For Use Only in 2013 – 2014 Pilot Program 2 Limits 2.1 The Idea of Limits 2.2 Definitions of Limits 2.3 Techniques for Computing Limits 2.4 Infinite Limits 2.5 Limits at Infinity 2.6 Continuity 2.7 Precise Definitions of Limits Biologists often use mathematical models to describe communities of organisms— anything from a culture of cells to a herd of zebras. These models result in functions that approximate the population of the community as it changes in time. An important question concerns the long-term behavior of these population functions. Does the population approach a steady-state level, which corresponds to a stable community (as in the first figure)? Does the population oscillate continually without settling at some fixed level (as in the second figure)? Or does the population decrease to zero, indicating that the organism becomes extinct (as in the third figure)? Such questions about the behavior of functions are answered using the powerful idea of a limit. Population increases and approches a steady state. 300 200 100 Population Population 400 500 500 400 400 Population 500 300 200 Population oscillates without approaching a steady state. 100 0 0 0 20 40 60 Time (months) 80 Population decays to zero (extinction). 300 200 100 0 0 20 40 60 80 Time (months) 60 Copyright © 2014 Pearson Education, Inc. 0 20 40 60 Time (months) 80 For Use Only in 2013 – 2014 Pilot Program 2.1 The Idea of Limits 61 Chapter Preview All of calculus is based on the idea of a limit. Not only are limits important in their own right, but they underlie the two fundamental operations of calculus: differentiation (calculating derivatives) and integration (evaluating integrals). Derivatives enable us to talk about the instantaneous rate of change of a function, which, in turn, leads to concepts such as velocity and acceleration, population growth rates, marginal cost, and flow rates. Integrals enable us to compute areas under curves, surface areas, and volumes. Because of the incredible reach of this single idea, it is essential to develop a solid understanding of limits. We first present limits intuitively by showing how they arise in computing instantaneous velocities and finding slopes of tangent lines. As the chapter progresses, we build more rigor into the definition of the limit, and we examine the different ways in which limits exist or fail to exist. The chapter concludes by introducing the important property called continuity and by giving the formal definition of a limit. 2.1 The Idea of Limits This brief opening section illustrates how limits arise in two seemingly unrelated problems: finding the instantaneous velocity of a moving object and finding the slope of a line tangent to a curve. These two problems provide important insights into limits, and they reappear in various forms throughout the book. Average Velocity Suppose you want to calculate your average velocity as you travel along a straight highway. If you pass milepost 100 at noon and milepost 130 at 12:30 p.m., you travel 30 miles in a halfhour, so your average velocity over this time interval is 130 mi2>10.5 hr2 = 60 mi>hr. By contrast, even though your average velocity may be 60 mi >hr, it’s almost certain that your instantaneous velocity, the speed indicated by the speedometer, varies from one moment to the next. EXAMPLE 1 Average velocity A rock is launched vertically upward from the ground with a speed of 96 ft>s. Neglecting air resistance, a well-known formula from physics states that the position of the rock after t seconds is given by the function s1t2 = -16t 2 + 96t. The position s is measured in feet with s = 0 corresponding to the ground. Find the average velocity of the rock between each pair of times. a. t = 1 s and t = 3 s b. t = 1 s and t = 2 s SOLUTION Figure 2.1 shows the position of the rock on the time interval 0 … t … 3. a. The average velocity of the rock over any time interval 3t0, t14 is the change in position divided by the elapsed time: vav = s1t12 - s1t02 t1 - t0 Therefore, the average velocity over the interval 31, 34 is vav = s132 - s112 144 ft - 80 ft 64 ft = = = 32 ft>s. 3 - 1 3s - 1s 2s Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 62 Limits Height above ground (ft) s s(t) ⫽ ⫺16t2 ⫹ 96t (3, s(3)) s(3) ⫽ 144 ft 144 s(2) ⫽ 128 ft 128 s(1) ⫽ 80 ft 80 (2, s(2)) (1, s(1)) Position of rock 0 1 2 t 3 Time (s) FIGURE 2.1 Here is an important observation: As shown in Figure 2.2a, the average velocity is simply the slope of the line joining the points 11, s1122 and 13, s1322 on the graph of the position function. b. The average velocity of the rock over the interval 31, 24 is vav = s122 - s112 128 ft - 80 ft 48 ft = = = 48 ft>s. 2 - 1 2s - 1s 1s Again, the average velocity is the slope of the line joining the points 11, s1122 and 12, s1222 on the graph of the position function (Figure 2.2b). 64 ft ⫽ 32 ft/s 2s vav ⫽ slope ⫽ s vav ⫽ slope ⫽ s 48 ft ⫽ 48 ft/s 1s (3, 144) (1, 80) Height above ground (ft) Change in position ⫽ s(3) ⫺ s(1) ⫽ 64 ft 80 Elapsed time ⫽3s⫺1s⫽2s 0 1 FIGURE 2.2 2 3 t 128 (2, 128) Change in position ⫽ s(2) ⫺ s(1) ⫽ 48 ft (1, 80) 80 Elapsed time ⫽2s⫺1s⫽1s 0 1 2 3 t Time (s) (b) Time (s) (a) Related Exercises 7–14 ➤ See Section 1.1 for a discussion of secant lines. In Example 1, what is the average velocity between t = 2 and t = 3? ➤ QUICK CHECK 1 ➤ Height above ground (ft) 144 In Example 1, we computed slopes of lines passing through two points on a curve. Any such line joining two points on a curve is called a secant line. The slope of the secant line, denoted msec, for the position function in Example 1 on the interval 3t0, t14 is msec = s1t12 - s1t02 t1 - t0 Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.1 The Idea of Limits Height above ground (ft) Example 1 demonstrates that the average velocity is the slope of a secant line on the graph of the position function; that is, vav = msec (Figure 2.3). s(t1) ⫺ s(t0 ) t 1 ⫺ t0 Instantaneous Velocity Change in position ⫽ s(t1) ⫺ s(t0 ) s s(t1) s(t0) Change in time ⫽ t1 ⫺ t 0 0 t0 t1 Time (s) t To compute the average velocity, we use the position of the object at two distinct points in time. How do we compute the instantaneous velocity at a single point in time? As illustrated in Example 2, the instantaneous velocity at a point t = t0 is determined by computing average velocities over intervals 3t0, t14 that decrease in length. As t1 approaches t0, the average velocities typically approach a unique number, which is the instantaneous velocity. This single number is called a limit. QUICK CHECK 2 velocity. FIGURE 2.3 ➤ vav ⫽ msec ⫽ 63 Explain the difference between average velocity and instantaneous EXAMPLE 2 Instantaneous velocity Estimate the instantaneous velocity of the rock in Example 1 at the single point t = 1. SOLUTION We are interested in the instantaneous velocity at t = 1, so we compute the average velocity over smaller and smaller time intervals 31, t4 using the formula Notice that these average velocities are also slopes of secant lines, several of which are shown in Table 2.1. We see that as t approaches 1, the average velocities appear to approach 64 ft>s. In fact, we could make the average velocity as close to 64 ft>s as we like by taking t sufficiently close to 1. Therefore, 64 ft>s is a reasonable estimate of the instantaneous velocity at t = 1. Table 2.1 Average velocity 31, 24 31, 1.54 31, 1.14 31, 1.014 31, 1.0014 31, 1.00014 48 ft>s 56 ft>s 62.4 ft>s 63.84 ft>s 63.984 ft>s 63.998 ft>s Related Exercises 15–20 In language to be introduced in Section 2.2, we say that the limit of vav as t approaches 1 equals the instantaneous velocity vinst , which is 64 ft >s. This statement is written compactly as vinst = lim vav = lim t S1 ➤ The same instantaneous velocity is Figure 2.4 gives a graphical illustration of this limit. obtained as t approaches 1 from the left (with t 6 1) and as t approaches 1 from the right (with t 7 1). 0 t 1 t 2 t S1 s1t2 - s112 = 64 ft>s. t - 1 Position of rock at various times t 128 ft t ⫽ 2 sec 108 ft t ⫽ 1.5 sec 86.24 ft t ⫽ 1.1 sec 80 ft vav ⫽ vav ⫽ ... vav ⫽ s(2) ⫺ s(1) 128 ⫺ 80 ⫽ 48 ft/s ⫽ 2⫺1 1 s(1.5) ⫺ s(1) 108 ⫺ 80 ⫽ 56 ft/s ⫽ 0.5 1.5 ⫺ 1 s(1.1) ⫺ s(1) 86.24 ⫺ 80 ⫽ ⫽ 62.4 ft/s 1.1 ⫺ 1 0.1 t ⫽ 1 sec As these intervals shrink... t ⫽ 0 (rock thrown at 96 ft/s) FIGURE 2.4 Copyright © 2014 Pearson Education, Inc. ... Time interval s1t2 - s112 . t - 1 ➤ vav = vinst ⫽ 64 ft/s ... the average velocities approach 64 ft/s— the instantaneous velocity at t ⫽ 1. For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 64 Limits Slope of the Tangent Line ➤ We define tangent lines carefully in Section 3.1. For the moment, imagine zooming in on a point P on a smooth curve. As you zoom in, the curve appears more and more like a line passing through P. This line is the tangent line at P. Because a smooth curve approaches a line as we zoom in on a point, a smooth curve is said to be locally linear at any given point. Several important conclusions follow from Examples 1 and 2. Each average velocity in Table 2.1 corresponds to the slope of a secant line on the graph of the position function (Figure 2.5). Just as the average velocities approach a limit as t approaches 1, the slopes of the secant lines approach the same limit as t approaches 1. Specifically, as t approaches 1, two things happen: 1. The secant lines approach a unique line called the tangent line. 2. The slopes of the secant lines msec approach the slope of the tangent line mtan at the point 11, s1122. Thus, the slope of the tangent line is also expressed as a limit: mtan = lim msec = lim tS1 y tS1 s1t2 - s112 = 64. t - 1 This limit is the same limit that defines the instantaneous velocity. Therefore, the instantaneous velocity at t = 1 is the slope of the line tangent to the position curve at t = 1. P s Height above ground (ft) x 128 Slopes of the secant lines approach slope of the tangent line. mtan ⫽ 64 The secant lines approach the tangent line. msec ⫽ 62.4 msec ⫽ 56 msec ⫽ 48 (2, 128) 108 (1.5, 108) (1, 80) 80 (1.1, 86.24) s(t) ⫽ ⫺16t2 ⫹ 96t 0 0.5 1 1.1 1.5 2.0 t Time (s) FIGURE 2.5 QUICK CHECK 3 In Figure 2.5, is mtan at t = 2 greater than or less than mtan at t = 1? ➤ P The parallels between average and instantaneous velocities, on one hand, and between slopes of secant lines and tangent lines, on the other, illuminate the power behind the idea of a limit. As t S 1, slopes of secant lines approach the slope of a tangent line. And as t S 1, average velocities approach an instantaneous velocity. Figure 2.6 summarizes these two parallel limit processes. These ideas lie at the foundation of what follows in the coming chapters. