙❙ 107 is continuous except where cos x 0. This happens when x is an odd integer multiple of 2, so y tan x has infinite discontinuities when x 2, 3 2, 5 2, and so on (see Figure 6). 1 3π _π _2 _ π 2 0 π 2 π 3π 2 x 7 Theorem The following types of functions are continuous at every number in their domains: polynomials root functions FIGURE 6 rational functions trigonometric functions y=tan x EXAMPLE 6 On what intervals is each function continuous? (a) f x x 100 2 x 37 (c) h x sx x x 75 1 1 x x2 x2 (b) t x 2 x 17 x2 1 1 1 SOLUTION (a) f is a polynomial, so it is continuous on , by Theorem 5(a). (b) t is a rational function, so by Theorem 5(b), it is continuous on its domain, which is D x x2 1 0 xx 1 . Thus, t is continuous on the intervals , 1, 1, 1 , and 1, . (c) We can write h x Fx Gx H x , where Fx sx Gx x x 1 1 x x2 Hx 1 1 F is continuous on 0, by Theorem 7. G is a rational function, so it is continuous everywhere except when x 1 0, that is, x 1. H is also a rational function, but its denominator is never 0, so H is continuous everywhere. Thus, by parts 1 and 2 of Theorem 4, h is continuous on the intervals 0, 1 and 1, . Another way of combining continuous functions f and t to get a new continuous function is to form the composite function f t. This fact is a consequence of the following theorem. |||| This theorem says that a limit symbol can be moved through a function symbol if the function is continuous and the limit exists. In other words, the order of these two symbols can be reversed. 8 Theorem If f is continuous at b and lim t x x la In other words, lim f t x xla ( b, then lim f t x f lim t x xla x la f b. ) Intuitively, Theorem 8 is reasonable because if x is close to a, then t x is close to b, and since f is continuous at b, if t x is close to b, then f t x is close to f b . A proof of Theorem 8 is given in Appendix F. n Let’s now apply Theorem 8 in the special case where f x sx , with n being a positive integer. Then f tx n st x 5E-02(pp 098-109) 108 ❙❙❙❙ 1/19/06 4:21 PM Page 108 CHAPTER 2 LIMITS AND RATES OF CHANGE ( f lim t x and xla ) n s xlim t x la If we put these expressions into Theorem 8, we get n lim st x n s xlim t x la xla and so Limit Law 11 has now been proved. (We assume that the roots exist.) 9 Theorem If t is continuous at a and f is continuous at t a , then the composite f t x is continuous at a. function f t given by f t x This theorem is often expressed informally by saying “a continuous function of a continuous function is a continuous function.” Proof Since t is continuous at a, we have lim t x ta xla Since f is continuous at b t a , we can apply Theorem 8 to obtain lim f t x f ta xla which is precisely the statement that the function h x is, f t is continuous at a. f t x is continuous at a; that EXAMPLE 7 Where are the following functions continuous? sin x 2 (a) h x (b) F x 1 sx 2 7 4 SOLUTION (a) We have h x f t x , where tx x2 and fx sin x Now t is continuous on since it is a polynomial, and f is also continuous everywhere. Thus, h f t is continuous on by Theorem 9. (b) Notice that F can be broken up as the composition of four continuous functions: F where fx fthk 1 x tx x Fx or 4 hx f thkx sx kx x2 7 We know that each of these functions is continuous on its domain (by Theorems 5 and 7), so by Theorem 9, F is continuous on its domain, which is {x sx 2 7 4} xx 3 , 3 3, 3 3, An important property of continuous functions is expressed by the following theorem, whose proof is found in more advanced books on calculus. 5E-02(pp 098-109) 1/19/06 4:21 PM Page 109 SECTION 2.5 CONTINUITY ❙❙❙❙ 109 10 The Intermediate Value Theorem Suppose that f is continuous on the closed interval a, b and let N be any number between f a and f b , where f a f b. Then there exists a number c in a, b such that f c N. The Intermediate Value Theorem states that a continuous function takes on every intermediate value between the function values f a and f b . It is illustrated by Figure 7. Note that the value N can be taken on once [as in part (a)] or more than once [as in part (b)]. y y f(b) f(b) y=ƒ N N y=ƒ f(a) 0 a f(a) y=ƒ y=N N f(b) 0 a b x 0 x cb FIGURE 7 y f(a) a c¡ c™ (a) c£ b x (b) If we think of a continuous function as a function whose graph has no hole or break, then it is easy to believe that the Intermediate Value Theorem is true. In geometric terms it says that if any horizontal line y N is given between y f a and y f b as in Figure 8, then the graph of f can’t jump over the line. It must intersect y N somewhere. It is important that the function f in Theorem 10 be continuous. The Intermediate Value Theorem is not true in general for discontinuous functions (see Exercise 42). One use of the Intermediate Value Theorem is in locating roots of equations as in the following example. FIGURE 8 EXAMPLE 8 Show that there is a root of the equation 4x 3 6x 2 3x 2 0 between 1 and 2. 