Limits. or Use Only in Pilot Program F The Idea of Limits 2.2 Definitions of Limits 2.3 Techniques for Computing.

Transcription

1 Limits or Use Onl in Pilot Program F he Idea of Limits. Definitions of Limits.3 echniques for Computing Limits.4 Infinite Limits.5 Limits at Infinit.6 Continuit.7 Precise Definitions of Limits Biologists often use mathematical models to describe communities of organisms anthing from a culture of cells to a herd of zebras. hese models result in functions that approimate the population of the communit as it changes in time. An important question concerns the long-term behavior of these population functions. Does the population approach a stead-state level, which corresponds to a stable communit (as in the first figure)? Does the population oscillate continuall without settling at some fied level (as in the second figure)? Or does the population decrease to zero, indicating that the organism becomes etinct (as in the third figure)? Such questions about the behavior of functions are answered using the powerful idea of a it Population Population increases and approches a stead state. Population Population oscillates without approaching a stead state. Population Population decas to zero (etinction) ime (months) ime (months) ime (months) Copright 04 Pearson Education, Inc.

2 or Use Onl in Pilot Program F he Idea of Limits 6 Chapter Preview All of calculus is based on the idea of a it. Not onl are its important in their own right, but the 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 velocit 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 its. We first present its intuitivel b showing how the arise in computing instantaneous velocities and finding slopes of tangent lines. As the chapter progresses, we build more rigor into the definition of the it, and we eamine the different was in which its eist or fail to eist. he chapter concludes b introducing the important propert called continuit and b giving the formal definition of a it.. he Idea of Limits his brief opening section illustrates how its arise in two seemingl unrelated problems: finding the instantaneous velocit of a moving object and finding the slope of a line tangent to a curve. hese two problems provide important insights into its, and the reappear in various forms throughout the book. Average Velocit Suppose ou want to calculate our average velocit as ou travel along a straight highwa. If ou pass milepost 00 at noon and milepost 30 at :30 p.m., ou travel 30 miles in a halfhour, so our average velocit over this time interval is 30 mi>0.5 hr = 60 mi>hr. B contrast, even though our average velocit ma be 60 mi>hr, it s almost certain that our instantaneous velocit, the speed indicated b the speedometer, varies from one moment to the net. EXAMPLE Average velocit A rock is launched verticall upward from the ground with a speed of 96 ft>s. Neglecting air resistance, a well-known formula from phsics states that the position of the rock after t seconds is given b the function st = -6t + 96t. he position s is measured in feet with s = 0 corresponding to the ground. Find the average velocit of the rock between each pair of times. a. t = s and t = 3 s b. t = s and t = s SOLUION Figure. shows the position of the rock on the time interval 0 t 3. a. he average velocit of the rock over an time interval 3t 0, t 4 is the change in position divided b the elapsed time: v av = st - st 0 t - t 0 herefore, the average velocit over the interval 3, 34 is v av = s3 - s 3 - = 44 ft - 80 ft 3 s - s = 64 ft s = 3 ft>s. Copright 04 Pearson Education, Inc.

3 or Use Onl in Pilot Program F Chapter Limits Height above ground (ft) s(3) 44 ft s() 8 ft s() 80 ft s s(t) 6t 96t (3, s(3)) (, s()) (, s()) Position of rock FIGURE. 0 ime (s) 3 t Here is an important observation: As shown in Figure.a, the average velocit is simpl the slope of the line joining the points, s and 3, s3 on the graph of the position function. b. he average velocit of the rock over the interval 3, 4 is s - s v av = = - 8 ft - 80 ft s - s = 48 ft s = 48 ft>s. Again, the average velocit is the slope of the line joining the points, s and, s on the graph of the position function (Figure.b). s v av slope 64 ft s 3 ft/s s v av slope 48 ft s 48 ft/s Height above ground (ft) (, 80) (3, 44) Elapsed time 3 s s s Change in position s(3) s() 64 ft Height above ground (ft) 8 80 (, 80) (, 8) Change in position s() s() 48 ft Elapsed time s s s 0 FIGURE. ime (s) (a) 3 t 0 3 t ime (s) (b) Related Eercises 7 4 QUICK CHECK In Eample, what is the average velocit between t = and t = 3? See Section. for a discussion of secant lines. In Eample, we computed slopes of lines passing through two points on a curve. An such line joining two points on a curve is called a secant line. he slope of the secant line, denoted m sec, for the position function in Eample on the interval 3t 0, t 4 is m sec = st - st 0 t - t 0 Copright 04 Pearson Education, Inc.