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.1 The Idea of Limits AVERAGE VELOCITY s(t) 16t2 96t Average velocity is the change in position divided by the change in time: t2s 128 s(t ) s(t0) . vav 1 t1 t0 80 t1s SECANT LINE s(t) 16t2 96t s 128 (2, 128) 80 Slope of the secant line is the change in s divided by the change in t: s(t1) s(t0) . msec t1 t0 (1, 80) msec vav 48 0 0.5 1.0 1.5 2.0 t s 108 80 t 1.5 s t1s 108 (1.5, 108) 80 (1, 80) msec vav 56 As the time interval shrinks, the average velocity approaches the instantaneous velocity at t 1. 0 0.5 1.0 1.5 2.0 t s 86.24 80 t 1.1 s t1s 86.24 As the interval on the t-axis shrinks, the slope of the secant line approaches the slope of the tangent line through (1, 80). (1.1, 86.24) (1, 80) 80 msec vav 62.4 0 INSTANTANEOUS VELOCITY 80 t 1 1.0 1.5 2.0 t TANGENT LINE s The instantaneous velocity at t 1 is the limit of the average velocities as t approaches 1. vinst lim 0.5 t1s The slope of the tangent line at (1, 80) is the limit of the slopes of the secant lines as t approaches 1. 80 (1, 80) s(t) s(1) 64 ft/s t1 mtan lim 0 Instantaneous velocity 64 ft/s t 1 0.5 1.0 1.5 2.0 Slope of the tangent line 64 FIGURE 2.6 Copyright © 2014 Pearson Education, Inc. t s(t) s(1) 64 t1 65 For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 66 Limits SECTION 2.1 EXERCISES Review Questions 1. Suppose s1t2 is the position of an object moving along a line at time t Ú 0. What is the average velocity between the times t = a and t = b? 2. Suppose s1t2 is the position of an object moving along a line at time t Ú 0. Describe a process for finding the instantaneous velocity at t = a. 3. What is the slope of the secant line between the points 1a, f 1a22 and 1b, f 1b22 on the graph of f ? 4. Describe a process for finding the slope of the line tangent to the graph of f at 1a, f 1a22. 5. 6. Describe the parallels between finding the instantaneous velocity of an object at a point in time and finding the slope of the line tangent to the graph of a function at a point on the graph. 7. Average velocity The function s1t2 represents the position of an object at time t moving along a line. Suppose s122 = 136 and s132 = 156. Find the average velocity of the object over the interval of time 32, 34 . 8. Average velocity The function s1t2 represents the position of an object at time t moving along a line. Suppose s112 = 84 and s142 = 144. Find the average velocity of the object over the interval of time 31, 44 . 9. Average velocity The position of an object moving along a line is given by the function s1t2 = -16t 2 + 128t. Find the average velocity of the object over the following intervals. 10. Average velocity The position of an object moving along a line is given by the function s1t2 = - 4.9t 2 + 30t + 20. Find the average velocity of the object over the following intervals. 30, 24 30, 1.54 30, 14 30, 0.54 0 0.5 1 1.5 2 s1t2 0 30 52 66 72 s(t) 46 0.5 1 1.5 2 2.5 t T 13. Average velocity Consider the position function s1t2 = -16t 2 + 100t representing the position of an object moving along a line. Sketch a graph of s with the secant line passing through 10.5, s10.522 and 12, s1222. Determine the slope of the secant line and explain its relationship to the moving object. T 14. Average velocity Consider the position function s1t2 = sin pt representing the position of an object moving along a line on the end of a spring. Sketch a graph of s together with a secant line passing through 10, s1022 and 10.5, s10.522. Determine the slope of the secant line and explain its relationship to the moving object. T 15. Instantaneous velocity Consider the position function s1t2 = - 16t 2 + 128t (Exercise 9). Complete the following table with the appropriate average velocities. Then make a conjecture about the value of the instantaneous velocity at t = 1. Time interval 31, 24 31, 1.54 31, 1.14 31, 1.014 31, 1.0014 Average velocity T 16. Instantaneous velocity Consider the position function s1t2 = -4.9t 2 + 30t + 20 (Exercise 10). Complete the following table with the appropriate average velocities. Then make a conjecture about the value of the instantaneous velocity at t = 2. Time interval 32, 34 32, 2.54 32, 2.14 32, 2.014 32, 2.0014 Average velocity T t d. 30.5, 14 84 11. Average velocity The table gives the position s1t2 of an object moving along a line at time t, over a two-second interval. Find the average velocity of the object over the following intervals. a. b. c. d. c. 30.5, 1.54 s b. 30, 24 d. 30, h4, where h 7 0 is a real number a. 30, 34 c. 30, 14 b. 30.5, 24 150 136 114 b. 31, 34 d. 31, 1 + h4, where h 7 0 is a real number a. 31, 44 c. 31, 24 T a. 30.5, 2.54 Graph the parabola f 1x2 = x 2. Explain why the secant lines between the points 1- a, f 1- a22 and 1a, f 1a22 have zero slope. What is the slope of the tangent line at x = 0? Basic Skills T 12. Average velocity The graph gives the position s1t2 of an object moving along a line at time t, over a 2.5-second interval. Find the average velocity of the object over the following intervals. 17. Instantaneous velocity The following table gives the position s1t2 of an object moving along a line at time t. Determine the average velocities over the time intervals 31, 1.014 , 31, 1.0014 , and 31, 1.00014 . Then make a conjecture about the value of the instantaneous velocity at t = 1. t 1 1.0001 1.001 1.01 s1t2 64 64.00479984 64.047984 64.4784 Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.1 The Idea of Limits T 18. Instantaneous velocity The following table gives the position s1t2 of an object moving along a line at time t. Determine the average velocities over the time intervals 32, 2.014, 32, 2.0014, and 32, 2.00014. Then make a conjecture about the value of the instantaneous velocity at t = 2. t 2 2.0001 2.001 2.01 s1t2 56 55.99959984 55.995984 55.9584 T 67 29. Tangent lines with zero slope a. Graph the function f 1x2 = x 2 - 4x + 3. b. Identify the point 1a, f 1a22 at which the function has a tangent line with zero slope. c. Confirm your answer to part (b) by making a table of slopes of secant lines to approximate the slope of the tangent line at this point. 30. Tangent lines with zero slope T a. Graph the function f 1x2 = 4 - x 2. b. Identify the point 1a, f 1a22 at which the function has a tangent line with zero slope. c. Consider the point 1a, f 1a22 found in part (b). Is it true that the secant line between 1a - h, f 1a - h22 and 1a + h, f 1a + h22 has slope zero for any value of h ≠ 0? 19. Instantaneous velocity Consider the position function s1t2 = -16t 2 + 100t. Complete the following table with the appropriate average velocities. Then make a conjecture about the value of the instantaneous velocity at t = 3. Time interval Average velocity T 32, 34 32.9, 34 a. Graph the position function, for 0 … t … 9. b. From the graph of the position function, identify the time at which the projectile has an instantaneous velocity of zero; call this time t = a. c. Confirm your answer to part (b) by making a table of average velocities to approximate the instantaneous velocity at t = a. d. For what values of t on the interval 30, 94 is the instantaneous velocity positive (the projectile moves upward)? e. For what values of t on the interval 30, 94 is the instantaneous velocity negative (the projectile moves downward)? 32.99, 34 32.999, 34 32.9999, 34 T 20. Instantaneous velocity Consider the position function s1t2 = 3 sin t that describes a block bouncing vertically on a spring. Complete the following table with the appropriate average velocities. Then make a conjecture about the value of the instantaneous velocity at t = p>2. Time interval T Average velocity 3p>2, p4 3p>2, p>2 + 0.14 3p>2, p>2 + 0.0014 T 3p>2, p>2 + 0.00014 Further Explorations 21–24. Instantaneous velocity For the following position functions, make a table of average velocities similar to those in Exercises 19 and 20 and make a conjecture about the instantaneous velocity at the indicated time. 21. s1t2 = - 16t 2 + 80t + 60 22. s1t2 = 20 cos t at t = p>2 23. s1t2 = 40 sin 2t at t = 0 24. s1t2 = 20>1t + 12 at t = 3 33. Slope of tangent line Given the function f 1x2 = 1 - cos x and the points A1p>2, f 1p>222, B1p>2 + 0.05, f 1p>2 + 0.0522, C1p>2 + 0.5, f 1p>2 + 0.522, and D1p, f 1p22 (see figure), find the slopes of the secant lines through A and D, A and C, and A and B. Then use your calculations to make a conjecture about the slope of the line tangent to the graph of f at x = p>2. y D 2 y ⫽ 1 ⫺ cos x C A 1 B at t = 0 25–28. Slopes of tangent lines For the following functions, make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at the indicated point. 25. f 1x2 = 2x 2 at x = 2 26. f 1x2 = 3 cos x at x = p>2 27. f 1x2 = e x at x = 0 28. f 1x2 = x 3 - x at x = 1 0 q q ⫹ 0.5 x q ⫹ 0.05 QUICK CHECK ANSWERS 1. 16 ft>s 2. Average velocity is the velocity over an interval of time. Instantaneous velocity is the velocity at one point of time. 3. Less than ➤ T 32. Impact speed A rock is dropped off the edge of a cliff, and its distance s (in feet) from the top of the cliff after t seconds is s1t2 = 16t 2. Assume the distance from the top of the cliff to the ground is 96 ft. a. When will the rock strike the ground? b. Make a table of average velocities and approximate the velocity at which the rock strikes the ground. 3p>2, p>2 + 0.014 T 31. Zero velocity A projectile is fired vertically upward and has a position given by s1t2 = -16t 2 + 128t + 192, for 0 … t … 9. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 68 Limits 2.2 Definitions of Limits Computing tangent lines and instantaneous velocities (Section 2.1) are just two of many important calculus problems that rely on limits. We now put these two problems aside until Chapter 3 and begin with a preliminary definition of the limit of a function. DEFINITION Limit of a Function (Preliminary) Suppose the function f is defined for all x near a except possibly at a. If f 1x2 is arbitrarily close to L (as close to L as we like) for all x sufficiently close (but not equal) to a, we write ➤ The terms arbitrarily close and sufficiently close will be made precise when rigorous definitions of limits are given in Section 2.7. lim f 1x2 = L xSa and say the limit of f 1x2 as x approaches a equals L. Informally, we say that lim f 1x2 = L if f 1x2 gets closer and closer to L as x gets xSa closer and closer to a from both sides of a. The value of lim f 1x2 (if it exists) depends xSa upon the values of f near a, but it does not depend on the value of f 1a2. In some cases, the limit lim f 1x2 equals f 1a2. In other instances, lim f 1x2 and f 1a2 differ, or f 1a2 may not xSa xSa even be defined. EXAMPLE 1 Finding limits from a graph Use the graph of f (Figure 2.7) to determine the following values, if possible. y 6 a. f 112 and lim f 1x2 y ⫽ f (x) 5 xS1 b. f 122 and lim f 1x2 xS2 c. f 132 and lim f 1x2 xS3 4 SOLUTION 3 a. We see that f 112 = 2. As x approaches 1 from either side, the values of f 1x2 approach 2 (Figure 2.8). Therefore, lim f 1x2 = 2. xS1 2 1 0 1 2 3 4 5 6 x b. We see that f 122 = 5. However, as x approaches 2 from either side, f 1x2 approaches 3 because the points on the graph of f approach the open circle at 12, 32 (Figure 2.9). Therefore, lim f 1x2 = 3 even though f 122 = 5. xS2 c. In this case, f 132 is undefined. We see that f 1x2 approaches 4 as x approaches 3 from either side (Figure 2.10). Therefore, lim f 1x2 = 4 even though f 132 does not exist. FIGURE 2.7 xS3 y y y f (1) ⫽ 2 f (2) ⫽ 5 6 f (3) undefined 6 6 y ⫽ f (x) y ⫽ f (x) y ⫽ f (x) 4 ... f (x) approaches 2. 