4 x 3 6 x 2 3x 2. We are looking for a solution of the given equation, that is, a number c between 1 and 2 such that f c 0. Therefore, we take a 1, b 2, and N 0 in Theorem 10. We have SOLUTION Let f x f1 f2 and 4 6 32 24 3 2 6 1 2 12 0 0 Thus, f 1 0 f 2 ; that is, N 0 is a number between f 1 and f 2 . Now f is continuous since it is a polynomial, so the Intermediate Value Theorem says there is a number c between 1 and 2 such that f c 0. In other words, the equation 4 x 3 6 x 2 3x 2 0 has at least one root c in the interval 1, 2 . In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f 1.2 0.128 0 and f 1.3 0.548 0 5E-02(pp 110-121) 110 ❙❙❙❙ 1/17/06 12:29 PM Page 110 CHAPTER 2 LIMITS AND RATES OF CHANGE a root must lie between 1.2 and 1.3. A calculator gives, by trial and error, f 1.22 0.007008 0 and f 1.23 0.056068 0 so a root lies in the interval 1.22, 1.23 . We can use a graphing calculator or computer to illustrate the use of the Intermediate Value Theorem in Example 8. Figure 9 shows the graph of f in the viewing rectangle 1, 3 by 3, 3 and you can see that the graph crosses the x-axis between 1 and 2. Figure 10 shows the result of zooming in to the viewing rectangle 1.2, 1.3 by 0.2, 0.2 . 3 0.2 3 _1 1.3 1.2 _3 _0.2 FIGURE 9 FIGURE 10 In fact, the Intermediate Value Theorem plays a role in the very way these graphing devices work. A computer calculates a finite number of points on the graph and turns on the pixels that contain these calculated points. It assumes that the function is continuous and takes on all the intermediate values between two consecutive points. The computer therefore connects the pixels by turning on the intermediate pixels. |||| 2.5 Exercises 1. Write an equation that expresses the fact that a function f y is continuous at the number 4. 2. If f is continuous on , , what can you say about its graph? 3. (a) From the graph of f , state the numbers at which f is discontinuous and explain why. (b) For each of the numbers stated in part (a), determine whether f is continuous from the right, or from the left, or neither. y _4 _2 2 4 6 8 x 5. Sketch the graph of a function that is continuous everywhere except at x 3 and is continuous from the left at 3. 6. Sketch the graph of a function that has a jump discontinuity at x 2 and a removable discontinuity at x ous elsewhere. _4 _2 0 2 4 6 4. From the graph of t, state the intervals on which t is continuous. x 4, but is continu- 7. A parking lot charges $3 for the first hour (or part of an hour) and $2 for each succeeding hour (or part), up to a daily maximum of $10. (a) Sketch a graph of the cost of parking at this lot as a function of the time parked there. (b) Discuss the discontinuities of this function and their significance to someone who parks in the lot. 5E-02(pp 110-121) 1/17/06 12:29 PM Page 111 ❙❙❙❙ S ECTION 2.5 CONTINUITY 8. Explain why each function is continuous or discontinuous. (a) The temperature at a specific location as a function of time (b) The temperature at a specific time as a function of the distance due west from New York City (c) The altitude above sea level as a function of the distance due west from New York City (d) The cost of a taxi ride as a function of the distance traveled (e) The current in the circuit for the lights in a room as a function of time 9. If f and t are continuous functions with f 3 lim x l 3 2 f x 5 and 10–12 |||| Use the definition of continuity and the properties of limits to show that the function is continuous at the given number. 10. f x x2 11. f x x 12. t x s7 x, a 25. h x cos 1 27. F x 1 2 x 3 4, x1 ,a 2x 2 1 ■ ; 29–30 ■ 13–14 ■ ■ ■ ■ ■ ■ Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval. 2x x 3 , 2 14. t x 2 s3 x, ■ ■ ■ ■ ■ 31–32 ■ ■ ■ ■ ■ ■ ■ 15–20 |||| Explain why the function is discontinuous at the given number a. Sketch the graph of the function. 1 1 x 2 a 2 1 16. f x if x 1 1 x2 1x 18. f x x2 x2 1 19. f x if x if x if x ■ ■ ■ 1 x2 x a 1 ■ ■ ■ tan sx ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5 sx s5 x sin x ■ ■ ■ ■ ■ ■ 33–34 |||| ■ Show that f is continuous on 2 33. f x x if x sx if x sin x if x cos x if x , . 1 1 ■ ■ ■ 12 3 if x 3 3 x 2 if x x if x ■ ■ 6 ■ ■ ■ 1 2 x x2 x 2 2 if x if 0 if x 0 x 2 2 1 x 3 3 x 2 if x 2x 2 if 0 2 x if x 0 x 1 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 3 38. The gravitational force exerted by Earth on a unit mass at a dis- tance r from the center of the planet is 1 1 ■ ■ x1 if x 1x if 1 sx 3 if x ■ a ■ 4 4 37. f x 1 if x x 5x ■ 36. f x a ■ ■ ■ 1 ■ ■ ■ 21–28 |||| Explain, using Theorems 4, 5, 7, and 9, why the function is continuous at every number in its domain. State the domain. 21. F x ■ Use continuity to evaluate the limit. 35. f x 1 a 1 5 1 4 ■ 35–37 |||| Find the numbers at which f is discontinuous. At which of these numbers is f continuous from the right, from the left, or neither? Sketch the graph of f . 1 a 1 1 if x x |||| 1 x 1 x2 20. f x 1 if x 17. f x ■ 30. y 34. f x ,3 ■ x ■ 2, ■ 15. f x ■ ■ ■ |||| 13. f x sin cos sin x ■ ■ 1 sin x 1 xl ■ ■ Locate the discontinuities of the function and illustrate by |||| x l4 ■ tan 2 x 28. F x ■ 32. lim sin x ■ x2 graphing. ■ ■ 1 sin x x1 26. h x s2 x ■ 31. lim 4 24. h x sx sin x ■ 4 a x2 29. y 4, find t 3 . tx 23. R x 111 22. G x 3 sx 1 x3 Fr GMr R3 GM r2 if r R if r R where M is the mass of Earth, R is its radius, and G is the gravitational constant. Is F a continuous function of r ? 5E-02(pp 110-121) 112 ❙❙❙❙ 1/17/06 12:30 PM Page 112 CHAPTER 2 LIMITS AND RATES OF CHANGE ; 51–52 39. For what value of the constant c is the function f continuous , on |||| (a) Prove that the equation has at least one real root. (b) Use your graphing device to find the root correct to three decimal places. 