4 or Use Onl in Pilot Program F he Idea of Limits 63 Height above ground (ft) v av m sec s(t ) s(t 0 ) t t 0 s s(t ) s(t 0 ) 0 t 0 t ime (s) FIGURE.3 able. ime interval Change in position s(t ) s(t 0 ) t Change in time t t 0 Average velocit 3, 4 48 ft>s 3, ft>s 3, ft>s 3, ft>s 3, ft>s 3, ft>s he same instantaneous velocit is obtained as t approaches from the left (with t 6 ) and as t approaches from the right (with t 7 ). 0 t t t Eample demonstrates that the average velocit is the slope of a secant line on the graph of the position function; that is, v av = m sec (Figure.3). Instantaneous Velocit o compute the average velocit, we use the position of the object at two distinct points in time. How do we compute the instantaneous velocit at a single point in time? As illustrated in Eample, the instantaneous velocit at a point t = t 0 is determined b computing average velocities over intervals 3t 0, t 4 that decrease in length. As t approaches t 0, the average velocities tpicall approach a unique number, which is the instantaneous velocit. his single number is called a it. QUICK CHECK Eplain the difference between average velocit and instantaneous velocit. EXAMPLE Instantaneous velocit Estimate the instantaneous velocit of the rock in Eample at the single point t =. SOLUION We are interested in the instantaneous velocit at t =, so we compute the average velocit over smaller and smaller time intervals 3, t4 using the formula 8 ft Position of rock at various times t sec v av = v av st - s. t - Notice that these average velocities are also slopes of secant lines, several of which are shown in able.. We see that as t approaches, the average velocities appear to approach 64 ft>s. In fact, we could make the average velocit as close to 64 ft>s as we like b taking t sufficientl close to. herefore, 64 ft>s is a reasonable estimate of the instantaneous velocit at t =. Related Eercises 5 0 In language to be introduced in Section., we sa that the it of v av as t approaches equals the instantaneous velocit v inst, which is 64 ft>s. his statement is written compactl as v inst = t S v av = t S st - s t - Figure.4 gives a graphical illustration of this it. s() s() = 64 ft>s ft/s 08 ft t.5 sec v av s(.5) s() ft/s 86.4 ft 80 ft t. sec t sec... v av As these intervals shrink... t 0 (rock thrown at 96 ft/s) s(.) s() ft/s... v inst 64 ft/s... the average velocities approach 64 ft/s the instantaneous velocit at t. FIGURE.4 Copright 04 Pearson Education, Inc.

5 or Use Onl in Pilot Program F Chapter Limits We define tangent lines carefull in Section 3.. For the moment, imagine zooming in on a point P on a smooth curve. As ou zoom in, the curve appears more and more like a line passing through P. his 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 locall linear at an given point. P P Slope of the angent Line Several important conclusions follow from Eamples and. Each average velocit in able. corresponds to the slope of a secant line on the graph of the position function (Figure.5). Just as the average velocities approach a it as t approaches, the slopes of the secant lines approach the same it as t approaches. Specificall, as t approaches, two things happen:. he secant lines approach a unique line called the tangent line.. he slopes of the secant lines m sec approach the slope of the tangent line m tan at the point, s. hus, the slope of the tangent line is also epressed as a it: st - s m tan = m sec = = 64. ts ts t - his it is the same it that defines the instantaneous velocit. herefore, the instantaneous velocit at t = is the slope of the line tangent to the position curve at t =. Height above ground (ft) s Slopes of the secant lines approach slope of the tangent line. he secant lines approach the tangent line. (, 80) (., 86.4) (.5, 08) m tan 64 (, 8) m sec 6.4 m sec 56 m sec 48 s(t) 6t 96t t ime (s) FIGURE.5 QUICK CHECK 3 In Figure.5, is m tan at t = greater than or less than m tan at t =? he 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 it. As t S, slopes of secant lines approach the slope of a tangent line. And as t S, average velocities approach an instantaneous velocit. Figure.6 summarizes these two parallel it processes. hese ideas lie at the foundation of what follows in the coming chapters. Copright 04 Pearson Education, Inc.

6 or Use Onl in Pilot Program F he Idea of Limits 65 AVERAGE VELOCIY SECAN LINE Average velocit is the change in position divided b the change in time: v av s(t ) s(t 0 ). t t 0 s(t) 6t 96t 8 t s s 8 s(t) 6t 96t (, 8) Slope of the secant line is the change in s divided b the change in t: m sec s(t ) s(t 0 ). t t 0 80 t s 80 (, 80) m sec v av t s 08 t.5 s 80 t s (, 80) (.5, 08) m sec v av 56 As the time interval shrinks, the average velocit approaches the instantaneous velocit at t t s As the interval on the t-ais shrinks, the slope of the secant line approaches the slope of the tangent line through (, 80) t. s 80 t s (., 86.4) (, 80) m sec v av t INSANANEOUS VELOCIY ANGEN LINE he instantaneous velocit at t is the it of the average velocities as t approaches. s he slope of the tangent line at (, 80) is the it of the slopes of the secant lines as t approaches. 80 t s 80 (, 80) s(t) s() s(t) s() v inst 64 ft/s m t t t t tan t FIGURE.6 Instantaneous velocit 64 ft/s Slope of the tangent line 64 Copright 04 Pearson Education, Inc.