1 As x approaches 1... FIGURE 2.8 6 x 0 2 6 x As x approaches 2... FIGURE 2.9 0 3 6 x As x approaches 3... FIGURE 2.10 Related Exercises 7–10 Copyright © 2014 Pearson Education, Inc. ➤ 2 0 ... f (x) approaches 4. ... f (x) approaches 3. 3 For Use Only in 2013 – 2014 Pilot Program 2.2 Definitions of Limits 69 QUICK CHECK 1 In Example 1, suppose we redefine the function at one point so that f 112 = 1. Does this change the value of lim f 1x2? ➤ xS1 In Example 1, we worked with the graph of a function. Let’s now work with tabulated values of a function. 1x - 1 x - 1 corresponding to values of x near 1. Then make a conjecture about the value of lim f 1x2. EXAMPLE 2 Finding limits from a table Create a table of values of f 1x2 = ➤ In Example 2, we have not stated with certainty that lim f 1x2 = 0.5. But this is xS1 our best guess based upon the numerical evidence. Methods for calculating limits precisely are introduced in Section 2.3. xS1 SOLUTION Table 2.2 lists values of f corresponding to values of x approaching 1 from both sides. The numerical evidence suggests that f 1x2 approaches 0.5 as x approaches 1. Therefore, we make the conjecture that lim f 1x2 = 0.5. xS1 x f 1x2 = 1x − 1 x − 1 S 1 d 0.9 0.99 0.999 0.9999 1.0001 1.001 1.01 1.1 0.5131670 0.5012563 0.5001251 0.5000125 0.4999875 0.4998751 0.4987562 0.4880885 Related Exercises 11–14 ➤ Table 2.2 One-Sided Limits The limit lim f 1x2 = L is referred to as a two-sided limit because f 1x2 approaches L as x xSa approaches a for values of x less than a and for values of x greater than a. For some functions, it makes sense to examine one-sided limits called left-sided and right-sided limits. DEFINITION One-Sided Limits ➤ As with two-sided limits, the value of a one-sided limit (if it exists) depends on the values of f 1x2 near a but not on the value of f 1a2. 1. Right-sided limit Suppose f is defined for all x near a with x 7 a. If f 1x2 is arbitrarily close to L for all x sufficiently close to a with x 7 a, we write lim f 1x2 = L x S a+ and say the limit of f 1x2 as x approaches a from the right equals L. 2. Left-sided limit Suppose f is defined for all x near a with x 6 a. If f 1x2 is arbitrarily close to L for all x sufficiently close to a with x 6 a, we write lim f 1x2 = L x S a- and say the limit of f 1x2 as x approaches a from the left equals L. x3 - 8 . 41x - 22 Use tables and graphs to make a conjecture about the values of lim+ f 1x2, lim- f 1x2, and EXAMPLE 3 Examining limits graphically and numerically Let f 1x2 = ➤ Computer-generated graphs and tables help us understand the idea of a limit. Keep in mind, however, that computers are not infallible and they may produce incorrect results, even for simple functions (see Example 5). xS2 xS2 lim f 1x2, if they exist. xS2 SOLUTION Figure 2.11a shows the graph of f obtained with a graphing utility. The graph is misleading because f 122 is undefined, which means there should be a hole in the graph at 12, 32 (Figure 2.11b). Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 70 Limits y f (x) ⫽ y x3 ⫺ 8 4(x ⫺ 2) x3 ⫺ 8 4(x ⫺ 2) f (x) ⫽ 3 3 2 2 The hole in the graph at x ⫽ 2 indicates that the function is undefined at this point. This computergenerated graph is inaccurate because f is undefined at x ⫽ 2. 0 1 x 2 FIGURE 2.11 0 1 x 2 (a) (b) The graph in Figure 2.12a and the function values in Table 2.3 suggest that f 1x2 approaches 3 as x approaches 2 from the right. Therefore, we write lim f 1x2 = 3, x S 2+ which says the limit of f 1x2 as x approaches 2 from the right equals 3. y f (x) ⫽ y x3 ⫺ 8 4(x ⫺ 2) f (x) ⫽ x3 ⫺ 8 4(x ⫺ 2) ... f (x) approaches 3. 3 3 2 2 ... f (x) approaches 3. 0 1 2 x x 0 As x approaches 2 from the right... FIGURE 2.12 ➤ Remember that the value of the limit does not depend upon the value of f 122. In this case, lim f 1x2 = 3 despite the fact that 1 x x 2 As x approaches 2 from the left... (a) (b) Similarly, Figure 2.12b and Table 2.3 suggest that as x approaches 2 from the left, f 1x2 approaches 3. So, we write lim f 1x2 = 3, xS2 f 122 is undefined. x S 2- which says the limit of f 1x2 as x approaches 2 from the left equals 3. Because f 1x2 approaches 3 as x approaches 2 from either side, we write lim f 1x2 = 3. xS2 Table 2.3 S x x − 8 41x − 22 d 1.99 1.999 1.9999 2.0001 2.001 2.01 2.1 2.8525 2.985025 2.99850025 2.99985000 3.00015000 3.00150025 3.015025 3.1525 3 Related Exercises 15–18 Copyright © 2014 Pearson Education, Inc. ➤ f 1x2 = 2 1.9 For Use Only in 2013 – 2014 Pilot Program 2.2 Definitions of Limits 71 Based upon the previous example, you might wonder whether the limits lim- f 1x2, lim+ f 1x2, and lim f 1x2 always exist and are equal. The remaining examples xSa xSa xSa demonstrate that these limits may have different values, and in other cases, some or all of these limits may not exist. The following theorem is useful when comparing one-sided and two-sided limits. ➤ If P and Q are statements, we write P if THEOREM 2.1 Relationship Between One-Sided and Two-Sided Limits Assume f is defined for all x near a except possibly at a. Then lim f 1x2 = L if xSa and only if lim+ f 1x2 = L and lim- f 1x2 = L. and only if Q when P implies Q and Q implies P. xSa xSa A proof of Theorem 2.1 is outlined in Exercise 44 of Section 2.7. Using this theorem, it follows that lim f 1x2 ≠ L if either lim+ f 1x2 ≠ L or lim- f 1x2 ≠ L (or both). xSa xSa xSa Furthermore, if either lim+ f 1x2 or lim- f 1x2 does not exist, then lim f 1x2 does not exist. xSa xSa xSa We put these ideas to work in the next two examples. EXAMPLE 4 A function with a jump Given the graph of g in Figure 2.13, find the y following limits, if they exist. 6 a. lim- g1x2 xS2 b. lim+ g1x2 c. lim g1x2 xS2 xS2 SOLUTION 4 a. As x approaches 2 from the left, g1x2 approaches 4. Therefore, lim- g1x2 = 4. y g(x) xS2 b. Because g1x2 = 1, for all x Ú 2, lim+ g1x2 = 1. 2 xS2 c. By Theorem 2.1, lim g1x2 does not exist because lim- g1x2 ≠ lim+ g1x2. 0 2 4 xS2 xS2 Related Exercises 19–24 x ➤ xS2 EXAMPLE 5 Some strange behavior Examine lim cos 11>x2. FIGURE 2.13 xS0 SOLUTION From the first three values of cos 11>x2 in Table 2.4, it is tempting to conclude that lim+ cos 11>x2 = -1. But this conclusion is not confirmed when we evaluate cos 11>x2 xS0 for values of x closer to 0. Table 2.4 cos 11 , x2 x 0.001 0.0001 0.00001 0.000001 0.0000001 0.00000001 0.56238 -0.95216 ¶ -0.99936 0.93675 -0.90727 -0.36338 We might incorrectly conclude that cos 11>x2 approaches - 1 as x approaches 0 from the right. The behavior of cos 11>x2 near 0 is better understood by letting x = 1>1np2, where n is a positive integer. By making this substitution, we can sample the function at discrete points that approach zero. In this case cos QUICK CHECK 2 ➤ Why is the graph of y = cos 11>x2 difficult to plot near x = 0, as suggested by Figure 2.14? 1 1 if n is even = cos np = b x -1 if n is odd. As n increases, the values of x = 1>1np2 approach zero, while the values of cos 11>x2 oscillate between -1 and 1 (Figure 2.14). Therefore, cos 11>x2 does not approach a single number as x approaches 0 from the right. We conclude that lim+ cos 11>x2 does xS0 not exist, which implies that lim cos 11>x2 does not exist. xS0 Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 72 Limits The values of cos (1/x) oscillate between 1 and 1, over shorter and shorter intervals, as x approaches 0 from the right. y 1 y cos(1/x) 0 1 6 1 5 1 4 1 3 x 1 2 1 Related Exercises 25–26 ➤ FIGURE 2.14 Using tables and graphs to make conjectures for the values of limits worked well until Example 5. The limitation of technology in this example is not an isolated incident. For this reason, analytical techniques (paper-and-pencil methods) for finding limits are developed in the next section. SECTION 2.2 EXERCISES Review Questions 8. 1. Explain the meaning of lim f 1x2 = L. 2. True or false: When lim f 1x2 exists, it always equals f 1a2. Explain. xSa Finding limits from a graph Use the graph of g in the figure to find the following values, if they exist. a. g102 b. lim g1x2 c. g112 d. lim g1x2 xS0 xSa xS1 y 3. Explain the meaning of lim+ f 1x2 = L. 4. Explain the meaning of lim- f 1x2 = L. 5. If lim- f 1x2 = L and lim+ f 1x2 = M, where L and M are finite xSa xSa xSa xSa real numbers, then how are L and M related if lim f 1x2 exists? xSa 6. What are the potential problems of using a graphing utility to estimate lim f 1x2? 2 y g(x) xSa Basic Skills 7. Finding limits from a graph Use the graph of h in the figure to find the following values, if they exist. a. h122 b. lim h1x2 xS2 c. h142 d. lim h1x2 e. lim h1x2 xS4 xS5 1 9. Finding limits from a graph Use the graph of f in the figure to find the following values, if they exist. a. f 112 b. lim f 1x2 c. f 102 d. lim f 1x2 xS1 y x 1 xS0 y 5 y f (x) y h(x) 3 1 1 1 1 0 1 2 4 x Copyright © 2014 Pearson Education, Inc. x For Use Only in 2013 – 2014 Pilot Program 2.2 Definitions of Limits 10. Finding limits from a graph Use the graph of f in the figure to find the following values, if they exist. a. f 122 b. lim f 1x2 c. lim f 1x2 xS2 T xS5 y 6 xS0 c. What mathematical constant does lim 11 + x21>x appear to xS0 equal? y f (x) 4 T 2 xS2 b. Evaluate f 1x2 for values of x near 2 to support your conjecture in part (a). 0 1 2 4 6 11. Estimating a limit from tables Let f 1x2 = x T x2 - 4 . x - 2 a. Calculate f 1x2 for each value of x in the following table. x2 - 4 b. Make a conjecture about the value of lim . xS2 x - 2 x f 1x2 = T 1.9 1.99 1.999 1.9999 2.1 2.01 2.001 2.0001 xS0 T x − 4 x − 2 b. Evaluate f 1x2 for values of x near 1 to support your conjecture in part (a). T 12. Estimating a limit from tables Let f 1x2 = x3 - 1 . x - 1 xS0 T 0.99 0.999 0.9999 x3 − 1 x − 1 x f 1x2 = b. Evaluate g1x2 for values of x near 0 to support your conjecture in part (a). T 0.9 18. Estimating a limit graphically and numerically 3 sin x - 2 cos x + 2 Let g1x2 = . x a. Plot a graph of g to estimate lim g1x2. a. Calculate f 1x2 for each value of x in the following table. x3 - 1 b. Make a conjecture about the value of lim . xS1 x - 1 f 1x2 = 17. Estimating a limit graphically and numerically 1 - cos12x - 22 Let f 1x2 = . 1x - 122 a. Plot a graph of f to estimate lim f 1x2. xS1 x2 − 4 x − 2 x 16. Estimating a limit graphically and numerically e 2x - 2x - 1 Let g1x2 = . x2 a. Plot a graph of g to estimate lim g1x2. b. Evaluate g1x2 for values of x near 0 to support your conjecture in part (a). 