1 51. x 5 x2 4 0 52. s x 5 x3 ? cx cx 2 fx 1 1 if x if x 3 3 40. Find the constant c that makes t continuous on , . ■ 2 x cx tx 2 c 20 if x if x 4 4 ■ ■ ■ ■ lim f a lim sin a 47. cos x ■ ■ x, 0, ■ 1 ■ ■ ■ 2 x, ■ ■ ■ ■ 2 ■ |||| 2.6 50. x 5 x ■ ■ ■ ■ ■ x2 ■ 2x ■ 3 ■ 0 ■ if x is rational if x is irrational 58. For what values of x is t continuous? 0 x if x is rational if x is irrational 60. (a) Show that the absolute value function F x 0, 1.4 ■ 0 1 59. Is there a number that is exactly 1 more than its cube? 49–50 |||| (a) Prove that the equation has at least one real root. (b) Use your calculator to find an interval of length 0.01 that contains a root. 49. sin x fx tx 0, 1 x, 48. tan x ■ sin a 57. For what values of x is f continuous? 2. (This proves the existence 3 46. sx 0, 1 ■ fa (b) Prove Theorem 4, part 5. x, show that there is a number c such 1, 2 ■ 56. (a) Prove Theorem 4, part 3. 45–48 |||| Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. 3 ■ 55. Prove that cosine is a continuous function. positive number c such that c 2 of the number s2.) x ■ Use (6) to show that this is true. 44. Use the Intermediate Value Theorem to prove that there is a 45. x 4 h hl0 0.25 and that f 0 1 and f 1 3. Let N 2. Sketch two possible graphs of f , one showing that f might not satisfy the conclusion of the Intermediate Value Theorem and one showing that f might still satisfy the conclusion of the Intermediate Value Theorem (even though it doesn’t satisfy the hypothesis). x3 x2 10. ■ lim x l a sin x sin a for every real number a. By Exercise 53 an equivalent statement is that 42. Suppose that a function f is continuous on [0, 1] except at that f c ■ 54. To prove that sine is continuous, we need to show that x 3 64 ,a 4 x4 3 sx ,a 9 9x 43. If f x h hl0 nuity at a ? If the discontinuity is removable, find a function t that agrees with f for x a and is continuous on . x 2 2x 8 (a) f x ,a 2 x2 x7 (b) f x ,a 7 x7 (d) f x ■ 53. Prove that f is continuous at a if and only if 41. Which of the following functions f has a removable disconti- (c) f x ■ ■ x is continuous everywhere. (b) Prove that if f is a continuous function on an interval, then so is f . (c) Is the converse of the statement in part (b) also true? In other words, if f is continuous, does it follow that f is continuous? If so, prove it. If not, find a counterexample. 61. A Tibetan monk leaves the monastery at 7:00 A.M. and takes his usual path to the top of the mountain, arriving at 7:00 P.M. The following morning, he starts at 7:00 A.M. at the top and takes the same path back, arriving at the monastery at 7:00 P.M. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days. Tangents, Velocities, and Other Rates of Change In Section 2.1 we guessed the values of slopes of tangent lines and velocities on the basis of numerical evidence. Now that we have defined limits and have learned techniques for computing them, we return to the tangent and velocity problems with the ability to calculate slopes of tangents, velocities, and other rates of change. 5E-02(pp 110-121) 1/17/06 12:30 PM Page 113 ❙❙❙❙ S ECTION 2.6 TANGENTS, VELOCITIES, AND OTHER RATES OF CHANGE 113 Tangents If a curve C has equation y f x and we want to find the tangent line to C at the point P a, f a , then we consider a nearby point Q x, f x , where x a, and compute the slope of the secant line PQ : fx x mPQ fa a Then we let Q approach P along the curve C by letting x approach a. If mPQ approaches a number m, then we define the tangent t to be the line through P with slope m. (This amounts to saying that the tangent line is the limiting position of the secant line PQ as Q approaches P. See Figure 1.) y y t Q Q{ x, ƒ } Q ƒ-f(a) Q P P { a, f(a)} x-a 0 a x x x 0 FIGURE 1 1 Definition The tangent line to the curve y line through P with slope m lim xla f x at the point P a, f a is the fx x fa a provided that this limit exists. In our first example we confirm the guess we made in Example 1 in Section 2.1. x 2 at the EXAMPLE 1 Find an equation of the tangent line to the parabola y point P 1, 1 . SOLUTION Here we have a m lim x l1 fx x lim x x l1 lim x x l1 |||| Point-slope form for a line through the point x1 , y1 with slope m : y y1 mx x 2, so the slope is 1 and f x lim x l1 f1 1 1x x1 1 x2 x 1 2 1 1 1 1 Using the point-slope form of the equation of a line, we find that an equation of the tangent line at 1, 1 is x1 y 1 2x 1 or y 2x 1 5E-02(pp 110-121) 114 ❙❙❙❙ 1/17/06 12:30 PM Page 114 CHAPTER 2 LIMITS AND RATES OF CHANGE We sometimes refer to the slope of the tangent line to a curve at a point as the slope of the curve at the point. The idea is that if we zoom in far enough toward the point, the curve looks almost like a straight line. Figure 2 illustrates this procedure for the curve y x 2 in Example 1. The more we zoom in, the more the parabola looks like a line. In other words, the curve becomes almost indistinguishable from its tangent line. 2 1.5 1.1 (1, 1) 0 (1, 1) 2 (1, 1) 1.5 0.5 1.1 0.9 FIGURE 2 Zooming in toward the point (1, 1) on the parabola y=≈ There is another expression for the slope of a tangent line that is sometimes easier to use. Let h a x Then x a h so the slope of the secant line PQ is fa mPQ h h fa (See Figure 3 where the case h 0 is illustrated and Q is to the right of P. If it happened that h 0, however, Q would be to the left of P.) y t Q{ a+h, f(a+h)} f(a+h)-f(a) P { a, f(a)} h 0 a+h a x FIGURE 3 Notice that as x approaches a, h approaches 0 (because h sion for the slope of the tangent line in Definition 1 becomes 2 m lim hl0 fa h h x a) and so the expres- fa EXAMPLE 2 Find an equation of the tangent line to the hyperbola y point 3, 1 . 