7 or Use Onl in Pilot Program F Chapter Limits SECION. EXERCISES Review Questions. Suppose st is the position of an object moving along a line at time t Ú 0. What is the average velocit between the times t = a and t = b?. Suppose st is the position of an object moving along a line at time t Ú 0. Describe a process for finding the instantaneous velocit at t = a. 3. What is the slope of the secant line between the points a, f a and b, f b on the graph of f? 4. Describe a process for finding the slope of the line tangent to the graph of f at a, f a. 5. Describe the parallels between finding the instantaneous velocit 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. 6. Graph the parabola f =. Eplain wh the secant lines between the points -a, f -a and a, f a have zero slope. What is the slope of the tangent line at = 0? Basic Skills 7. Average velocit he function st represents the position of an object at time t moving along a line. Suppose s = 36 and s3 = 56. Find the average velocit of the object over the interval of time 3, Average velocit he function st represents the position of an object at time t moving along a line. Suppose s = 84 and s4 = 44. Find the average velocit of the object over the interval of time 3, Average velocit he position of an object moving along a line is given b the function st = -6t + 8t. Find the average velocit of the object over the following intervals. a. 3, 44 b. 3, 34 c. 3, 4 d. 3, + h4, where h 7 0 is a real number 0. Average velocit he position of an object moving along a line is given b the function st = -4.9t + 30t + 0. Find the average velocit of the object over the following intervals. a. 30, 34 b. 30, 4 c. 30, 4 d. 30, h4, where h 7 0 is a real number. Average velocit he table gives the position st of an object moving along a line at time t, over a two-second interval. Find the average velocit of the object over the following intervals. a. 30, 4 b. 30,.54 c. 30, 4 d. 30, 0.54 t st Average velocit he graph gives the position st of an object moving along a line at time t, over a.5-second interval. Find the average velocit of the object over the following intervals. a. 30.5,.54 b. 30.5, 4 c. 30.5,.54 d. 30.5, 4 s Average velocit Consider the position function st = -6t + 00t representing the position of an object moving along a line. Sketch a graph of s with the secant line passing through 0.5, s0.5 and, s. Determine the slope of the secant line and eplain its relationship to the moving object. 4. Average velocit Consider the position function st = 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 0, s0 and 0.5, s0.5. Determine the slope of the secant line and eplain its relationship to the moving object. 5. Instantaneous velocit Consider the position function st = -6t + 8t (Eercise 9). Complete the following table with the appropriate average velocities. hen make a conjecture about the value of the instantaneous velocit at t =. ime interval Average velocit.5 s(t) 3, 4 3,.54 3,.4 3,.04 3, Instantaneous velocit Consider the position function st = -4.9t + 30t + 0 (Eercise 0). Complete the following table with the appropriate average velocities. hen make a conjecture about the value of the instantaneous velocit at t =. ime interval Average velocit 3, 34 3,.54 3,.4 3,.04 3, Instantaneous velocit he following table gives the position st of an object moving along a line at time t. Determine the average velocities over the time intervals 3,.04, 3,.004, and 3, hen make a conjecture about the value of the instantaneous velocit at t =. t st t Copright 04 Pearson Education, Inc.

8 or Use Onl in Pilot Program F he Idea of Limits Instantaneous velocit he following table gives the position st of an object moving along a line at time t. Determine the average velocities over the time intervals 3,.04, 3,.004, and 3, hen make a conjecture about the value of the instantaneous velocit at t =. t st Instantaneous velocit Consider the position function st = -6t + 00t. Complete the following table with the appropriate average velocities. hen make a conjecture about the value of the instantaneous velocit at t = 3. ime interval 3, , , , , 34 Average velocit 0. Instantaneous velocit Consider the position function st = 3 sin t that describes a block bouncing verticall on a spring. Complete the following table with the appropriate average velocities. hen make a conjecture about the value of the instantaneous velocit at t = p>. ime interval 3p>, p4 3p>, p> p>, p> p>, p> p>, p> Average velocit Further Eplorations 4. Instantaneous velocit For the following position functions, make a table of average velocities similar to those in Eercises 9 and 0 and make a conjecture about the instantaneous velocit at the indicated time.. st = -6t + 80t + 60 at t = 3. st = 0 cos t at t = p> 3. st = 40 sin t at t = 0 4. st = 0>t + at t = 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. 5. f = at = 6. f = 3 cos at = p> 7. f = e at = 0 8. f = 3 - at = 9. angent lines with zero slope a. Graph the function f = b. Identif the point a, f a at which the function has a tangent line with zero slope. c. Confirm our answer to part (b) b making a table of slopes of secant lines to approimate the slope of the tangent line at this point. 30. angent lines with zero slope a. Graph the function f = 4 -. b. Identif the point a, f a at which the function has a tangent line with zero slope. c. Consider the point a, f a found in part (b). Is it true that the secant line between a - h, f a - h and a + h, f a + h has slope zero for an value of h 0? 3. Zero velocit A projectile is fired verticall upward and has a position given b st = -6t + 8t + 9, for 0 t 9. a. Graph the position function, for 0 t 9. b. From the graph of the position function, identif the time at which the projectile has an instantaneous velocit of zero; call this time t = a. c. Confirm our answer to part (b) b making a table of average velocities to approimate the instantaneous velocit at t = a. d. For what values of t on the interval 30, 94 is the instantaneous velocit positive (the projectile moves upward)? e. For what values of t on the interval 30, 94 is the instantaneous velocit negative (the projectile moves downward)? 3. 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 st = 6t. 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 approimate the velocit at which the rock strikes the ground. 33. Slope of tangent line Given the function f = - cos and the points Ap>, f p>, Bp> , f p> , Cp> + 0.5, f p> + 0.5, and Dp, f p (see figure), find the slopes of the secant lines through A and D, A and C, and A and B. hen use our calculations to make a conjecture about the slope of the line tangent to the graph of f at = p>. QUICK CHECK ANSWERS 0 A B q C q 0.5 q 0.05 D cos. 6 ft>s. Average velocit is the velocit over an interval of time. Instantaneous velocit is the velocit at one point of time. 3. Less than Copright 04 Pearson Education, Inc.