2 x f 1x2 = 15. Estimating a limit graphically and numerically x - 2 Let f 1x2 = . ln x - 2 a. Plot a graph of f to estimate lim f 1x2. 1 T 14. Estimating a limit of a function Let f 1x2 = 11 + x21>x. a. Make two tables, one showing the values of f for x = 0.01, 0.001, 0.0001, and 0.00001 and one showing values of f for x = - 0.01, -0.001, -0.0001, and -0.00001. Round your answers to five digits. b. Estimate the value of lim 11 + x21>x. d. lim f 1x2 xS4 73 x 2 - 25 . Use tables x - 5 and graphs to make a conjecture about the values of lim+ f 1x2, xS5 lim- f 1x2, and lim f 1x2, if they exist. 19. One-sided and two-sided limits Let f 1x2 = xS5 1.1 1.01 1.001 1.0001 x3 − 1 x − 1 T xS5 x - 100 . Use 1x - 10 tables and graphs to make a conjecture about the values of lim + g1x2, lim - g1x2, and lim g1x2, if they exist. 20. One-sided and two-sided limits Let g1x2 = x S 100 t - 9 . 1t - 3 a. Make two tables, one showing the values of g for t = 8.9, 8.99, and 8.999 and one showing values of g for t = 9.1, 9.01, and 9.001. t - 9 . b. Make a conjecture about the value of lim S t 9 1t - 3 13. Estimating a limit of a function Let g1t2 = Copyright © 2014 Pearson Education, Inc. x S 100 x S 100 For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 74 Limits 21. One-sided and two-sided limits Use the graph of f in the figure to find the following values, if they exist. If a limit does not exist, explain why. a. f 112 b. lim- f 1x2 c. lim+ f 1x2 xS1 y 5 y f (x) d. lim f 1x2 xS1 xS1 3 y 2 1 1 y f (x) 2 3 x 5 24. Finding limits from a graph Use the graph of g in the figure to find the following values, if they exist. If a limit does not exist, explain why. 1 0 x 1 a. g1- 12 b. d. lim g1x2 e. g112 f. lim g1x2 g. lim g1x2 h. g152 i. xS - 1 22. One-sided and two-sided limits Use the graph of g in the figure to find the following values, if they exist. If a limit does not exist, explain why. a. g122 b. lim- g1x2 c. lim+ g1x2 d. lim g1x2 e. g132 f. lim- g1x2 g. lim+ g1x2 h. g142 i. lim g1x2 xS2 xS2 xS3 xS3 lim g1x2 x S - 1- c. lim g1x2 x S - 1+ xS1 lim g1x2 x S 5- y 6 xS2 5 xS3 4 xS4 y g(x) 2 y 1 5 3 y g(x) T 25. Strange behavior near x=0 x 2 2 2 , , , p 3p 5p 2 2 2 2 , , and . Describe the pattern of values you observe. 7p 9p 11p b. Why does a graphing utility have difficulty plotting the graph of y = sin 11>x2 near x = 0 (see figure)? c. What do you conclude about lim sin 11>x2? 1 2 3 x 5 xS0 y 23. Finding limits from a graph Use the graph of f in the figure to find the following values, if they exist. If a limit does not exist, explain why. a. f 112 b. lim- f 1x2 c. lim+ f 1x2 d. lim f 1x2 e. f 132 f. lim- f 1x2 g. lim+ f 1x2 h. lim f 1x2 i. f 122 j. lim- f 1x2 k. lim+ f 1x2 l. lim f 1x2 xS2 5 a. Create a table of values of sin 1 1>x 2 , for x = 0 xS3 1 3 1 xS1 1 xS1 xS3 xS2 1 xS1 0 xS3 xS2 2 9 2 7 2 5 x 1 T 26. Strange behavior near x=0 a. Create a table of values of tan 13>x2 for x = 12>p, 12>13p2, 12>15p2, c, 12>111p2. Describe the general pattern in the values you observe. b. Use a graphing utility to graph y = tan 13>x2. Why does a graphing utility have difficulty plotting the graph near x = 0? c. What do you conclude about lim tan 13>x2? xS0 Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.2 Definitions of Limits 75 e. For what values of a does lim : x ; exist? Explain. Further Explorations xSa 27. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. y x2 - 9 does not exist. a. The value of lim xS3 x - 3 b. The value of lim f 1x2 is always found by computing f 1a2. 3 xSa c. The value of lim f 1x2 does not exist if f 1a2 is undefined. d. lim 1x = 0 1 xSa xS0 e. 3 lim cot x = 0 1 x S p>2 3 x y :x; 28–29. Sketching graphs of functions Sketch the graph of a function with the given properties. You do not need to find a formula for the function. 3 28. f 112 = 0, f 122 = 4, f 132 = 6, lim- f 1x2 = - 3, lim + f 1x2 = 5 xS2 xS2 29. g112 = 0, g122 = 1, g132 = - 2, lim g1x2 = 0, 38. The ceiling function For any real number x, the ceiling function < x = is the least integer greater than or equal to x. xS2 lim g1x2 = -1, lim+ g1x2 = - 2 x S 3- a. Graph the ceiling function y = < x = , for -2 … x … 3. xS3 30. h1- 12 = 2, lim - h1x2 = 0, lim + h1x2 = 3, x S -1 b. Evaluate lim- < x = , lim+ < x = , and lim < x = . x S -1 h112 = lim- h1x2 = 1, lim+ h1x2 = 4 xS1 31. p102 = 2, lim p1x2 = 0, lim p1x2 does not exist, xS0 xSa Applications xS2 32–35. Calculator limits Estimate the value of the following limits by creating a table of function values for h = 0.01, 0.001, and 0.0001, and h = - 0.01, - 0.001, and - 0.0001. 32. lim 11 + 2h21>h hS0 2 - 1 h h hS0 35. lim ln 11 + h2 hS0 w S 3.3 h c. Interpret the limits lim+ f 1w2 and lim- f 1w2. wS1 x , for x ≠ 0. x 36. A step function Let f 1x2 = 39. Postage rates Assume that postage for sending a first-class letter in the United States is $0.44 for the first ounce (up to and including 1 oz) plus $0.17 for each additional ounce (up to and including each additional ounce). a. Graph the function p = f 1w2 that gives the postage p for sending a letter that weighs w ounces, for 0 6 w … 5. b. Evaluate lim f 1w2. 33. lim 11 + 3h22>h hS0 34. lim wS4 xS0 40. The Heaviside function The Heaviside function is used in engineering applications to model flipping a switch. It is defined as examining lim- f 1x2 and lim+ f 1x2. xS0 H1x2 = b 37. The floor function For any real number x, the floor function (or greatest integer function) : x ; is the greatest integer less than or equal to x (see figure). xS - 1 x S 2.3 x S 2.3 xS2 b. Compute lim - : x ; , lim + : x ; , and lim : x ; . xS0 x S 2.3 xSa xSa d. In general, if a is not an integer, state the values of lim- : x ; xSa and lim+ : x ; . xSa examining lim- H1x2 and lim+ H1x2. xS2 c. For a given integer a, state the values of lim- : x ; and lim+ : x ; . 0 if x 6 0 1 if x Ú 0. a. Sketch a graph of H on the interval 3-1, 24. b. Does lim H1x2 exist? Explain your reasoning after first a. Compute lim - : x ; , lim + : x ; , lim- : x ; , and lim+ : x ; . xS - 1 wS1 d. Does lim f 1w2 exist? Explain. a. Sketch a graph of f on the interval 3-2, 24. b. Does lim f 1x2 exist? Explain your reasoning after first xS0 x S 1.5 xS2 p122 = lim+ p1x2 = 1 T xS1 xS2 c. For what values of a does lim < x = exist? Explain. xS1 xS0 xS0 Additional Exercises 41. Limits of even functions A function f is even if f 1-x2 = f 1x2, for all x in the domain of f. If f is even, with lim+ f 1x2 = 5 and xS2 lim- f 1x2 = 8, find the following limits. xS2 a. lim x S - 2+ f 1x2 b. lim x S - 2- f 1x2 42. Limits of odd functions A function g is odd if g1- x2 = -g1x2, for all x in the domain of g. If g is odd, with lim+ g1x2 = 5 and xS2 lim- g1x2 = 8, find the following limits. xS2 a. lim g1x2 x S - 2+ Copyright © 2014 Pearson Education, Inc. b. lim g1x2 x S - 2- For Use Only in 2013 – 2014 Pilot Program Limits Technology Excercises T 43. lim x sin xS0 45. lim xS1 T T 43–46. Limit by graphing Use the zoom and trace features of a graphing utility to approximate the following limits. 1 x 44. lim xS1 3 9 1 22x - x 4 - 2 x2 1 - x 3>4 3 1812x - 12 x3 - 1 6x - 3x x S 0 x ln 2 46. lim 47. Limits by graphs tan 2x , a. Use a graphing utility to estimate lim x S 0 sin x tan 3x tan 4x lim , and lim . x S 0 sin x x S 0 sin x tan px , for any real b. Make a conjecture about the value of lim S x 0 sin x constant p. T sin px for sin qx at least three different pairs of nonzero constants p and q of your sin px choice. Estimate lim in each case. Then use your work to x S 0 sin qx sin px make a conjecture about the value of lim for any nonzero x S 0 sin qx values of p and q. 49. Limits by graphs Use a graphing utility to plot y = sin nx , for n = 1, 2, 3, and 4 x (four graphs). Use the window 3-1, 14 * 30, 54. sin x sin 2x sin 3x sin 4x a. Estimate lim , lim , lim , and lim . x xS0 x x xS0 x xS0 xS0 sin px b. Make a conjecture about the value of lim , for any real x xS0 constant p. 48. Limits by graphs Graph f 1x2 = QUICK CHECK ANSWERS 1. The value of lim f 1x2 depends on the value of f only xS1 near 1, not at 1. Therefore, changing the value of f 112 will not change the value of lim f 1x2. 2. A graphing device xS1 has difficulty plotting y = cos 11>x2 near 0 because values of the function vary between -1 and 1 over shorter and shorter intervals as x approaches 0. ➤ Chapter 2 r 76 2.3 Techniques for Computing Limits Graphical and numerical techniques for estimating limits, like those presented in the previous section, provide intuition about limits. These techniques, however, occasionally lead to incorrect results. Therefore, we turn our attention to analytical methods for evaluating limits precisely. Limits of Linear Functions The graph of f 1x2 = mx + b is a line with slope m and y-intercept b. From Figure 2.15, we see that f 1x2 approaches f 1a2 as x approaches a. Therefore, if f is a linear function we have lim f 1x2 = f 1a2. It follows that for linear functions, lim f 1x2 is found by direct xSa xSa substitution of x = a into f 1x2. This observation leads to the following theorem, which is proved in Exercise 28 of Section 2.7. y y y f (x) y f (x) f (x) f (a) (a, f (a)) ... f (x) approaches f (a) f (a) (a, f (a)) ... f (x) approaches f (a) f (x) O x a As x approaches a from the left... x lim f (x) f (a) because f (x) ᠬ x a FIGURE 2.15 O a x As x approaches a from the right... f (a) as x a from both sides of a. Copyright © 2014 Pearson Education, Inc. x For Use Only in 2013 – 2014 Pilot Program 2.3 Techniques for Computing Limits 77 THEOREM 2.2 Limits of Linear Functions Let a, b, and m be real numbers. For linear functions f 1x2 = mx + b, lim f 1x2 = f 1a2 = ma + b. xSa EXAMPLE 1 Limits of linear functions Evaluate the following limits. a. lim f 1x2, where f 1x2 = 12 x - 7 b. lim g1x2, where g1x2 = 6 xS3 xS2 SOLUTION xS3 xS3 1 12 x - 7 2 = f 132 = - 11 2. b. lim g1x2 = lim 6 = g122 = 6. xS2 xS2 Related Exercises 11–16 ➤ a. lim f 1x2 = lim Limit Laws The following limit laws greatly simplify the evaluation of many limits. THEOREM 2.3 Limit Laws Assume lim f 1x2 and lim g1x2 exist. The following properties hold, where c is a xSa xSa real number, and m 7 0 and n 7 0 are integers. 