3 x at the 5E-02(pp 110-121) 1/17/06 12:30 PM Page 115 S ECTION 2.6 TANGENTS, VELOCITIES, AND OTHER RATES OF CHANGE SOLUTION Let f x f3 lim h h hl0 3 3 lim lim hl0 f3 3 1 h h hl0 h h 1 lim h h 3 lim hl0 h h3 3 3 hl0 h 1 3 y y= 3 x Therefore, an equation of the tangent at the point 3, 1 is (3, 1) 0 115 3 x. Then the slope of the tangent at 3, 1 is m x+3y-6=0 ❙❙❙❙ y x which simplifies to FIGURE 4 1 3 1 x 3y x 3 6 0 The hyperbola and its tangent are shown in Figure 4. EXAMPLE 3 Find the slopes of the tangent lines to the graph of the function f x sx at the points (1, 1), (4, 2), and (9, 3). SOLUTION Since three slopes are requested, it is efficient to start by finding the slope at the general point (a, sa ): m lim lim Rationalize the numerator fa hl0 sa hl0 lim hl0 lim Continuous function of h hl0 h h h h a h(sa sa At the point (1, 1), we have a At (4, 2), we have m 1 (2 s4 ) sa lim h h hl0 sa h h 1 h fa sa sa a sa ) h h lim hl0 sa sa h(sa 1 sa sa sa sa h h sa ) 1 2 sa 1, so the slope of the tangent is m 1 1 (2 s9 ) 1 . 4 ; at (9, 3), m 6 1 (2 s1 ) 1 2 . Velocities Learn about average and instantaneous velocity by comparing falling objects. Resources / Module 3 / Derivative at a Point / The Falling Robot In Section 2.1 we investigated the motion of a ball dropped from the CN Tower and defined its velocity to be the limiting value of average velocities over shorter and shorter time periods. In general, suppose an object moves along a straight line according to an equation of motion s f t , where s is the displacement (directed distance) of the object from the 5E-02(pp 110-121) 116 ❙❙❙❙ 1/17/06 12:31 PM Page 116 CHAPTER 2 LIMITS AND RATES OF CHANGE origin at time t. The function f that describes the motion is called the position function of the object. In the time interval from t a to t a h the change in position is fa h f a . (See Figure 5.) The average velocity over this time interval is displacement time average velocity fa h h fa which is the same as the slope of the secant line PQ in Figure 6. s Q { a+h, f(a+h)} P { a, f(a)} position at time t=a position at time t=a+h h s 0 0 f(a+h)-f(a) a f(a) mPQ= f(a+h) a+h t f(a+h)- (a) f h average velocity FIGURE 5 FIGURE 6 Now suppose we compute the average velocities over shorter and shorter time intervals a, a h . In other words, we let h approach 0. As in the example of the falling ball, we define the velocity (or instantaneous velocity) v a at time t a to be the limit of these average velocities: va 3 fa lim hl0 h h fa This means that the velocity at time t a is equal to the slope of the tangent line at P (compare Equations 2 and 3). Now that we know how to compute limits, let’s reconsider the problem of the falling ball. EXAMPLE 4 Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground. (a) What is the velocity of the ball after 5 seconds? (b) How fast is the ball traveling when it hits the ground? |||| Recall from Section 2.1: The distance (in meters) fallen after t seconds is 4.9t 2. SOLUTION We first use the equation of motion s ft 4.9t 2 to find the velocity v a after a seconds: va lim fa hl0 4.9 a 2 h h fa hl0 lim 4.9 2a h hl0 (a) The velocity after 5 s is v 5 4.9 a hl0 h2 2ah h lim lim a2 lim hl0 9.8a 9.8 5 49 m s. h2 h 4.9a 2 4.9 2ah h h2 5E-02(pp 110-121) 1/17/06 12:31 PM Page 117 S ECTION 2.6 TANGENTS, VELOCITIES, AND OTHER RATES OF CHANGE ❙❙❙❙ 117 (b) Since the observation deck is 450 m above the ground, the ball will hit the ground at the time t1 when s t1 450, that is, 4.9t 2 1 450 This gives t2 1 450 4.9 and 450 4.9 t1 9.6 s The velocity of the ball as it hits the ground is therefore v t1 9.8t1 9.8 450 4.9 94 m s Other Rates of Change Suppose y is a quantity that depends on another quantity x. Thus, y is a function of x and we write y f x . If x changes from x 1 to x 2 , then the change in x (also called the increment of x) is x x2 x1 and the corresponding change in y is y f x2 f x1 y x f x2 x2 f x1 x1 The difference quotient is called the average rate of change of y with respect to x over the interval x 1, x 2 and can be interpreted as the slope of the secant line PQ in Figure 7. y Q{ ¤, ‡} Îy P { ⁄, fl} Îx 0 FIGURE 7 ⁄ ¤ x average rate of change mPQ instantaneous rate of change slope of tangent at P By analogy with velocity, we consider the average rate of change over smaller and smaller intervals by letting x 2 approach x 1 and therefore letting x approach 0. The limit of these average rates of change is called the (instantaneous) rate of change of y with respect to x at x x 1 , which is interpreted as the slope of the tangent to the curve y f x at P x 1, f x 1 : 5E-02(pp 110-121) 118 ❙❙❙❙ 1/17/06 12:31 PM Page 118 CHAPTER 2 LIMITS AND RATES OF CHANGE instantaneous rate of change 4 xh TC xh TC 0 1 2 3 4 5 6 7 8 9 10 11 12 . 