9 or Use Onl in Pilot Program F Chapter Limits. Definitions of Limits Computing tangent lines and instantaneous velocities (Section.) are just two of man important calculus problems that rel on its. We now put these two problems aside until Chapter 3 and begin with a preinar definition of the it of a function. he terms arbitraril close and sufficientl close will be made precise when rigorous definitions of its are given in Section.7. DEFINIION Limit of a Function (Preinar) Suppose the function f is defined for all near a ecept possibl at a. If f is arbitraril close to L (as close to L as we like) for all sufficientl close (but not equal) to a, we write f = L Sa and sa the it of f as approaches a equals L. 6 5 f () Informall, we sa that f = L if f gets closer and closer to L as gets Sa closer and closer to a from both sides of a. he value of f (if it eists) depends Sa upon the values of f near a, but it does not depend on the value of f a. In some cases, the it f equals f a. In other instances, f and f a differ, or f a ma not Sa Sa even be defined. EXAMPLE Finding its from a graph Use the graph of f (Figure.7) to determine the following values, if possible. a. f and S f b. f and S f c. f 3 and S3 f FIGURE.7 SOLUION a. We see that f =. As approaches from either side, the values of f approach (Figure.8). herefore, f =. S b. We see that f = 5. However, as approaches from either side, f approaches 3 because the points on the graph of f approach the open circle at, 3 (Figure.9). herefore, S f = 3 even though f = 5. c. In this case, f 3 is undefined. We see that f approaches 4 as approaches 3 from either side (Figure.0). herefore, S3 f = 4 even though f 3 does not eist. 6 f () 6 f () 5 6 f (3) undefined f () f () f () f () approaches f () approaches f () approaches. 0 6 As approaches... FIGURE As approaches... FIGURE As approaches 3... FIGURE.0 Related Eercises 7 0 Copright 04 Pearson Education, Inc.

10 or Use Onl in Pilot Program F Definitions of Limits 69 In Eample, we have not stated with certaint that S f = 0.5. But this is our best guess based upon the numerical evidence. Methods for calculating its precisel are introduced in Section.3. QUICK CHECK In Eample, suppose we redefine the function at one point so that f =. Does this change the value of f? S In Eample, we worked with the graph of a function. Let s now work with tabulated values of a function. EXAMPLE Finding its from a table Create a table of values of f = - - corresponding to values of near. hen make a conjecture about the value of S f. SOLUION able. lists values of f corresponding to values of approaching from both sides. he numerical evidence suggests that f approaches 0.5 as approaches. herefore, we make the conjecture that S f = 0.5. able. S d f = Related Eercises 4 One-Sided Limits he it f = L is referred to as a two-sided it because f approaches L as Sa approaches a for values of less than a and for values of greater than a. For some functions, it makes sense to eamine one-sided its called left-sided and right-sided its. As with two-sided its, the value of a one-sided it (if it eists) depends on the values of f near a but not on the value of f a. DEFINIION One-Sided Limits. Right-sided it Suppose f is defined for all near a with 7 a. If f is arbitraril close to L for all sufficientl close to a with 7 a, we write f = L Sa + and sa the it of f as approaches a from the right equals L.. Left-sided it Suppose f is defined for all near a with 6 a. If f is arbitraril close to L for all sufficientl close to a with 6 a, we write f = L Sa - and sa the it of f as approaches a from the left equals L. Computer-generated graphs and tables help us understand the idea of a it. Keep in mind, however, that computers are not infallible and the ma produce incorrect results, even for simple functions (see Eample 5). EXAMPLE 3 Eamining its graphicall and numericall Let f = Use tables and graphs to make a conjecture about the values of f, f, and S + S - f, if the eist. S SOLUION Figure.a shows the graph of f obtained with a graphing utilit. he graph is misleading because f is undefined, which means there should be a hole in the graph at, 3 (Figure.b). Copright 04 Pearson Education, Inc.

11 or Use Onl in Pilot Program F Chapter Limits f () 3 8 4( ) f () 3 8 4( ) 3 3 his computergenerated graph is inaccurate because f is undefined at. he hole in the graph at indicates that the function is undefined at this point. 0 FIGURE. (a) 0 (b) he graph in Figure.a and the function values in able.3 suggest that f approaches 3 as approaches from the right. herefore, we write f = 3, S + which sas the it of f as approaches from the right equals 3. f () 3 8 4( ) f () 3 8 4( ) 3... f () approaches f () approaches 3. 0 As approaches from the right... 0 As approaches from the left... FIGURE. (a) (b) Remember that the value of the it does not depend upon the value of f. In this case, f = 3 despite the fact that S f is undefined. Similarl, Figure.b and able.3 suggest that as approaches from the left, f approaches 3. So, we write f = 3, S - which sas the it of f as approaches from the left equals 3. Because f approaches 3 as approaches from either side, we write S f = 3. able.3 S d f = Related Eercises 5 8 Copright 04 Pearson Education, Inc.

12 or Use Onl in Pilot Program F Definitions of Limits 7 Based upon the previous eample, ou might wonder whether the its f, f, and f alwas eist and are equal. he remaining eamples Sa - Sa + Sa demonstrate that these its ma have different values, and in other cases, some or all of these its ma not eist. he following theorem is useful when comparing one-sided and two-sided its. If P and Q are statements, we write P if and onl if Q when P implies Q and Q implies P. HEOREM. Relationship Between One-Sided and wo-sided Limits Assume f is defined for all near a ecept possibl at a. hen f = L if and onl if f = L and f = L. Sa Sa + Sa - A proof of heorem. is outlined in Eercise 44 of Section.7. Using this theorem, it follows that f L if either f L or f L (or both). Sa Sa + Sa - Furthermore, if either f or f does not eist, then f does not eist. Sa + Sa - Sa We put these ideas to work in the net two eamples. 6 4 g() 0 4 FIGURE.3 EXAMPLE 4 A function with a jump Given the graph of g in Figure.3, find the following its, if the eist. a. S - g SOLUION b. S + g c. g S a. As approaches from the left, g approaches 4. herefore, g = 4. S - b. Because g =, for all Ú, g =. S + c. B heorem., g does not eist because g g. S S - S + EXAMPLE 5 Some strange behavior Eamine S0 cos >. Related Eercises 9 4 SOLUION From the first three values of cos > in able.4, it is tempting to conclude that S0 +cos > = -. But this conclusion is not confirmed when we evaluate cos > for values of closer to 0. able.4 cos, We might incorrectl conclude that cos > approaches - as approaches 0 from the right. QUICK CHECK Wh is the graph of = cos > difficult to plot near = 0, as suggested b Figure.4? he behavior of cos > near 0 is better understood b letting = >np, where n is a positive integer. B making this substitution, we can sample the function at discrete points that approach zero. In this case cos if n is even = cos np = b - if n is odd. As n increases, the values of = >np approach zero, while the values of cos > oscillate between - and (Figure.4). herefore, cos > does not approach a single number as approaches 0 from the right. We conclude that S0 +cos > does not eist, which implies that cos > does not eist. S0 Copright 04 Pearson Education, Inc.