1. Sum lim 3 f 1x2 + g1x24 = lim f 1x2 + lim g1x2 xSa xSa xSa 2. Difference lim 3 f 1x2 - g1x24 = lim f 1x2 - lim g1x2 xSa xSa xSa 3. Constant multiple lim 3cf 1x24 = c lim f 1x2 xSa xSa 3 43 4 4. Product lim 3 f 1x2g1x24 = lim f 1x2 lim g1x2 xSa 5. Quotient lim c xSa ➤ Law 6 is a special case of Law 7. Letting m = 1 in Law 7 gives Law 6. xSa xSa lim f 1x2 f 1x2 xSa d = , provided lim g1x2 ≠ 0 g1x2 lim g1x2 xSa xSa 3 4 6. Power lim 3 f 1x24 = lim f 1x2 n xSa xSa n 3 4 7. Fractional power lim 3 f 1x24 n>m = lim f 1x2 xSa xSa n>m , provided f 1x2 Ú 0, for x near a, if m is even and n>m is reduced to lowest terms A proof of Law 1 is outlined in Section 2.7. Laws 2–5 are proved in Appendix B. Law 6 is proved from Law 4 as follows. For a positive integer n, if lim f 1x2 exists, we xSa have lim 3 f 1x24 n = lim 3 f 1x2 f 1x2 g f 1x24 x S a (+++)+++* xSa 3 n factors of f 1x2 43 4 3 4 = lim f 1x2 lim f 1x2 g lim f 1x2 xSa xSa xSa (++ ++++++)++++++ ++* n factors of lim f 1x2 3 4 n = lim f 1x2 . xSa Copyright © 2014 Pearson Education, Inc. xSa Repeated use of Law 4 For Use Only in 2013 – 2014 Pilot Program Chapter 2 r Limits ➤ Recall that to take even roots of a number (for example, square roots or fourth roots), the number must be nonnegative if the result is to be real. In Law 7, the limit of 3 f 1x24 n>m involves the mth root of f 1x2 when x is near a. If the fraction n>m is in lowest terms and m is even, this root is undefined unless f 1x2 is nonnegative for all x near a, which explains the restrictions shown. EXAMPLE 2 Evaluating limits Suppose lim f 1x2 = 4, lim g1x2 = 5, and xS2 xS2 lim h1x2 = 8. Use the limit laws in Theorem 2.3 to compute each limit. xS2 a. lim xS2 f 1x2 - g1x2 h1x2 b. lim 36f 1x2g1x2 + h1x24 c. lim 3g1x24 3 xS2 xS2 SOLUTION a. lim xS2 lim 3 f 1x2 - g1x24 f 1x2 - g1x2 xS2 = h1x2 lim h1x2 Law 5 xS2 lim f 1x2 - lim g1x2 xS2 = xS2 Law 2 lim h1x2 xS2 4 - 5 1 = - . 8 8 = b. lim 36 f 1x2g1x2 + h1x24 = lim 36 f 1x2g1x24 + lim h1x2 xS2 xS2 = 6 # lim 3 f 1x2g1x24 + lim h1x2 Law 3 = 6 # lim f 1x2 Law 4 xS2 3 xS2 4 # 3 lim g1x24 + lim h1x2 xS2 xS2 xS2 = 6 # 4 # 5 + 8 = 128. 3 4 c. lim 3g1x24 3 = lim g1x2 xS2 xS2 Law 1 xS2 = 53 = 125. 3 Law 6 Related Exercises 17–24 ➤ 78 Limits of Polynomial and Rational Functions The limit laws are now used to find the limits of polynomial and rational functions. For example, to evaluate the limit of the polynomial p1x2 = 7x 3 + 3x 2 + 4x + 2 at an arbitrary point a, we proceed as follows: lim p1x2 = lim 17x 3 + 3x 2 + 4x + 22 xSa xSa = lim 17x 32 + lim 13x 22 + lim 14x + 22 Law 1 = 7 lim 1x 32 + 3 lim 1x 22 + lim 14x + 22 Law 3 xSa xSa xSa xSa xSa 3 1 2 + 3 lim x xSa s a 2 + lim 14x + 22 xSa u 2 = 7 lim x xSa s 1 xSa a 4a + 2 = 7a + 3a + 4a + 2 = p1a2. 3 2 Law 6 Theorem 2.2 As in the case of linear functions, the limit of a polynomial is found by direct substitution; that is, lim p1x2 = p1a2 (Exercise 91). xSa It is now a short step to evaluating limits of rational functions of the form f 1x2 = p1x2>q1x2, where p and q are polynomials. Applying Law 5, we have lim xSa lim p1x2 p1x2 p1a2 xSa = = , provided q1a2 ≠ 0, q1x2 lim q1x2 q1a2 xSa which shows that limits of rational functions are also evaluated by direct substitution. Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.3 Techniques for Computing Limits xSa 2 used to evaluate a limit become clear in Section 2.6, when the important property of continuity is discussed. THEOREM 2.4 Limits of Polynomial and Rational Functions Assume p and q are polynomials and a is a constant. a. Polynomial functions: b. Rational functions: lim p1x2 = p1a2 xSa p1a2 p1x2 = , provided q1a2 ≠ 0 x S a q1x2 q1a2 lim 3x 2 - 4x . 5x 3 - 36 QUICK CHECK 1 EXAMPLE 3 Limit of a rational function Evaluate lim xS2 SOLUTION Notice that the denominator of this function is nonzero at x = 2. Using x S -1 x - 1 . x ➤ lim xS2 Theorem 2.4b, lim xS2 31222 - 4122 3x 2 - 4x = = 1. 5x 3 - 36 51232 - 36 Related Exercises 25–27 QUICK CHECK 2 Use Theorem 2.4b to compute lim xS1 5x 4 - 3x 2 + 8x - 6 . x + 1 EXAMPLE 4 An algebraic function Evaluate lim xS2 SOLUTION Using Theorems 2.3 and 2.4, we have 22x 3 + 9 + 3x - 1 lim = xS2 4x + 1 = ➤ Evaluate lim 12x 4 - 8x - 162 and 22x 3 + 9 + 3x - 1 . 4x + 1 lim 1 22x 3 + 9 + 3x - 1 2 xS2 Law 5 lim 14x + 12 xS2 2lim 12x 3 + 92 + lim 13x - 12 xS2 xS2 Laws 1 and 7 lim 14x + 12 xS2 212122 + 92 + 13122 - 12 14122 + 12 125 + 5 10 = = . 9 9 3 = Theorem 2.4 Notice that the limit at x = 2 equals the value of the function at x = 2. Related Exercises 28–32 ➤ 1 substitution lim f 1x2 = f 1a2 can be ➤ ➤ The conditions under which direct 79 One-Sided Limits Theorem 2.2, Limit Laws 1–6, and Theorem 2.4 also hold for left-sided and right-sided limits. In other words, these laws remain valid if we replace lim with lim+ or lim-. Law 7 xSa xSa must be modified slightly for one-sided limits, as shown in the next theorem. Copyright © 2014 Pearson Education, Inc. xSa For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 80 Limits THEOREM 2.3 (CONTINUED) Limit Laws for One-Sided Limits Laws 1–6 hold with lim replaced by lim+ or lim-. Law 7 is modified as follows. xSa xSa xSa Assume m 7 0 and n 7 0 are integers. 7. Fractional power 4 a. lim+ 3 f 1x24 n>m = 3 lim f 1x2 n>m, provided f 1x2 Ú 0, for x near a with x 7 a, if m is even and n>m is reduced to lowest terms. b. lim- 3 f 1x24 n>m = 3 lim f 1x2 n>m, xSa x S a+ x S a- xSa 4 provided f 1x2 Ú 0, for x near a with x 6 a, if m is even and n>m is reduced to lowest terms. EXAMPLE 5 Calculating left- and right-sided limits Let y 2x 4 if x 1 x 1 if x 1 f (x) 4 f 1x2 = b Find the values of lim- f 1x2, lim+ f 1x2, and lim f 1x2, or state that they do not exist. xS1 lim f (x) 2 xS1 xS1 xS1 We verify this observation analytically by applying the limit laws. For x … 1, f 1x2 = -2x + 4; therefore, lim f (x) 0 x 1 lim f 1x2 = lim- 1-2x + 42 = 2. x S 1- x xS1 Theorem 2.2 For x 7 1, note that x - 1 7 0; it follows that lim f 1x2 = lim+ 1x - 1 = 0. x S 1+ xS1 Law 7 Because lim- f 1x2 = 2 and lim+ f 1x2 = 0, lim f 1x2 does not exist by Theorem 2.1. xS1 xS1 xS1 Related Exercises 33–38 ➤ 1 FIGURE 2.16 xS1 SOLUTION The graph of f (Figure 2.16) suggests that lim- f 1x2 = 2 and lim+ f 1x2 = 0. x 1 2 -2x + 4 if x … 1 1x - 1 if x 7 1. Other Techniques So far, we have evaluated limits by direct substitution. A more challenging problem is finding lim f 1x2 when the limit exists, but lim f 1x2 ≠ f 1a2. Two typical cases are xSa xSa shown in Figure 2.17. In the first case, f 1a2 is defined, but it is not equal to lim f 1x2; in xSa the second case, f 1a2 is not defined at all. y y lim f (x) f (a) ᠬ lim f (x) exists, but f (a) is undefined. ᠬ x a x a y f (x) f (a) O a x FIGURE 2.17 Copyright © 2014 Pearson Education, Inc. y f (x) O a x For Use Only in 2013 – 2014 Pilot Program 2.3 Techniques for Computing Limits 81 EXAMPLE 6 Other techniques Evaluate the following limits. a. lim xS2 x 2 - 6x + 8 x2 - 4 b. lim xS1 1x - 1 x - 1 SOLUTION a. Factor and Cancel This limit cannot be found by direct substitution because the denominator is zero when x = 2. Instead, the numerator and denominator are factored; then, assuming x ≠ 2, we cancel like factors: 1x - 221x - 42 x 2 - 6x + 8 x - 4 = = . 2 1x - 221x + 22 x + 2 x - 4 ➤ The argument used in this example is common. In the limit process, x approaches 2, but x ≠ 2. Therefore, we may cancel like factors. x - 4 x 2 - 6x + 8 = whenever x ≠ 2, the two functions have the same 2 x + 2 x - 4 limit as x approaches 2 (Figure 2.18). Therefore, Because x 2 - 6x + 8 x - 4 2 - 4 1 = lim = = - . 2 xS2 xS2 x + 2 2 + 2 2 x - 4 lim y y 1 1 1 1 4 y x x2 6x 8 x2 4 2 1 1 4 y x x4 x2 2 2 lim x 6x 8 lim x 4 q x 2 x 2 x2 x2 4 ᠬ ᠬ FIGURE 2.18 b. Use Conjugates This limit was approximated numerically in Example 2 of Section 2.2; we conjectured that the value of the limit is 12. Direct substitution fails in this case because the denominator is zero at x = 1. Instead, we first simplify the function by multiplying the numerator and denominator by the algebraic conjugate of the numerator. The conjugate of 1x - 1 is 1x + 1; therefore, ➤ We multiply the given function by 1 = 1x + 1 1x + 1 . 11x - 1211x + 12 1x - 1 = x - 1 1x - 1211x + 12 = x + 1x - 1x - 1 1x - 1211x + 12 x - 1 1x - 1211x + 12 1 . = 1x + 1 = Copyright © 2014 Pearson Education, Inc. Rationalize the numerator; multiply by 1. Expand the numerator. Simplify. Cancel like factors when x ≠ 1. For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 82 Limits The limit can now be evaluated: Related Exercises 39–52 QUICK CHECK 3 ➤ 1 1x - 1 1 1 = = lim = . x - 1 x S 1 1x + 1 1 + 1 2 lim xS1 An Important Limit Evaluate x 2 - 7x + 10 lim . xS5 x - 5 Despite our success in evaluating limits using direct substitution, algebraic manipulation, and the limit laws, there are important limits for which these techniques do not work. One such limit arises when investigating the slope of a line tangent to the graph of an exponential function. ➤ EXAMPLE 7 Slope of a line tangent to f 1 x2 = 2x Estimate the slope of the line tangent to the graph of f 1x2 = 2x at the point P10, 12. SOLUTION In Section 2.1, the slope of a tangent line was obtained by finding the limit of slopes of secant lines; the same strategy is employed here. We begin by selecting a point Q near P on the graph of f with coordinates 1x, 2x2. The secant line joining the points P10, 12 and Q1x, 2x2 is an approximation to the tangent line. To compute the slope of the tangent line (denoted by mtan) at x = 0, we look at the slope of the secant line msec = 12x - 12>x and take the limit as x approaches 0. y y f (x) ⫽ 2x f (x) ⫽ 2x Q(x, 2x ) As x 0⫹, Q approaches P, ... ... the secant lines approach the tangent line, and msec mtan. tangent line tangent line P(0, 1) 0 As x 0⫺, Q approaches P, ... Approaching P from the right x ... the secant lines approach the tangent line, and msec mtan. P(0, 1) Q(x, 2x ) x x x 0 (a) Approaching P from the left (b) FIGURE 2.19 2x - 1 exists only if it has the same value as x S 0 + (Figure 2.19a) x xS0 and as x S 0- (Figure 2.19b). Because it is not an elementary limit, it cannot be evaluated using the limit laws of this section. Instead, we investigate the limit using numerical evidence. Choosing positive values of x near 0 results in Table 2.5. The limit lim Table 2.5 x msec 2x − 1 = x 1.0 0.1 0.01 0.001 0.0001 0.00001 1.000000 0.7177 0.6956 0.6934 0.6932 0.6931 Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.3 Techniques for Computing Limits 2x - 1 lim ≈ 0.693, which is xS0 x approximately ln 2. The connection between the natural logarithm and slopes of lines tangent to exponential curves is made clear in Chapters 3 and 6. We see that as x approaches 0 from the right, the slopes of the secant lines approach the slope of the tangent line, which is approximately 0.693. A similar calculation (Exercise 53) gives the same approximation for the limit as x approaches 0 from the left. Because the left-sided and right-sided limits are the same, we conclude that lim 12x - 12>x ≈ 0.693 (Theorem 2.1). Therefore, the slope of the line tangent to xS0 f 1x2 = 2x at x = 0 is approximately 0.693. Related Exercises 53–54 ➤ The Squeeze Theorem is also called The Squeeze Theorem the Pinching Theorem or the Sandwich Theorem. y f (x) ⱕ g(x) ⱕ h(x) ➤ ➤ Example 7 shows that 83 y ⫽ h(x) The Squeeze Theorem provides another useful method for calculating limits. Suppose the functions f and h have the same limit L at a and assume the function g is trapped between f and h (Figure 2.20). The Squeeze Theorem says