6.5 6.1 5.6 4.9 4.2 4.0 4.0 4.8 6.1 8.3 10.0 12.1 14.3 13 14 15 16 17 18 19 20 21 22 23 24 16.0 17.3 18.2 18.8 17.6 16.0 14.1 11.5 10.2 9.0 7.9 7.0 . xl0 lim x2 l x1 f x2 x2 f x1 x1 E XAMPLE 5 Temperature readings T (in degrees Celsius) were recorded every hour starting at midnight on a day in April in Whitefish, Montana. The time x is measured in hours from midnight. The data are given in the table at the left. (a) Find the average rate of change of temperature with respect to time (i) from noon to 3 P.M. (ii) from noon to 2 P.M. (iii) from noon to 1 P.M. (b) Estimate the instantaneous rate of change at noon. SOLUTION (a) (i) From noon to 3 P.M. the temperature changes from 14.3°C to 18.2°C, so T T 15 T 12 18.2 14.3 3.9 C while the change in time is x 3 h. Therefore, the average rate of change of temperature with respect to time is T x |||| A NOTE ON UNITS The units for the average rate of change T x are the units for T divided by the units for x, namely, degrees Celsius per hour. The instantaneous rate of change is the limit of the average rates of change, so it is measured in the same units: degrees Celsius per hour. y x lim 3.9 3 1.3 C h (ii) From noon to 2 P.M. the average rate of change is T x T 14 14 T 12 12 17.3 14.3 2 1.5 C h (iii) From noon to 1 P.M. the average rate of change is T x T 13 13 T 12 12 16.0 14.3 1 1.7 C h (b) We plot the given data in Figure 8 and use them to sketch a smooth curve that approximates the graph of the temperature function. T B 18 16 P 14 12 10 A 8 6 4 2 FIGURE 8 0 1 C 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Then we draw the tangent at the point P where x x 12 and, after measuring the sides of 5E-02(pp 110-121) 1/17/06 12:31 PM Page 119 SECTION 2.6 TANGENTS, VELOCITIES, AND OTHER RATES OF CHANGE ❙❙❙❙ 119 triangle ABC, we estimate that the slope of the tangent line is BC AC |||| Another method is to average the slopes of two secant lines. See Example 2 in Section 2.1. 10.3 5.5 1.9 Therefore, the instantaneous rate of change of temperature with respect to time at noon is about 1.9°C h. The velocity of a particle is the rate of change of displacement with respect to time. Physicists are interested in other rates of change as well—for instance, the rate of change of work with respect to time (which is called power). Chemists who study a chemical reaction are interested in the rate of change in the concentration of a reactant with respect to time (called the rate of reaction). A steel manufacturer is interested in the rate of change of the cost of producing x tons of steel per day with respect to x (called the marginal cost). A biologist is interested in the rate of change of the population of a colony of bacteria with respect to time. In fact, the computation of rates of change is important in all of the natural sciences, in engineering, and even in the social sciences. Further examples will be given in Section 3.4. All these rates of change can be interpreted as slopes of tangents. This gives added significance to the solution of the tangent problem. Whenever we solve a problem involving tangent lines, we are not just solving a problem in geometry. We are also implicitly solving a great variety of problems involving rates of change in science and engineering. |||| 2.6 Exercises 1. A curve has equation y 5. (a) Find the slope of the tangent line to the parabola f x. (a) Write an expression for the slope of the secant line through the points P 3, f 3 and Q x, f x . (b) Write an expression for the slope of the tangent line at P. 2. Suppose an object moves with position function s f t. (a) Write an expression for the average velocity of the object in the time interval from t a to t a h. (b) Write an expression for the instantaneous velocity at time t a. y x 2 2 x at the point 3, 3 (i) using Definition 1 (ii) using Equation 2 (b) Find an equation of the tangent line in part (a). (c) Graph the parabola and the tangent line. As a check on your work, zoom in toward the point ( 3, 3) until the parabola and the tangent line are indistinguishable. ; 6. (a) Find the slope of the tangent line to the curve y point 1, 1 (i) using Definition 1 (ii) using Equation 2 (b) Find an equation of the tangent line in part (a). (c) Graph the curve and the tangent line in successively smaller viewing rectangles centered at ( 1, 1) until the curve and the line appear to coincide. 3. Consider the slope of the given curve at each of the five points shown. List these five slopes in decreasing order and explain your reasoning. y ; A E B 7–10 |||| Find an equation of the tangent line to the curve at the given point. D 7. y ; 4. Graph the curve y sin x in the viewing rectangles 2, 2 by 2, 2 , 1, 1 by 1, 1 , and 0.5, 0.5 by 0.5, 0.5 . What do you notice about the curve as you zoom in toward the origin? s2 x 9. y x 1 8. y C 0 x 3 at the x 10. y ■ 1, 1 2x x ■ x 3, 2x ■ x 1, 2 4, 3 2, 1 2, ■ 3, 2 0, 0 ■ ■ ■ ■ ■ ■ ■ ■ 5E-02(pp 110-121) 120 ❙❙❙❙ 1/17/06 12:32 PM Page 120 CHAPTER 2 LIMITS AND RATES OF CHANGE 11. (a) Find the slope of the tangent to the curve y 2 x 3 at the point where x a. (b) Find the slopes of the tangent lines at the points whose x-coordinates are (i) 1, (ii) 0, and (iii) 1. 12. (a) Find the slope of the tangent to the parabola ; y 1 x x 2 at the point where x a. (b) Find the slopes of the tangent lines at the points whose x-coordinates are (i) 1, (ii) 1, and (iii) 1. 2 (c) Graph the curve and the three tangents on a common screen. x 3 4x 1 at the point where x a. (b) Find equations of the tangent lines at the points 1, 2 and 2, 1 . (c) Graph the curve and both tangents on a common screen. 13. (a) Find the slope of the tangent to the curve y ; 14. (a) Find the slope of the tangent to the curve y ; 1 sx at the point where x a. (b) Find equations of the tangent lines at the points 1, 1 and (4, 1 ). 