13 or Use Onl in Pilot Program F Chapter Limits he values of cos (/) oscillate between and, over shorter and shorter intervals, as approaches 0 from the right. cos(/) FIGURE.4 Related Eercises 5 6 Using tables and graphs to make conjectures for the values of its worked well until Eample 5. he itation of technolog in this eample is not an isolated incident. For this reason, analtical techniques (paper-and-pencil methods) for finding its are developed in the net section. SECION. EXERCISES Review Questions 8. Finding its from a graph Use the graph of g in the figure to. Eplain the meaning of f = L. find the following values, if the eist. Sa. rue or false: When f eists, it alwas equals f a. Eplain. a. g0 b. g c. g d. g S0 S Sa 3. Eplain the meaning of f = L. Sa + 4. Eplain the meaning of f = L. Sa - 5. If Sa - f = L and f = M, where L and M are finite Sa + f eists? real numbers, then how are L and M related if Sa 6. What are the potential problems of using a graphing utilit to estimate Sa f? Basic Skills 7. Finding its from a graph Use the graph of h in the figure to find the following values, if the eist. a. h b. S h c. h4 d. S4 h e. S5 h g() 9. Finding its from a graph Use the graph of f in the figure to find the following values, if the eist. a. f b. f c. f 0 d. f S S0 5 3 h() f () 0 4 Copright 04 Pearson Education, Inc.

14 or Use Onl in Pilot Program F Definitions of Limits Finding its from a graph Use the graph of f in the figure to find the following values, if the eist. a. f b. f S 6 4 c. f S4 f () Estimating a it from tables Let f = d. f S5 a. Calculate f for each value of in the following table. - 4 b. Make a conjecture about the value of S f = f = 4. Estimating a it from tables Let f = a. Calculate f for each value of in the following table. 3 - b. Make a conjecture about the value of S f = f = 3 3. Estimating a it of a function Let gt = t - 9 t - 3. a. Make two tables, one showing the values of g for t = 8.9, 8.99, and and one showing values of g for t = 9., 9.0, and b. Make a conjecture about the value of ts9 t - 9 t Estimating a it of a function Let f = + >. a. Make two tables, one showing the values of f for = 0.0, 0.00, 0.000, and and one showing values of f for = -0.0, -0.00, , and Round our answers to five digits. b. Estimate the value of S0 + >. c. What mathematical constant does S0 + > appear to equal? 5. Estimating a it graphicall and numericall - Let f = ln -. a. Plot a graph of f to estimate S f. b. Evaluate f for values of near to support our conjecture in part (a). 6. Estimating a it graphicall and numericall Let g = e - -. a. Plot a graph of g to estimate S0 g. b. Evaluate g for values of near 0 to support our conjecture in part (a). 7. Estimating a it graphicall and numericall - cos - Let f = -. a. Plot a graph of f to estimate f. S b. Evaluate f for values of near to support our conjecture in part (a). 8. Estimating a it graphicall and numericall 3 sin - cos + Let g =. a. Plot a graph of g to estimate g. S0 b. Evaluate g for values of near 0 to support our conjecture in part (a). 9. One-sided and two-sided its Let f = - 5. Use tables - 5 and graphs to make a conjecture about the values of f, S5 f, and f, if the eist. + S5 - S5 0. One-sided and two-sided its Let g = Use tables and graphs to make a conjecture about the values of g, g, and g, if the eist. + - S00 S00 S00 Copright 04 Pearson Education, Inc.

15 or Use Onl in Pilot Program F Chapter Limits. One-sided and two-sided its Use the graph of f in the figure to find the following values, if the eist. If a it does not eist, eplain wh. a. f b. S - f c. S + f d. S f 5 3 f () f () One-sided and two-sided its Use the graph of g in the figure to find the following values, if the eist. If a it does not eist, eplain wh. a. g b. S - c. S + d. g S e. g3 f. S3 - g. g h. g4 i. g + S3 S Finding its from a graph Use the graph of g in the figure to find the following values, if the eist. If a it does not eist, eplain wh. a. g- b. g c. g - S - S - + d. g e. g f. g S - S g. g h. g5 i. g S3 S g() g() Strange behavior near =0 a. Create a table of values of sin >, for = p, 3p, 5p, 7p, 9p, and. Describe the pattern of values ou observe. p b. Wh does a graphing utilit have difficult plotting the graph of = sin > near = 0 (see figure)? c. What do ou conclude about sin >? S0 3. Finding its from a graph Use the graph of f in the figure to find the following values, if the eist. If a it does not eist, eplain wh. a. f b. f c. f S - S + d. f e. f 3 f. f S S3 - g. f h. f i. f S3 + S3 j. f k. f l. f S - S + S Strange behavior near =0 a. Create a table of values of tan 3> for = >p, >3p, >5p, c, >p. Describe the general pattern in the values ou observe. b. Use a graphing utilit to graph = tan 3>. Wh does a graphing utilit have difficult plotting the graph near = 0? c. What do ou conclude about tan 3>? S0 Copright 04 Pearson Education, Inc.