that g must also have the limit L at a. A proof of this theorem is outlined in Exercise 54 of Section 2.7. y ⫽ g(x) L y ⫽ f (x) O FIGURE 2.20 xSa xSa xSa x a Squeeze Theorem: As x a, h(x) L and f (x) Therefore, g(x) L. THEOREM 2.5 The Squeeze Theorem Assume the functions f, g, and h satisfy f 1x2 … g1x2 … h1x2 for all values of x near a, except possibly at a. If lim f 1x2 = lim h1x2 = L, then lim g1x2 = L. EXAMPLE 8 Sine and cosine limits A geometric argument (Exercise 90) may be used to show that for -p>2 6 x 6 p>2, L. - x … sin x … x and 0 … 1 - cos x … x . Use the Squeeze Theorem to confirm the following limits. a. lim sin x = 0 b. lim cos x = 1 xS0 xS0 SOLUTION a. Letting f 1x2 = - x , g1x2 = sin x, and h1x2 = x , we see that g is trapped between f and h on -p>2 6 x 6 p>2 (Figure 2.21a). Because lim f 1x2 = lim h1x2 = 0 xS0 xS0 (Exercise 37), the Squeeze Theorem implies that lim g1x2 = lim sin x = 0. xS0 y 兩x兩 sin x 兩x兩 y 兩x兩 y y sin x 1 xS0 y 兩x兩 1 on q x q y 1 cos x ➤ The two limits in Example 8 play a q crucial role in establishing fundamental properties of the trigonometric functions. The limits reappear in Section 2.6. q 1 x y 兩x兩 (a) FIGURE 2.21 Copyright © 2014 Pearson Education, Inc. y0 q q x 0 1 cos x 兩x兩 1 on q x q (b) For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 84 Limits b. In this case, we let f 1x2 = 0, g1x2 = 1 - cos x, and h1x2 = x (Figure 2.21b). Because lim f 1x2 = lim h1x2 = 0, the Squeeze Theorem implies that xS0 xS0 lim g1x2 = lim 11 - cos x2 = 0. By the limit laws, it follows that lim 1 - lim cos x = 0, or lim cos x = 1. xS0 xS0 y EXAMPLE 9 Squeeze Theorem for an important limit 1 y cos x 1 y Related Exercises 55–58 xS0 ➤ xS0 xS0 a. Use a graphing utility to confirm that sin x x cos x … sin x 1 , for 0 6 x … 1. … x cos x b. Use part (a) and the Squeeze Theorem to prove that y cos x lim 0.5 xS0 sin x = 1. x SOLUTION a. Figure 2.22 shows the graphs of y = cos x (lower curve), y = 1 0 1 x sin x (middle curve), x 1 (upper curve) on the interval -1 … x … 1. These graphs confirm the cos x given inequalities. and y = FIGURE 2.22 b. By part (a), cos x … QUICK CHECK 4 Suppose f satisfies x2 1 … f 1x2 … 1 + for all values of x 6 near zero. Find lim f 1x2, if possible. ➤ xS0 lim cos x = lim xS0 xS0 sin x 1 , for x near 0, except at 0. Furthermore, … x cos x 1 = 1. The conditions of the Squeeze Theorem are satisfied and cos x we conclude that lim xS0 sin x = 1. x This important limit is used in Chapter 3 to discover derivative rules for trigonometric functions. Related Exercises 55–58 ➤ SECTION 2.3 EXERCISES 10. Suppose Review Questions 1. How is lim f 1x2 calculated if f is a polynomial function? 2. How are lim- f 1x2 and lim+ f 1x2 calculated if f is a polynomial xSa xSa f 1x2 = b xSa function? 4 if x … 3 x + 2 if x 7 3. Compute lim- f 1x2 and lim+ f 1x2. 3. For what values of a does lim r1x2 = r1a2 if r is a rational xSa function? 4. Assume lim g1x2 = 4 and f 1x2 = g1x2 whenever x ≠ 3. 11–16. Limits of linear functions Evaluate the following limits. Evaluate lim f 1x2, if possible. 11. lim 13x - 72 12. lim 1-2x + 52 13. 14. lim 1-3x2 15. lim 4 16. xS3 x 2 - 7x + 12 = lim 1x - 42. xS3 x - 3 xS3 5. Explain why lim 6. If lim f 1x2 = -8, find lim 3 f 1x24 2>3. 7. 8. xS0 xS2 Evaluate lim 2x 2 - 9. lim 5x x S -9 lim p x S -5 17–24. Applying limit laws Assume lim f 1x2 = 8, lim g1x2 = 3, xS1 xS1 and lim h1x2 = 2. Compute the following limits and state the limit p1x2 q1x2 = 10 and Suppose lim f 1x2 = lim h1x2 = 5. Find lim g1x2, where xS5 xS6 laws used to justify your computations. xS2 f 1x2 17. lim 34f 1x24 18. lim c 19. lim 3 f 1x2 - g1x24 20. lim 3 f 1x2h1x24 xS1 f 1x2 … g1x2 … h1x2, for all x. 9. xS1 xS1 Suppose p and q are polynomials. If lim xS2 Basic Skills xS2 xS2 q102 = 2, find p102. xS3 xS4 xS3 xS2 xS3 xS1 xS1 21. lim c xS1 f 1x2g1x2 h1x2 d 23. lim 3h1x24 5 xS1 Copyright © 2014 Pearson Education, Inc. h1x2 d xS1 22. lim c xS1 f 1x2 g1x2 - h1x2 d 3 24. lim 2 f 1x2g1x2 + 3 xS1 For Use Only in 2013 – 2014 Pilot Program 2.3 Techniques for Computing Limits 25–32. Evaluating limits Evaluate the following limits. 25. lim 12x 3 - 3x 2 + 4x + 52 xS1 5x 2 + 6x + 1 xS1 8x - 4 45. 26. lim 1t 2 + 5t + 72 lim x S -1 t S -2 27. lim 3 2 28. lim 2 t - 10 47. lim 3b 29. lim b S 2 14b + 1 - 1 30. lim 1x - x2 49. lim 31. lim xS3 xS9 tS3 2 5 xS2 - 5x 14x - 3 32. lim hS0 3 116 + 3h + 4 lim f 1x2 b. lim + f 1x2 x S -1 34. One-sided limits Let lim f 1x2 b. lim + f 1x2 c. lim f 1x2 e. lim+ f 1x2 f. lim f 1x2 x S -5 d. lim- f 1x2 xS5 x S -5 xS5 x T xSa xS2 xS2 36. One-sided limits x - 3 . A2 - x x - 3 b. Explain why lim+ does not exist. xS3 A 2 - x a. Evaluate lim- x xS3 -1 -0.1 -0.01 -0.001 -0.1 -0.01 - 0.001 -0.0001 0.0001 0.001 T 55. Applying the Squeeze Theorem 1 … x , for x ≠ 0. x b. Illustrate the inequalities in part (a) with a graph. a. Show that - x … x sin c. Use the Squeeze Theorem to show that lim x sin 38. Absolute value limit Show that lim x = a , for any real xS0 xSa T 39–52. Other techniques Evaluate the following limits, where a and b are fixed real numbers. x2 - 1 xS1 x - 1 x 2 - 16 41. lim S x 4 4 - x 43. lim xSb 1x - b250 - x + b x - b x 2 - 2x - 3 xS3 x - 3 3t 2 - 7t + 2 42. lim tS2 2 - t 40. lim 44. lim x S -b 1x + b27 + 1x + b210 41x + b2 0.01 3 − 1 x x if x Ú 0 x = b - x if x 6 0. 39. lim - 0.0001 - 0.00001 x xS0 number. (Hint: Consider the cases a 6 0 and a Ú 0.) x3 - a x - a a. Sketch a graph of y = 3x and carefully draw four secant lines connecting the points P10, 12 and Q1x, 3x2, for x = - 2, - 1, 1, and 2. b. Find the slope of the line that joins P10, 12 and Q1x, 3x2, for x ≠ 0. c. Complete the table and make a conjecture about the value 3x - 1 of lim . S x x 0 b. Explain why lim- 1x - 2 does not exist. xS0 x 2 - a2 ,a 7 0 x S a 1x - 1a 54. Slope of a tangent line a. Evaluate lim+ 1x - 2. xS0 2 b16 + t - t 22 t - 3 2 − 1 x 35. One-sided limits lim- x and lim+ x . Recall that tS3 x xS5 37. Absolute value limit Show that lim x = 0 by first evaluating 1 5 h 48. lim a4t - 52. lim Compute the following limits, or state that they do not exist. x S -5- 1x - 3 x - 9 116 + h - 4 h 0 if x … - 5 f 1x2 = c 225 - x 2 if - 5 6 x 6 5 3x if x Ú 5. a. hS0 - a. Sketch a graph of y = 2x and carefully draw three secant lines connecting the points P10, 12 and Q1x, 2x2, for x = - 3, - 2, and -1. b. Find the slope of the line that joins P10, 12 and Q1x, 2x2, for x ≠ 0. c. Complete the table and make a conjecture about the value of 2x - 1 lim. x xS0 c. lim f 1x2 x S -1 1 5 + h 53. Slope of a tangent line Compute the following limits or state that they do not exist. x S -1- x + 1 50. lim hS0 x 2 + 1 if x 6 - 1 f 1x2 = b 1x + 1 if x Ú - 1. a. 46. lim x - a ,a 7 0 x S a 1x - 1a 51. lim T 33. One-sided limits Let 12x - 122 - 9 85 1 = 0. x 56. A cosine limit by the Squeeze Theorem It can be shown that x2 … cos x … 1, for x near 0. 1 2 a. Illustrate these inequalities with a graph. b. Use these inequalities to find lim cos x. xS0 T 57. A sine limit by the Squeeze Theorem It can be shown that sin x x2 … 1 … 1, for x near 0. x 6 a. Illustrate these inequalities with a graph. sin x . b. Use these inequalities to find lim xS0 x Copyright © 2014 Pearson Education, Inc. 0.1 For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 86 T Limits 58. A logarithm limit by the Squeeze Theorem a. Draw a graph to verify that - x … x ln x … x , for - 1 … x … 1, where x ≠ 0. b. Use the Squeeze Theorem to determine lim x 2 ln x 2. 2 2 xS0 x5 xSa x xn 74. lim S x a x 73. lim Further Explorations 75. lim 59. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume a and L are finite numbers. 76. a. If lim f 1x2 = L, then f 1a2 = L. xSa xSa c. If lim f 1x2 = L and lim g1x2 = L, then f 1a2 = g1a2. xSa xSa f 1x2 77. lim xS1 does not exist if g1a2 = 0. 79. lim e. If lim+ 2f 1x2 = 4 lim+ f 1x2, it follows that xS1 xS1 f 1x2. lim 2f 1x2 = 4xlim S1 xS1 80. lim d. The limit lim xSa g1x2 xS4 xS0 60–67. Evaluating limits Evaluate the following limits, where c and k are constants. 100 h S 0 110h - 1211 + 2 61. lim 15x - 623>2 60. lim xS2 1 1 15 x 2 + 2x 62. lim xS3 x - 3 64. lim a xS2 63. lim xS1 1 2 b - 2 x - 2 x - 2x 65. lim 110x - 9 - 1 x - 1 15 + h22 - 25 hS0 - 11232 4 2 x - 2 x - 16 x - 1 1x - 1 78. lim xS1 x - 1 14x + 5 - 3 31x - 421x + 5 3 - 1x + 5 x , where c is a nonzero constant 1cx + 1 - 1 81. Creating functions satisfying given limit conditions Find functions f and g such that lim f 1x2 = 0 and lim 1 f 1x2 g1x22 = 5. xS1 xS1 82. Creating functions satisfying given limit conditions Find a f 1x2 function f satisfying lim a b = 2. xS1 x - 1 83. Finding constants Find constants b and c in the polynomial p1x2 p1x2 = x 2 + bx + c such that lim = 6. Are the xS2 x - 2 constants unique? h Applications x 2 - 2cx + c 2 x - c xSc 84. A problem from relativity theory Suppose a spaceship of length L 0 is traveling at a high speed v relative to an observer. To the observer, the ship appears to have a smaller length given by the Lorentz contraction formula 66. lim 67. x S 16 3 12 x23 77–80. Limits involving conjugates Evaluate the following limits. b. If lim- f 1x2 = L, then lim+ f 1x2 = L. xSa a5 a an , for any positive integer n a 3 2 x - 1 (Hint: x - 1 = x - 1 xS1 lim - w 2 + 5kw + 4k 2 , for k ≠ 0 wS - k w 2 + kw lim 68. Finding a constant Suppose f 1x2 = b L = L0 3x + b if x … 2 x - 2 if x 7 2. B 1 - v2 , c2 where c is the speed of light. Determine a value of the constant a for which lim g1x2 exists xS - 1 and state the value of the limit, if possible. a. What is the observed length L of the ship if it is traveling at 50% of the speed of light? b. What is the observed length L of the ship if it is traveling at 75% of the speed of light? c. In parts (a) and (b), what happens to L as the speed of the ship increases? v2 d. Find lim- L 0 1 - 2 and explain the significance of this vSc B c limit. 70–76. Useful factorization formula Calculate the following limits using the factorization formula 85. Limit of the radius of a cylinder A right circular cylinder with a height of 10 cm and a surface area of S cm2 has a radius given by Determine a value of the constant b for which lim f 1x2 exists and xS2 state the value of the limit, if possible. 69. Finding a constant Suppose g1x2 = b x 2 - 5x if x … -1 ax 3 - 7 if x 7 - 1. x n - a n = 1x - a21x n - 1 + x n - 2a + x n - 3a 2 + g + xa n - 2 + a n - 12, r1S2 = where n is a positive integer and a is a real number. x 5 - 32 S x 2 x - 2 70. lim 72. 2S 1 a 100 + - 10b. p 2 A Find lim+ r1S2 and interpret your result. x6 - 1 S x 1 x - 1 SS0 71. lim x7 + 1 (Hint: Use the formula for x 7 - a 7 with a = -1.) xS - 1 x + 1 lim Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.4 Infinite Limits b. Show that sin u 6 u , for - p>2 6 u 6 p>2. 1Hint: The length of arc AB is u, if 0 … u 6 p>2, and -u, if - p>2 6 u 6 0.2 c. Conclude that - u … sin u … u , for - p>2 6 u 6 p>2. d. Show that 0 … 1 - cos u … u , for -p>2 6 u 6 p>2. 