2 (c) Graph the curve and both tangents on a common screen. 15. The graph shows the position function of a car. Use the shape of the graph to explain your answers to the following questions. (a) What was the initial velocity of the car? (b) Was the car going faster at B or at C ? (c) Was the car slowing down or speeding up at A, B, and C ? (d) What happened between D and E ? s D E C B A 18. If an arrow is shot upward on the moon with a velocity of 58 m s, its height (in meters) after t seconds is given by H 58 t 0.83t 2. (a) Find the velocity of the arrow after one second. (b) Find the velocity of the arrow when t a. (c) When will the arrow hit the moon? (d) With what velocity will the arrow hit the moon? 19. The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s 4 t 3 6 t 2, where t is measured in seconds. Find the velocity of the particle at times t a, t 1, t 2, and t 3. 20. The displacement (in meters) of a particle moving in a straight line is given by s t 2 8 t 18, where t is measured in seconds. (a) Find the average velocity over each time interval: (i) 3, 4 (ii) 3.5, 4 (iii) 4, 5 (iv) 4, 4.5 (b) Find the instantaneous velocity when t 4. (c) Draw the graph of s as a function of t and draw the secant lines whose slopes are the average velocities in part (a) and the tangent line whose slope is the instantaneous velocity in part (b). 21. A warm can of soda is placed in a cold refrigerator. Sketch the graph of the temperature of the soda as a function of time. Is the initial rate of change of temperature greater or less than the rate of change after an hour? 22. A roast turkey is taken from an oven when its temperature has reached 185°F and is placed on a table in a room where the temperature is 75°F. The graph shows how the temperature of the turkey decreases and eventually approaches room temperature. (In Section 10.4 we will be able to use Newton’s Law of Cooling to find an equation for T as a function of time.) By measuring the slope of the tangent, estimate the rate of change of the temperature after an hour. T (°F) 0 t 200 P 16. Valerie is driving along a highway. Sketch the graph of the position function of her car if she drives in the following manner: At time t 0, the car is at mile marker 15 and is traveling at a constant speed of 55 mi h. She travels at this speed for exactly an hour. Then the car slows gradually over a 2-minute period as Valerie comes to a stop for dinner. Dinner lasts 26 min; then she restarts the car, gradually speeding up to 65 mi h over a 2-minute period. She drives at a constant 65 mi h for two hours and then over a 3-minute period gradually slows to a complete stop. 17. If a ball is thrown into the air with a velocity of 40 ft s, its height (in feet) after t seconds is given by y Find the velocity when t 2. 40 t 16 t 2. 100 0 30 60 90 120 150 t (min) 23. (a) Use the data in Example 5 to find the average rate of change of temperature with respect to time (i) from 8 P.M. to 11 P.M. (ii) from 8 P.M. to 10 P.M. (iii) from 8 P.M. to 9 P.M. (b) Estimate the instantaneous rate of change of T with respect to time at 8 P.M. by measuring the slope of a tangent. 5E-02(pp 110-121) 1/17/06 12:32 PM Page 121 C HAPTER 2 REVIEW 24. The population P (in thousands) of Belgium from 1992 to 2000 Year Year 1992 1994 1996 1998 2000 P 10,036 10,109 10,152 10,175 10,186 (a) Find the average rate of growth (i) from 1992 to 1996 (ii) from 1994 to 1996 (iii) from 1996 to 1998 In each case, include the units. (b) Estimate the instantaneous rate of growth in 1996 by taking the average of two average rates of change. What are its units? (c) Estimate the instantaneous rate of growth in 1996 by measuring the slope of a tangent. 1994 1995 1996 304 572 873 1513 2461 (a) Find the average rate of growth (i) from 1995 to 1997 (ii) from 1995 to 1996 (iii) from 1994 to 1995 In each case, include the units. (b) Estimate the instantaneous rate of growth in 1995 by taking the average of two average rates of change. What are its units? (c) Estimate the instantaneous rate of growth in 1995 by measuring the slope of a tangent. ■ sketch. (a) lim f x x la (c) lim f x x la L L (b) lim f x x la L (d) lim f x x la (e) lim f x 1886 2135 3300 100,000 1 t 60 2 0 t 60 Find the rate at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t ) as a function of t. What are its units? For times t 0, 10, 20, 30, 40, 50, and 60 min, find the flow rate and the amount of water remaining in the tank. Summarize your findings in a sentence or two. At what time is the flow rate the greatest? The least? ■ 4. State the following Limit Laws. (a) (c) (e) (g) Sum Law Constant Multiple Law Quotient Law Root Law (b) Difference Law (d) Product Law (f ) Power Law 5. What does the Squeeze Theorem say? xla 2. Describe several ways in which a limit can fail to exist. Illustrate with sketches. 3. What does it mean to say that the line x asymptote of the curve y various possibilities. 1412 Vt CONCEPT CHECK 1. Explain what each of the following means and illustrate with a 1015 be drained from the bottom of the tank in an hour, then Torricelli’s Law gives the volume V of water remaining in the tank after t minutes as given in the table. (The numbers of locations as of June 30 are given.) 