16 or Use Onl in Pilot Program F Definitions of Limits 75 Further Eplorations 7. Eplain wh or wh not Determine whether the following statements are true and give an eplanation or countereample. - 9 a. he value of does not eist. S3-3 b. he value of f is alwas found b computing f a. Sa c. he value of f does not eist if f a is undefined. Sa d. = 0 S0 e. cot = 0 Sp> 8 9. 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. 8. f = 0, f = 4, f 3 = 6, f = -3, f = 5 S - S + 9. g = 0, g =, g3 = -, g = 0, S g = -, g = S3 S3 30. h- =, h = 0, h = 3, S- - S- + h = h =, h = 4 S - S + 3. p0 =, p = 0, p does not eist, S0 S p = p = S Calculator its Estimate the value of the following its b creating a table of function values for h = 0.0, 0.00, and 0.000, and h = -0.0, -0.00, and hs0 + h >h 34. hs0 h - h 33. hs0 + 3h >h 35. hs0 ln + h h 36. A step function Let f =, for 0. a. Sketch a graph of f on the interval 3-, 4. b. Does f eist? Eplain our reasoning after first S0 eamining S0 - f and S0 + f. 37. he floor function For an real number, the floor function (or greatest integer function) :; is the greatest integer less than or equal to (see figure). a. Compute :;, :;, :;, and :;. S - - S - + S - S + b. Compute :;, :;, and :;. - S.3 S.3 + S.3 c. For a given integer a, state the values of :; and :;. - Sa Sa + d. In general, if a is not an integer, state the values of :; Sa and :;. - Sa + e. For what values of a does Sa :; eist? Eplain :; 38. he ceiling function For an real number, the ceiling function <= is the least integer greater than or equal to. a. Graph the ceiling function = <=, for - 3. b. Evaluate <=, <=, and <=. S - S + S.5 c. For what values of a does <= eist? Eplain. Sa Applications 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 oz) plus $0.7 for each additional ounce (up to and including each additional ounce). a. Graph the function p = f w that gives the postage p for sending a letter that weighs w ounces, for 0 6 w 5. b. Evaluate f w. ws3.3 c. Interpret the its ws + f w and f w. ws - d. Does f w eist? Eplain. ws4 40. he Heaviside function he Heaviside function is used in engineering applications to model flipping a switch. It is defined as H = b 0 if 6 0 if Ú 0. a. Sketch a graph of H on the interval 3-, 4. b. Does H eist? Eplain our reasoning after first S0 eamining H and H. S0 - S0 + Additional Eercises 4. Limits of even functions A function f is even if f - = f, for all in the domain of f. If f is even, with f = 5 and f S = 8, find the following its. + S - a. f b. f S - + S Limits of odd functions A function g is odd if g- = -g, for all in the domain of g. If g is odd, with g = 5 and + S g = 8, find the following its. - S a. g b. + S - S - -g 3 Copright 04 Pearson Education, Inc.

17 or Use Onl in Pilot Program F Chapter Limits echnolog Ecercises Limit b graphing Use the zoom and trace features of a graphing utilit to approimate the following its. 43. sin S0 S S - 3>4 S0 ln 47. Limits b graphs tan a. Use a graphing utilit to estimate S0 sin, tan 3 tan 4, and S0 sin S0 sin. tan p b. Make a conjecture about the value of, for an real S0 sin constant p. sin n 48. Limits b graphs Graph f =, for n =,, 3, and 4 (four graphs). Use the window 3-, 4 * 30, 54. sin a. Estimate S0, sin sin 3 sin 4,, and. S0 S0 S0 sin p b. Make a conjecture about the value of, for an real S0 constant p. sin p 49. Limits b graphs Use a graphing utilit to plot = sin q for at least three different pairs of nonzero constants p and q of our sin p choice. Estimate in each case. hen use our work to S0 sin q sin p make a conjecture about the value of for an nonzero S0 sin q values of p and q. QUICK CHECK ANSWERS. he value of S f depends on the value of f onl near, not at. herefore, changing the value of f will not change the value of S f.. A graphing device has difficult plotting = cos > near 0 because values of the function var between - and over shorter and shorter intervals as approaches 0..3 echniques for Computing Limits Graphical and numerical techniques for estimating its, like those presented in the previous section, provide intuition about its. hese techniques, however, occasionall lead to incorrect results. herefore, we turn our attention to analtical methods for evaluating its precisel. Limits of Linear Functions he graph of f = m + b is a line with slope m and -intercept b. From Figure.5, we see that f approaches f a as approaches a. herefore, if f is a linear function we have f = f a. It follows that for linear functions, f is found b direct Sa Sa substitution of = a into f. his observation leads to the following theorem, which is proved in Eercise 8 of Section.7. f () f () f () f (a) (a, f (a))... f () approaches f (a) f (a) (a, f (a))... f () approaches f (a) f () O a As approaches a from the left... O a As approaches a from the right... FIGURE.5 f () f (a) because f () f (a) as a Copright 04 Pearson Education, Inc. a from both sides of a.