86. Torricelli’s Law A cylindrical tank is filled with water to a depth of 9 meters. At t = 0, a drain in the bottom of the tank is opened and water flows out of the tank. The depth of water in the tank (measured from the bottom of the tank) t seconds after the drain is opened is approximated by d1t2 = 13 - 0.015t22, for 0 … t … 200. Evaluate and interpret lim - d1t2. t S 200 2 0 A 87. Electric field The magnitude of the electric field at a point x meters from the midpoint of a 0.1m line of charge is given by 4.35 E1x2 = (in units of newtons per coulomb, N>C). x2x 2 + 0.01 Evaluate lim E1x2. C 1 x S 10 O 1 B C 88–89. Limits of composite functions p1x2 = b n x n + b n - 1x n - 1 + g + b 1x + b 0, xS - 1 89. Suppose g1x2 = f 11 - x2, for all x, lim+ f 1x2 = 4, and prove that lim p1x2 = p1a2 for any value of a. xS1 xSa lim- f 1x2 = 6. Find lim+ g1x2 and lim- g1x2. xS0 A 91. Theorem 2.4a Given the polynomial 88. If lim f 1x2 = 4, find lim f 1x 22. xS1 B 0 2 Additional Exercises xS1 O xS0 90. Two trigonometric inequalities Consider the angle u in standard position in a unit circle, where 0 … u 6 p>2 or - p>2 6 u 6 0 (use both figures). a. Show that AC = sin u , for -p>2 6 u 6 p>2. 1Hint: Consider the cases 0 … u 6 p>2 and - p>2 6 u 6 0 separately.2 QUICK CHECK ANSWERS 1. 0, 2 2. 2 3. 3 4. 1 ➤ T 87 2.4 Infinite Limits Two more limit scenarios are frequently encountered in calculus and are discussed in this and the following section. An infinite limit occurs when function values increase or decrease without bound near a point. The other type of limit, known as a limit at infinity, occurs when the independent variable x increases or decreases without bound. The ideas behind infinite limits and limits at infinity are quite different. Therefore, it is important to distinguish these limits and the methods used to calculate them. An Overview To illustrate the differences between infinite limits and limits at infinity, consider the values of f 1x2 = 1>x 2 in Table 2.6. As x approaches 0 from either side, f 1x2 grows larger and larger. Because f 1x2 does not approach a finite number as x approaches 0, lim f 1x2 xS0 does not exist. Nevertheless, we use limit notation and write lim f 1x2 = ∞. The infinity xS0 symbol indicates that f 1x2 grows arbitrarily large as x approaches 0. This is an example of an infinite limit; in general, the dependent variable becomes arbitrarily large in magnitude as the independent variable approaches a finite number. y Table 2.6 y x f 1x2 = 1>x2 {0.1 {0.01 {0.001 T 0 100 10,000 1,000,000 T ∞ 0 Copyright © 2014 Pearson Education, Inc. lim 1 x 0 x2 ᠬ 1 x2 x For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 88 Limits With limits at infinity, the opposite occurs: The dependent variable approaches a finite number as the independent variable becomes arbitrarily large in magnitude. In Table 2.7, we see that f 1x2 = 1>x 2 approaches 0 as x increases. In this case, we write lim f 1x2 = 0. xS∞ y Table 2.7 x 10 100 1000 T ∞ y f 1x2 = 1 , x2 0.01 0.0001 0.000001 T 0 1 0 x2 O lim lim x x 1 x2 1 0 x2 x A general picture of these two limit scenarios is shown in Figure 2.23. y Infinite limit y as x a Limit at infinity y L as x L O x a M y Limit at infinity y M as x f (x) large and positive FIGURE 2.23 y f (x) O Infinite Limits x a The following definition of infinite limits is informal, but it is adequate for most functions encountered in this book. A precise definition is given in Section 2.7. x approaches a DEFINITION Infinite Limits (a) y O Suppose f is defined for all x near a. If f 1x2 grows arbitrarily large for all x sufficiently close (but not equal) to a (Figure 2.24a), we write x approaches a x a y f (x) lim f 1x2 = ∞. xSa We say the limit of f 1x2 as x approaches a is infinity. If f 1x2 is negative and grows arbitrarily large in magnitude for all x sufficiently close (but not equal) to a (Figure 2.24b), we write lim f 1x2 = - ∞. f (x) large and negative (b) xSa In this case, we say the limit of f 1x2 as x approaches a is negative infinity. In both cases, the limit does not exist. FIGURE 2.24 Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 2.4 Infinite Limits EXAMPLE 1 Infinite limits Evaluate lim xS1 89 x x and lim using the x S - 1 1x 2 - 122 1x 2 - 122 graph of the function. x (Figure 2.25) shows that as x approaches 1 1x 2 - 122 (from either side), the values of f grow arbitrarily large. Therefore, the limit does not exist and we write SOLUTION The graph of f 1x2 = lim lim f (x) 5 xS1 ᠬ x 1 As x approaches -1, the values of f are negative and grow arbitrarily large in magnitude; therefore, x f (x) 2 (x 1)2 lim 1 1 x = ∞. 1x - 122 2 xS - 1 x x = - ∞. 1x 2 - 122 Related Exercises 7–8 ➤ y Example 1 illustrates two-sided infinite limits. As with finite limits, we also need to work with right-sided and left-sided infinite limits. lim f (x) x 1 ᠬ FIGURE 2.25 5 DEFINITION One-Sided Infinite Limits Suppose f is defined for all x near a with x 7 a. If f 1x2 becomes arbitrarily large for all x sufficiently close to a with x 7 a, we write lim+ f 1x2 = ∞ (Figure 2.26a). xSa The one-sided infinite limits lim+ f 1x2 = - ∞ (Figure 2.26b), lim- f 1x2 = ∞ xSa xSa (Figure 2.26c), and lim- f 1x2 = - ∞ (Figure 2.26d) are defined analogously. xSa y y y f (x) O O a y f (x) x lim f (x) lim f (x) x a x a ᠬ ᠬ (a) y (b) y y f (x) O O x a a lim f (x) x a ᠬ (c) FIGURE 2.26 Copyright © 2014 Pearson Education, Inc. x a y f (x) lim f (x) x a ᠬ (d) x For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 90 Limits In all the infinite limits illustrated in Figure 2.26, the line x = a is called a vertical asymptote; it is a vertical line that is approached by the graph of f as x approaches a. QUICK CHECK 1 Sketch the graph of a function and its vertical asymptote that satisfies the conditions lim+ f 1x2 = - ∞ and lim- f 1x2 = ∞. xS2 DEFINITION Vertical Asymptote If lim f 1x2 = { ∞, lim+ f 1x2 = { ∞, or lim- f 1x2 = { ∞, the line x = a is xSa ➤ xS2 xSa EXAMPLE 2 Determining limits graphically The vertical lines x = 1 and lim g(x) ᠬ x 1 y xSa called a vertical asymptote of f . x = 3 are vertical asymptotes of the function g1x2 = lim g(x) ᠬ x 3 x - 2 . Use 1x - 122 1x - 32 Figure 2.27 to analyze the following limits. g(x) x2 (x 1)2(x 3) a. lim g1x2 b. lim- g1x2 xS1 c. lim g1x2 xS3 xS3 SOLUTION 1 a. The values of g grow arbitrarily large as x approaches 1 from either side. Therefore, lim g1x2 = ∞. x 3 xS1 b. The values of g are negative and grow arbitrarily large in magnitude as x approaches 3 from the left, so lim- g1x2 = - ∞. lim g(x) xS3 x 3 ᠬ c. Note that lim+ g1x2 = ∞ and lim- g1x2 = - ∞. Therefore, lim g1x2 does not FIGURE 2.27 xS3 xS3 xS3 exist. Because g behaves differently as x S 3- and as x S 3+ , we do not write lim g1x2 = ∞ . We simply say that the limit does not exist. Related Exercises 9–16 ➤ xS3 Finding Infinite Limits Analytically Table 2.8 x 0.01 0.001 0.0001 T 0+ 5 + x x 5.01 = 501 0.01 5.001 = 5001 0.001 5.0001 = 50,001 0.0001 T ∞ Many infinite limits are analyzed using a simple arithmetic property: The fraction a>b grows arbitrarily large in magnitude if b approaches 0 while a remains nonzero and relatively constant. For example, consider the fraction 15 + x2>x for values of x approaching 0 from the right (Table 2.8). 5 + x S ∞ as x S 0 + because the numerator 5 + x approaches 5 while We see that x 5 + x the denominator is positive and approaches 0. Therefore, we write lim+ = ∞. x xS0 5 + x Similarly, lim= - ∞ because the numerator approaches 5 while the denominator x xS0 approaches 0 through negative values. EXAMPLE 3 Evaluating limits analytically Evaluate the following limits. a. lim+ xS3 2 - 5x x - 3 b. limxS3 2 - 5x x - 3 SOLUTION QUICK CHECK 2 Evaluate lim+ xS0 x - 5 x approaches - 13 c x - 5 by determining the sign x xS0 of the numerator and denominator. and lim- a. As x S 3 + , the numerator 2 - 5x approaches 2 - 5132 = -13 while the denominator x - 3 is positive and approaches 0. Therefore, lim+ xS3 2 - 5x = - ∞. x - 3 s positive and approaches 0 Copyright © 2014 Pearson Education, Inc. ➤ For Use Only in 2013 – 2014 Pilot Program 2.4 Infinite Limits 91 b. As x S 3 -, 2 - 5x approaches 2 - 5132 = -13 while x - 3 is negative and approaches 0. Therefore, approaches - 13 c lim- xS3 s 2 - 5x = ∞. x - 3 negative and approaches 0 Related Exercises 17–28 EXAMPLE 4 Evaluating limits analytically Evaluate lim + xS - 4 ➤ We can assume that x ≠ 0 because we ➤ These limits imply that the given function has a vertical asymptote at x = 3. -x 3 + 5x 2 - 6x . -x 3 - 4x 2 SOLUTION First we factor and simplify, assuming x ≠ 0: are considering function values near x = - 4. 1x - 221x - 32 -x1x - 221x - 32 -x 3 + 5x 2 - 6x = = . 3 2 2 x1x + 42 -x - 4x -x 1x + 42 As x S -4 + , we find that approaches 42 f lim + xS - 4 1x - 221x - 32 -x 3 + 5x 2 - 6x = lim + = - ∞. 3 2 xS - 4 x1x + 42 -x - 4x u negative and approaches 0 Verify that x1x + 42 S 0 through negative values as x S -4 + . This limit implies that the given function has a vertical asymptote at x = -4. ➤ Related Exercises 17–28 ➤ Example 5 illustrates that f 1x2>g1x2 might not grow arbitrarily large in magnitude if both f 1x2 and g1x2 approach 0. Such limits are called indeterminate forms and are examined in detail in Section 4.7. x 2 - 4x + 3 . Evaluate x2 - 1 the following limits and find the vertical asymptotes of f . Verify your work with a graphing utility. EXAMPLE 5 Location of vertical asymptotes Let f 1x2 = a. lim f 1x2 b. xS1 lim x S - 1- f 1x2 c. lim x S - 1+ f 1x2 SOLUTION a. Notice that as x S 1, both the numerator and denominator of f approach 0, and the function is undefined at x = 1. To compute lim f 1x2, we first factor: xS1 ➤ It is permissible to cancel the x - 1 1x - 121x - 32 because 1x - 121x + 12 x approaches 1 but is not equal to 1. Therefore, x - 1 ≠ 0. factors in lim xS1 ➤ QUICK CHECK 3 x 2 - 4x + 3 lim f 1x2 = lim xS1 xS1 x2 - 1 1x - 121x - 32 = lim x S 1 1x - 121x + 12 1x - 32 = lim S x 1 1x + 12 = 1 - 3 = -1. 1 + 1 Factor. Cancel like factors, x ≠ 1. Substitute x = 1. Therefore, lim f 1x2 = -1 (even though f 112 is undefined). The line x = 1 is not a xS1 vertical asymptote of f . Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 – 2014 Pilot Program 92 Chapter 2 r Limits b. In part (a), we showed that f 1x2 = x - 3 x 2 - 4x + 3 = , provided x ≠ 1. x + 1 x2 - 1 We use this fact again. As x approaches -1 from the left, the one-sided limit is approaches - 4 c lim - f 1x2 = lim x S -1 x S -1 x-3 = ∞. x + 1 s negative and approaches 0 c. As x approaches -1 from the right, the one-sided limit is approaches - 4 c lim f 1x2 = lim + x S -1 + x S -1 x-3 = - ∞. x + 1 s positive and approaches 0 The infinite limits lim + f 1x2 = - ∞ and lim - f 1x2 = ∞ each imply that the line xS - 1 xS - 1 x = -1 is a vertical asymptote of f . The graph of f generated by a graphing utility may appear as shown in Figure 2.28a. If so, two corrections must be made. A hole should appear in the graph at 11, -12 because lim f 1x2 = -1, but f 112 is undefined. It is also a xS1 good idea to replace the solid vertical line with a dashed line to emphasize that the vertical asymptote is not a part of the graph of f (Figure 2.28b). ➤ Graphing utilities vary in how they Two versions of the graph of y display vertical asymptotes. The errors shown in Figure 2.28a do not occur on all graphing utilities. x2 4x 3 x2 1 y y 1 1 x 2 x 2 Calculator graph Correct graph (a) (b) FIGURE 2.28 QUICK CHECK 4 ➤ Why not? The line x = 2 is not a vertical asymptote of y = Copyright © 2014 Pearson Education, Inc. 1x - 121x - 22 . x - 2 ➤ Related Exercises 29–34 For Use Only in 2013 – 2014 Pilot Program 2.4 Infinite Limits y 93 EXAMPLE 6 Limits of trigonometric functions Evaluate the following limits. lim cot 0 ᠬ a. lim+ cot u b. lim- cot u uS0 uS0 SOLUTION y cot a. Recall that cot u = cos u>sin u. Furthermore (Example 8, Section 2.3), lim+ cos u = 1 1 uS0 d 1 and sin u is positive and approaches 0 as u S 0+. Therefore, as u S 0+, cot u becomes arbitrarily large and positive, which means lim+ cot u = ∞. This limit is con- q uS0 firmed by the graph of cot u (Figure 2.29), which has a vertical asymptote at u = 0. b. In this case, lim- cos u = 1 and as u S 0 -, sin u S 0 with sin u 6 0. Therefore, as uS0 lim cot u S 0 -, cot u is negative and becomes arbitrarily large in magnitude. It follows that lim- cot u = - ∞, as confirmed by the graph of cot u. 0 ᠬ FIGURE 2.29 uS0 Related Exercises 35–40 SECTION 2.4 EXERCISES Review Questions 8. Use a graph to explain the meaning of lim+ f 1x2 = - ∞ . 2. Use a graph to explain the meaning of lim f 1x2 = ∞ . y 3. What is a vertical asymptote? 1 4. Consider the function F 1x2 = f 1x2>g1x2 with g1a2 = 0. Does F necessarily have a vertical asymptote at x = a? Explain your reasoning. xSa xSa 5. Suppose f 1x2 S 100 and g1x2 S 0, with g1x2 6 0, as x S 2. f 1x2 Determine lim . x S 2 g1x2 6. Evaluate limxS3 1 1 and lim+ . S x - 3 x 3 x - 3 7. Analyzing infinite limits numerically Compute the values of x + 1 f 1x2 = in the following table and use them to discuss 1x - 122 lim f 1x2. x + 1 1x − 1 2 2 3 a. lim- f 1x2 b. lim+ f 1x2 c. lim f 1x2 d. lim- f 1x2 e. lim+ f 1x2 f. lim f 1x2 xS1 xS2 xS1 xS1 xS2 xS2 y x + 1 x 1.1 0.9 1.01 0.99 1.001 0.999 1.0001 0.9999 x Analyzing infinite limits graphically The graph of f in the figure has vertical asymptotes at x = 1 and x = 2. Analyze the following limits. xS1 x y f (x) 1 9. Basic Skills T Analyzing infinite limits graphically Use the graph of x to discuss lim f 1x2 and lim f 1x2. f 1x2 = 2 xS - 1 xS3 1x - 2x - 322 1. y f (x) 1x − 1 2 2 1 Copyright © 2014 Pearson Education, Inc. 2 x ➤ q For Use Only in 2013 – 2014 Pilot Program Chapter 2 r 94 Limits 10. Analyzing infinite limits graphically The graph of g in the figure has vertical asymptotes at x = 2 and x = 4. Analyze the following limits. a. lim- g1x2 b. lim+ g1x2 c. lim g1x2 d. lim- g1x2 e. lim+ g1x2 f. lim g1x2 xS2 xS2 xS4 T xS2 xS4 xS4 y x 4 a. lim - h1x2 xS - 2 d. lim- h1x2 xS3 b. a. lim + h1x2 c. lim h1x2 xS - 2 xS - 2 e. lim+ h1x2 f. lim h1x2 xS3 xS3 y b. lim+ f 1x2 x S 0- lim x S - 2+ f 1x2 b. lim f 1x2 f 112 = 0, lim f 1x2 = - ∞ , d. lim- p1x2 xS3 b. lim p1x2 f. lim p1x2 xS3 xS3 y x y p(x) lim f 1x2 = ∞ x S 4- xS2 g152 = -1, lim g1x2 = - ∞ , xS4 lim g1x2 = - ∞ x S 7+ b. lim- 1 x - 2 c. lim 1 x - 2 18. a. lim+ 2 1x - 323 b. lim- 2 1x - 323 c. lim 2 1x - 323 19. a. lim+ x - 5 1x - 422 b. lim- x - 5 1x - 422 c. lim x - 5 - 422 20. a. lim+ x - 2 1x - 123 b. lim- x - 2 1x - 123 c. lim x - 2 - 123 xS3 xS4 xS1 21. a. lim+ 3 lim f 1x2 = 1, xS3 1 x - 2 c. lim lim + xS - 2 xS0 25. lim+ xS1 xS4 xS1 xS3 x S 4 1x x S 1 1x b. lim- 1x - 32 1x - 121x - 22 xS3 1x - 32 1x - 32 1x - 42 x1x + 22 x - 5x x2 3 23. lim xS3 xS2 1x - 121x - 22 xS3 22. a. xS2 1x - 121x - 22 xS3 2 xS0 17. a. lim+ xS - 2 e. lim+ p1x2 d. lim+ f 1x2 16. Sketching graphs Sketch a possible graph of a function g, together with vertical asymptotes, satisfying all the following conditions. c. lim p1x2 x S - 2+ lim f 1x2 lim f 1x2 = ∞ , x S 0+ xS2 12. Analyzing infinite limits graphically The graph of p in the figure has vertical asymptotes at x = - 2 and x = 3. Investigate the following limits. lim p1x2 xS1 17–28. Evaluating limits analytically Evaluate the following limits or state that they do not exist. x y h(x) x S - 2- d. lim+ f 1x2 x S 0- f 132 is undefined, lim g1x2 = ∞ , a. c. xS - 2 x S 7- 3 lim f 1x2 x S 1- 15. Sketching graphs Sketch a possible graph of a function f , together with vertical asymptotes, satisfying all the following conditions on 30, 44 . g122 = 1, 2 c. xS0 14. Analyzing infinite limits graphically Graph the function e -x f 1x2 = using a graphing utility. (Experiment with x1x + 222 your choice of a graphing window.) Use your graph to discuss the following limits. y g(x) 11. Analyzing infinite limits graphically The graph of h in the figure has vertical asymptotes at x = - 2 and x = 3. Investigate the following limits. lim f 1x2 a. T 2 13. Analyzing infinite limits graphically Graph the function 1 using a graphing utility with the window f 1x2 = 2 x - x 3-1, 24 * 3-10, 104. Use your graph to discuss the following limits. b. lim xS - 2 1x - 42 x1x + 22 Copyright © 2014 Pearson Education, Inc. xS - 2 1x - 42 x1x + 22 24. lim 4t - 100 t - 5 26. lim z - 5 1z - 10z + 2422 2 x 2 - 5x + 6 x - 1 c. lim 2 tS5 zS4 2 For Use Only in 2013 – 2014 Pilot Program 2.4 Infinite Limits x 2 - 4x + 3 xS2 1x - 222 2 x - 4x + 3 c. lim xS2 1x - 222 b. lim- 27. a. lim+ xS2 x - 5x + 6x x 4 - 4x 2 x 3 - 5x 2 + 6x c. lim xS - 2 x 4 - 4x 2 3 28. a. 2 lim x S - 2+ x 2 - 4x + 3 1x - 222 x - 5x + 6x x 4 - 4x 2 x - 5x 2 + 6x d. lim xS2 x 4 - 4x 2 3 b. 2 lim x S - 23 29. Location of vertical asymptotes Analyze the following limits x - 5 . and find the vertical asymptotes of f 1x2 = 2 x - 25 b. lim - f 1x2 c. lim + f 1x2 a. lim f 1x2 xS5 x S -5 x S -5 30. Location of vertical asymptotes Analyze the following limits x + 7 and find the vertical asymptotes of f 1x2 = 4 . x - 49x 2 c. lim f 1x2 d. lim f 1x2 a. lim- f 1x2 b. lim+ f 1x2 xS7 xS7 xS - 7 42. Finding a function with vertical asymptotes Find polynomials p and q such that f = p>q is undefined at 1 and 2, but p>q has a vertical asymptote only at 2. Sketch a graph of your function. 43. Finding a function with infinite limits Give a formula for a function f that satisfies lim+ f 1x2 = ∞ and lim- f 1x2 = - ∞ . xS6 xSa x x2 + 1 1 c. f 1x2 = 2 x - 1 1 e. f 1x2 = 1x - 122 x x2 - 1 x d. f 1x2 = 1x - 122 x f. f 1x2 = x + 1 a. f 1x2 = A. b. f 1x2 = B. y xS0 y 1 1 1 1 x 1 1 x 1 x 1 x xSa x 2 - 9x + 14 31. f 1x2 = 2 x - 5x + 6 33. f 1x2 = xS6 44. Matching Match functions a–f with graphs A–F in the figure without using a graphing utility. 31–34. Finding vertical asymptotes Find all vertical asymptotes x = a of the following functions. For each value of a, discuss lim+ f 1x2, lim- f 1x2, and lim f 1x2. xSa 95 x + 1 x 3 - 4x 2 + 4x 32. f 1x2 = cos x x + 2x 34. f 1x2 = x 3 - 10x 2 + 16x x 2 - 8x 2 C. D. y y 35–38. Trigonometric limits Investigate the following limits. 35. lim csc u xS0 37. lim+ 1- 10 cot x2 xS0 T c. T 38. lim u S p>2 + lim + tan x x S p>2 lim x S - p>2 + tan x b. d. c. lim sec x tan x x S p>2+ lim x S - p>2+ sec x tan x b. d. E. 1 x F. y y lim - tan x x S p>2 lim x S - p>2- tan x 1 40. Analyzing infinite limits graphically Graph the function y = sec x tan x with the window 3-p, p4 * 3- 10, 104. Use the graph to analyze the following limits. a. 1 1 tan u 3 39. Analyzing infinite limits graphically Graph the function y = tan x with the window 3- p, p4 * 3- 10, 104. Use the graph to analyze the following limits. a. 1 36. lim- csc x u S 0+ 1 1 1 x 1 lim sec x tan x x S p>2- lim x S - p>2- sec x tan x Further Explorations Additional Exercises 41. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. 45. Limits with a parameter Let f 1x2 = a. The line x = 1 is a vertical asymptote of the function x 2 - 7x + 6 . f 1x2 = x2 - 1 b. The line x = -1 is a vertical asymptote of the function x 2 - 7x + 6 f 1x2 = . x2 - 1 c. If g has a vertical asymptote at x = 1 and lim+ g1x2 = ∞ , xS1 then lim- g1x2 = ∞ . x 2 - 7x + 12 . x - a a. For what values of a, if any, does lim+ f 1x2 equal a finite xSa number? b. For what values of a, if any, does lim+ f 1x2 = ∞ ? xSa c. For what values of a, if any, does lim+ f 1x2 = - ∞ ? xS1 Copyright © 2014 Pearson Education, Inc. xSa For Use Only in 2013 – 2014 Pilot Program 96 Chapter 2 r Limits b. Create a graph that gives a more complete representation of f. 46–47. Steep secant lines a. Given the graph of f in the following figures, find the slope of the secant line that passes through 10, 02 and 1h, f 1h22 in terms of h, for h 7 0 and h 6 0. b. Evaluate the limit of the slope of the secant line found in part (a) as h S 0+ and h S 0-. What does this tell you about the line tangent to the curve at 10, 02? 46. f 1x2 = x y 20 15 1>3 10 y y⫽ 2000 50 ⫹ 100 x2 2 4 5 (h, h1/3) (0, 0) ⫺4 x h T 47. f 1x2 = x 2>3 x2/3 (h, h2/3) 49. f 1x2 = x 2 - 3x + 2 x 10 - x 9 50. g1x2 = 2 - ln x 2 51. h1x2 = ex 1x + 123 px 52. p1x2 = sec a b , for x 6 2 2 53. g1u2 = tan a (0, 0) x h 55. f 1x2 = T 48. Care with graphing The figure shows the graph of the function 2000 graphed in the window 3 -4, 44 * 30, 204. f 1x2 = 50 + 100x 2 xS0 pu b 10 1 1x sec x 54. q1s2 = p s - sin s 56. g1x2 = e1>x 57. Can a graph intersect a vertical asymptote? A common misconception is that the graph of a function never intersects its vertical asymptotes. Let 4 f 1x2 = W x - 1 x2 a. Evaluate lim+ f 1x2, lim- f 1x2, and lim f 1x2. xS0 if x 6 1 if x Ú 1 . Explain why x = 1 is a vertical asymptote of the graph of f and show that the graph of f intersects the line x = 1. QUICK CHECK ANSWERS 1. Answers will vary, but all graphs should have a vertical asymptote at x = 2. 2. - ∞; ∞ 3. As x S -4 + , x 6 0 and 1x + 42 7 0, so x1x + 42 S 0 through negative values. 1x - 121x - 22 4. lim = lim 1x - 12 = 1, which is not xS2 x - 2 xS2 an infinite limit, so x = 2 is not a vertical asymptote. ➤ xS0 x 49–56. Asymptotes Use analytical methods and/or a graphing utility to identify the vertical asymptotes (if any) of the following functions. y T 0 Technology Exercises y ⫽ x1/3 y⫽ ⫺2 2.5 Limits at Infinity Limits at infinity—as opposed to infinite limits—occur when the independent variable becomes large in magnitude. For this reason, limits at infinity determine what is called the end behavior of a function. An application of these limits is to determine whether a system (such as an ecosystem or a large oscillating structure) reaches a steady state as time increases. Copyright © 2014 Pearson Education, Inc.

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