2 Review 2000 28. If a cylindrical tank holds 100,000 gallons of water, which can 26. The number N of locations of a popular coffeehouse chain is |||| 1999 modity is C x 5000 10 x 0.05x 2. (a) Find the average rate of change of C with respect to x when the production level is changed (i) from x 100 to x 105 (ii) from x 100 to x 101 (b) Find the instantaneous rate of change of C with respect to x when x 100. (This is called the marginal cost. Its significance will be explained in Section 3.4.) 1997 N 1998 27. The cost (in dollars) of producing x units of a certain com- Malaysia is shown in the table. (Midyear estimates are given.) 1993 1997 121 (a) Find the average rate of growth (i) from 1996 to 1998 (ii) from 1997 to 1998 (iii) from 1998 to 1999 In each case, include the units. (b) Estimate the instantaneous rate of growth in 1998 by taking the average of two average rates of change. What are its units? (c) Estimate the instantaneous rate of growth in 1998 by measuring the slope of a tangent. 25. The number N (in thousands) of cellular phone subscribers in Year 1996 N is shown in the table. (Midyear estimates are given.) ❙❙❙❙ a is a vertical f x ? Draw curves to illustrate the 6. (a) What does it mean for f to be continuous at a ? (b) What does it mean for f to be continuous on the interval , ? What can you say about the graph of such a function? 7. What does the Intermediate Value Theorem say? 5E-02(pp 122-123) 122 ❙❙❙❙ 1/17/06 12:23 PM Page 122 CHAPTER 2 LIMITS AND RATES OF CHANGE 8. Write an expression for the slope of the tangent line to the curve y 10. If y f x and x changes from x 1 to x 2 , write expressions for the following. (a) The average rate of change of y with respect to x over the interval x 1, x 2 . (b) The instantaneous rate of change of y with respect to x at x x 1. f x at the point a, f a . 9. Suppose an object moves along a straight line with position f t at time t. Write an expression for the instantaneous velocity of the object at time t a. How can you interpret this velocity in terms of the graph of f ? ■ TRUE-FALSE QUIZ 7. If p is a polynomial, then lim x l b p x 1. lim x l4 2. lim x l1 3. lim x l1 8 x 4 x2 x2 6x 5x x x2 x 7 6 lim 4 x lim x 2 6x lim x 5x 6 x l1 x l1 2 lim x x l1 4 x l4 x 4 2x tx and lim x l 0 t x 0. 9. If the line x 1 is a vertical asymptote of y not defined at 1. 0 and f 3 0, then there exists a number c between 1 and 3 such that f c 0. 11. If f is continuous at 5 and f 5 lim x l 2 f 4 x 2 4 2 and lim x l 5 t x f x t x does not exist. 5. If lim x l 5 f x 0 and lim x l 5 t x lim x l 5 f x t x does not exist. f x , then f is 10. If f 1 3 4. If lim x l 5 f x limx l 5 , then lim x l 0 f x 8 7 2 x l1 lim x 4 lim x l4 3 2x 2x pb. 8. If lim x l 0 f x Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 2x ■ 11 2 and f 4 12. If f is continuous on 1, 1 and f 1 then there exists a number r such that r 0, then 4 and f 1 1 and f r 13. Let f be a function such that lim x l 0 f x exists a number such that if 0 fx 6 1. 0, then 3, then 2. 3, . 6. Then there , then x 14. If f x 1 for all x and lim x l 0 f x exists, then lim x l 0 f x 1. 6. If lim x l 6 f x t x exists, then the limit must be f 6 t 6 . ■ EXERCISES ■ (b) State the equations of the vertical asymptotes. (c) At what numbers is f discontinuous? Explain. 1. The graph of f is given. y 2. Sketch the graph of an example of a function f that satisfies all of the following conditions: lim f x 2, lim f x x l0 1 x l0 , lim f x x l2 0 |||| (a) Find each limit, or explain why it does not exist. (i) lim f x (ii) lim f x xl 3 (iii) lim f x (iv) lim f x (v) lim f x (vi) lim f x x l0 lim f x x l2 Find the limit. 3. lim cos x xl 3 1, f0 x 1 3–16 x l2 1, xl0 5. lim xl 3 x l4 x l2 sin x 7. lim h l0 x x2 h 2 9 2x 13 h 3 1 4. lim x l3 6. lim x l1 8. lim t l2 x2 x 2 9 2x x 2 x2 t2 t3 9 2x 4 8 3 3 5E-02(pp 122-123) 1/17/06 12:24 PM Page 123 C HAPTER 2 REVIEW 9. lim r r l9 sr 9 4 11. lim s l16 s x l8 15. lim 12. lim 1 ■ 17. If 2 x x2 16. lim ■ ■ ■ 2 x for 0 fx 18. Prove that lim x l 0 x cos 1 x |||| 27 x l5 21. lim x 2 sx 2 x2 ■ x 1 ■ ■ ■ 20. lim sx s2x 2x ■ ■ ■ ■ sx xl4 ■ ■ ■ if x if 0 if x 2 ■ 4 ■ 0 x 3 (a) Evaluate each limit, if it exists. (i) lim f x (ii) lim f x x l0 ■ ■ ■ 3 ■ ■ (v) lim f x ■ 9 2x 2 30. Find equations of the tangent lines to the curve 2 1 3x at the points with x-coordinates 0 and 1. line is given by s 1 2 t t 2 4, where t is measured in seconds. (a) Find the average velocity over each time period. (i) 1, 3 (ii) 1, 2 (iii) 1, 1.5 (iv) 1, 1.1 (b) Find the instantaneous velocity when t 1. is held fixed, then the product of the pressure P and the volume V is a constant. Suppose that, for a certain gas, PV 800, where P is measured in pounds per square inch and V is measured in cubic inches. (a) Find the average rate of change of P as V increases from 200 in3 to 250 in3. (b) Express V as a function of P and show that the instantaneous rate of change of V with respect to P is inversely proportional to the square of P. x l0 (vi) lim f x x l3 ■ 32. According to Boyle’s Law, if the temperature of a confined gas (iii) lim f x x l0 (iv) lim f x ■ 31. The displacement (in meters) of an object moving in a straight 23. Let x l3 0, 1 ■ at the point 2, 1 . (b) Find an equation of this tangent line. 2 22. lim sx 3x x3 2 x, ■ 1 ) 0 xl0 fx 3 ■ 2, 0. 3 ■ 0, y 2 2 ■ 2 3, find lim x l1 f x . x 8 3x xl2 ■ 28. 2 sin x Prove the statement using the precise definition of a limit. 19. lim 7x ■ 9 x2 29. (a) Find the slope of the tangent line to the curve y x l2 2 19–22 27. 