18 or Use Onl in Pilot Program F echniques for Computing Limits 77 HEOREM. Limits of Linear Functions Let a, b, and m be real numbers. For linear functions f = m + b, f = f a = ma + b. Sa EXAMPLE Limits of linear functions Evaluate the following its. a. S3 f, where f = - 7 SOLUION a. S3 f = Limit Laws - 7 = f 3 = - S3 b. g, where g = 6 S. b. S he following it laws greatl simplif the evaluation of man its. g = 6 = g = 6. S Related Eercises 6 Law 6 is a special case of Law 7. Letting m = in Law 7 gives Law 6. HEOREM.3 Limit Laws Assume f and g eist. he following properties hold, where c is a Sa Sa real number, and m 7 0 and n 7 0 are integers.. Sum Sa 3 f + g4 = Sa f + Sa g. Difference Sa 3 f - g4 = Sa f - Sa g 3. Constant multiple Sa 3cf 4 = c Sa f 4. Product 3 f g4 = Sa 3 f Sa 43 g Sa 4 5. Quotient c f Sa g d = Sa f Sa 6. Power Sa 3 f 4 n = 3 Sa f 4 n g, provided Sa g 0 7. Fractional power 3 f 4 n>m = Sa 3 f Sa 4 n>m, provided f Ú 0, for near a, if m is even and n>m is reduced to lowest terms A proof of Law is outlined in Section.7. Laws 5 are proved in Appendi B. Law 6 is proved from Law 4 as follows. For a positive integer n, if f eists, we Sa have 3 f 4 n = 3 f f g f 4 Sa Sa (+++)+++* n factors of f = 3 f Sa 4 3 f Sa 4 g3 f Sa 4 Repeated use of Law 4 ( ) * = 3 Sa f 4 n. n factors of Sa f Copright 04 Pearson Education, Inc.

19 or Use Onl in Pilot Program F Chapter Limits Recall that to take even roots of a number (for eample, square roots or fourth roots), the number must be nonnegative if the result is to be real. In Law 7, the it of 3 f 4 n>m involves the mth root of f when is near a. If the fraction n>m is in lowest terms and m is even, this root is undefined unless f is nonnegative for all near a, which eplains the restrictions shown. EXAMPLE Evaluating its Suppose f = 4, g = 5, and S S h = 8. Use the it laws in heorem.3 to compute each it. S a. S f - g h b. S 36f g + h4 c. S 3g4 3 SOLUION a. S f - g h = = 3 f - g4 S h Law 5 S f - g S S = h Law S = - 8. b. S 36 f g + h4 = S 36 f g4 + S h Law = 6 # S 3 f g4 + S h Law 3 = 6 # 3 f S 4 # 3 g S 4 + h Law 4 S = 6 # 4 # = 8. c. S 3g4 3 = 3 S g 4 3 = 5 3 = 5. Law 6 Limits of Polnomial and Rational Functions Related Eercises 7 4 he it laws are now used to find the its of polnomial and rational functions. For eample, to evaluate the it of the polnomial p = at an arbitrar point a, we proceed as follows: p = Sa Sa = Law Sa Sa Sa = 7 Sa Sa + Sa 4 + Law 3 = 7 Sa Sa + Sa 4 + Law 6 s s u a a 4a + = 7a 3 + 3a + 4a + = pa. heorem. As in the case of linear functions, the it of a polnomial is found b direct substitution; that is, Sa p = pa (Eercise 9). It is now a short step to evaluating its of rational functions of the form f = p>q, where p and q are polnomials. Appling Law 5, we have p p Sa q = Sa q = Sa pa, provided qa 0, qa which shows that its of rational functions are also evaluated b direct substitution. Copright 04 Pearson Education, Inc.

20 or Use Onl in Pilot Program F echniques for Computing Limits 79 he conditions under which direct substitution Sa f = f a can be used to evaluate a it become clear in Section.6, when the important propert of continuit is discussed. HEOREM.4 Limits of Polnomial and Rational Functions Assume p and q are polnomials and a is a constant. a. Polnomial functions: Sa p = pa p b. Rational functions: Sa q = pa, provided qa 0 qa QUICK CHECK Evaluate and S S- -. EXAMPLE 3 Limit of a rational function Evaluate S SOLUION Notice that the denominator of this function is nonzero at =. Using heorem.4b, 3-4 S = =. Related Eercises QUICK CHECK Use heorem.4b to compute. S + EXAMPLE An algebraic function Evaluate. S 4 + SOLUION Using heorems.3 and.4, we have S 4 + = = S 4 + Law 5 S S S Laws and 7 S = = = 0 9. heorem.4 Notice that the it at = equals the value of the function at =. Related Eercises 8 3 One-Sided Limits heorem., Limit Laws 6, and heorem.4 also hold for left-sided and right-sided its. In other words, these laws remain valid if we replace with or Sa + Sa Sa-. Law 7 must be modified slightl for one-sided its, as shown in the net theorem. Copright 04 Pearson Education, Inc.