2 x 3 2v 8 4 v 16 xl9 ■ 1 v 14. lim (s x 123 27–28 |||| Use the Intermediate Value Theorem to show that there is a root of the equation in the given interval. v ■ s1 x ■ v2 v l2 8 8 x l0 ■ vl4 ss 16 x x 13. lim 4 4 10. lim 4 ❙❙❙❙ x l3 (b) Where is f discontinuous? (c) Sketch the graph of f . 24. Let 2x x 2 2x x4 tx if if if if 0 2 3 x x x x 4 ; 33. Use a graph to find a number such that 2 3 4 x x (a) For each of the numbers 2, 3, and 4, discover whether t is continuous from the left, continuous from the right, or continuous at the number. (b) Sketch the graph of t. 25–26 Show that the function is continuous on its domain. State the domain. |||| 25. h x 4 sx 26. t x sx 2 x2 ■ ■ ■ 0.2 whenever x y x x 1 1 and the tangent lines to this curve at the points 2, 3 and 1, 0 . 36. Let f x ■ ■ ■ ■ ■ ■ ■ 2 ; 34. Graph the curve t x for all x, where lim x l a t x Find lim x l a f x . 9 2 ■ 3 35. Suppose that f x x 3 cos x ■ 1 1 x x. (a) For what values of a does lim x l a f x exist? (b) At what numbers is f discontinuous? 0. 5E-02(pp 124-125) 1/17/06 12:21 PM PROBLEMS PLUS Page 124 In our discussion of the principles of problem solving we considered the problem-solving strategy of introducing something extra (see page 58). In the following example we show how this principle is sometimes useful when we evaluate limits. The idea is to change the variable—to introduce a new variable that is related to the original variable—in such a way as to make the problem simpler. Later, in Section 5.5, we will make more extensive use of this general idea. EXAMPLE 1 Evaluate lim 3 s1 cx x xl0 1 , where c is a constant. SOLUTION As it stands, this limit looks challenging. In Section 2.3 we evaluated several limits in which both numerator and denominator approached 0. There our strategy was to perform some sort of algebraic manipulation that led to a simplifying cancellation, but here it’s not clear what kind of algebra is necessary. So we introduce a new variable t by the equation 3 s1 t cx We also need to express x in terms of t, so we solve this equation: t3 1 cx t3 x 1 c Notice that x l 0 is equivalent to t l 1. This allows us to convert the given limit into one involving the variable t : 3 s1 cx x xl0 lim 1 t t3 lim lim ct t3 t l1 t l1 1 1c 1 1 The change of variable allowed us to replace a relatively complicated limit by a simpler one of a type that we have seen before. Factoring the denominator as a difference of cubes, we get lim t l1 ct t3 1 1 lim t l1 lim t l1 t t2 ct 1 1 t2 t c t 1 1 c 3 The following problems are meant to test and challenge your problem-solving skills. Some of them require a considerable amount of time to think through, so don’t be discouraged if you can’t solve them right away. If you get stuck, you might find it helpful to refer to the discussion of the principles of problem solving on page 58. 124 5E-02(pp 124-125) 1/17/06 12:22 PM P RO B L E M S Page 125 3 sx sx 1. Evaluate lim x l1 1 . 1 2. Find numbers a and b such that lim sax 2x 3. Evaluate lim 1 2x 1 x xl0 b 2 1. x xl0 . x 2 and the point Q where the perpendicular bisector of OP intersects the y-axis. As P approaches the origin along the parabola, what happens to Q ? Does it have a limiting position? If so, find it. 4. The figure shows a point P on the parabola y y y=≈ Q P 5. If x denotes the greatest integer function, find lim x l x x . 6. Sketch the region in the plane defined by each of the following equations. (a) x 0 x 2 y 2 1 (b) x 2 y 2 (c) x 3 7. Find all values of a such that f is continuous on FIGURE FOR PROBLEM 4 fx x x2 y 2 1 (d) x y 1 : 1 if x if x a a 8. A fixed point of a function f is a number c in its domain such that f c c. (The function doesn’t move c; it stays fixed.) (a) Sketch the graph of a continuous function with domain 0, 1 whose range also lies in 0, 1 . Locate a fixed point of f . (b) Try to draw the graph of a continuous function with domain 0, 1 and range in 0, 1 that does not have a fixed point. What is the obstacle? (c) Use the Intermediate Value Theorem to prove that any continuous function with domain 0, 1 and range a subset of 0, 1 must have a fixed point. 9. If lim x l a f x A 2 and lim x l a f x tx 1, find lim x l a f x t x . 10. (a) The figure shows an isosceles triangle ABC with B C. The bisector of angle B intersects the side AC at the point P. Suppose that the base BC remains fixed but the altitude AM of the triangle approaches 0, so A approaches the midpoint M of BC. What happens to P during this process? Does it have a limiting position? If so, find it. (b) Try to sketch the path traced out by P during this process. Then find the equation of this curve and use this equation to sketch the curve. P B tx M FIGURE FOR PROBLEM 10 C 11. (a) If we start from 0 latitude and proceed in a westerly direction, we can let T x denote the temperature at the point x at any given time. Assuming that T is a continuous function of x, show that at any fixed time there are at least two diametrically opposite points on the equator that have exactly the same temperature. (b) Does the result in part (a) hold for points lying on any circle on Earth’s surface? (c) Does the result in part (a) hold for barometric pressure and for altitude above sea level? 125 ...
View Full Document

This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas.

Ask a homework question - tutors are online