21 or Use Onl in Pilot Program F Chapter Limits HEOREM.3 (CONINUED) Limit Laws for One-Sided Limits Laws 6 hold with replaced b or Sa + Sa Sa -. Law 7 is modified as follows. Assume m 7 0 and n 7 0 are integers. 7. Fractional power a. 3 f 4 n>m = 3 4 n>m, provided f Ú 0, for near a with 7 a, if Sa + Sa + m is even and n>m is reduced to lowest terms. b. Sa - 3 f 4 n>m = 3 Sa - f 4 n>m, provided f Ú 0, for near a with 6 a, if m is even and n>m is reduced to lowest terms. 4 FIGURE.6 f () 4 if if f () f () 0 EXAMPLE 5 Calculating left- and right-sided its Let Find the values of S - f, S if f = b - if 7. f, and S f, or state that the do not eist. SOLUION he graph of f (Figure.6) suggests that f = and f = 0. S - S + We verif this observation analticall b appling the it laws. For, f = - + 4; therefore, f = =. heorem. S - S - For 7, note that - 7 0; it follows that S + Because S - f = and S + f = - = 0. Law 7 + S f = 0, S f does not eist b heorem.. Related Eercises Other echniques So far, we have evaluated its b direct substitution. A more challenging problem is finding f when the it eists, but f f a. wo tpical cases are Sa Sa shown in Figure.7. In the first case, f a is defined, but it is not equal to f ; in Sa the second case, f a is not defined at all. f () f (a) a f () eists, but f (a) is undefined. a f (a) f () f () O a O a FIGURE.7 Copright 04 Pearson Education, Inc.

22 or Use Onl in Pilot Program F echniques for Computing Limits 8 he argument used in this eample is common. In the it process, approaches, but. herefore, we ma cancel like factors. EXAMPLE 6 Other techniques Evaluate the following its. a. S SOLUION - b. S - a. Factor and Cancel his it cannot be found b direct substitution because the denominator is zero when =. Instead, the numerator and denominator are factored; then, assuming, we cancel like factors: = = Because = - 4 whenever, the two functions have the same it as approaches (Figure.8). herefore, S = S + = = FIGURE q We multipl the given function b = + +. b. Use Conjugates his it was approimated numericall in Eample of Section.; we conjectured that the value of the it is. Direct substitution fails in this case because the denominator is zero at =. Instead, we first simplif the function b multipling the numerator and denominator b the algebraic conjugate of the numerator. he conjugate of - is + ; therefore, = - + Rationalize the numerator; multipl b = - + Epand the numerator. - = - + Simplif. =. + Cancel like factors when. Copright 04 Pearson Education, Inc.

23 or Use Onl in Pilot Program F Chapter Limits he it can now be evaluated: - S - = S + = + =. Related Eercises 39 5 QUICK CHECK 3 Evaluate S5-5 An Important Limit Despite our success in evaluating its using direct substitution, algebraic manipulation, and the it laws, there are important its for which these techniques do not work. One such it arises when investigating the slope of a line tangent to the graph of an eponential function. EXAMPLE 7 Slope of a line tangent to f = Estimate the slope of the line tangent to the graph of f = at the point P0,. SOLUION In Section., the slope of a tangent line was obtained b finding the it of slopes of secant lines; the same strateg is emploed here. We begin b selecting a point Q near P on the graph of f with coordinates,. he secant line joining the points P0, and Q, is an approimation to the tangent line. o compute the slope of the tangent line (denoted b m tan ) at = 0, we look at the slope of the secant line m sec = - > and take the it as approaches 0. f () f () As 0, Q approaches P,... Q(, )... the secant lines approach the tangent line, and m sec m tan. tangent line P(0, ) tangent line Approaching P from the right As 0, Q approaches P,... Q(, ) P(0, ) Approaching P from the left... the secant lines approach the tangent line, and m sec m tan. 0 0 FIGURE.9 (a) (b) - he it eists onl if it has the same value as S 0 + (Figure.9a) S0 and as S 0 - (Figure.9b). Because it is not an elementar it, it cannot be evaluated using the it laws of this section. Instead, we investigate the it using numerical evidence. Choosing positive values of near 0 results in able.5. able m sec = Copright 04 Pearson Education, Inc.

24 or Use Onl in Pilot Program F echniques for Computing Limits 83 Eample 7 shows that , which is S0 approimatel ln. he connection between the natural logarithm and slopes of lines tangent to eponential curves is made clear in Chapters 3 and 6. he Squeeze heorem is also called the Pinching heorem or the Sandwich heorem. f () g() h() h() L O a Squeeze heorem: As a, h() L and f () L. herefore, g() L. FIGURE.0 g() f () We see that as approaches 0 from the right, the slopes of the secant lines approach the slope of the tangent line, which is approimatel A similar calculation (Eercise 53) gives the same approimation for the it as approaches 0 from the left. Because the left-sided and right-sided its are the same, we conclude that - > (heorem.). herefore, the slope of the line tangent to S0 f = at = 0 is approimatel Related Eercises he Squeeze heorem he Squeeze heorem provides another useful method for calculating its. Suppose the functions f and h have the same it L at a and assume the function g is trapped between f and h (Figure.0). he Squeeze heorem sas that g must also have the it L at a. A proof of this theorem is outlined in Eercise 54 of Section.7. HEOREM.5 he Squeeze heorem Assume the functions f, g, and h satisf f g h for all values of near a, ecept possibl at a. If f = h = L, then g = L. Sa Sa Sa EXAMPLE 8 Sine and cosine its A geometric argument (Eercise 90) ma be used to show that for -p> 6 6 p>, - sin and 0 - cos. Use the Squeeze heorem to confirm the following its. a. S0 sin = 0 SOLUION b. S0 cos = a. Letting f = -, g = sin, and h =, we see that g is trapped between f and h on -p> 6 6 p> (Figure.a). Because f = h = 0 S0 S0 (Eercise 37), the Squeeze heorem implies that g = sin = 0. S0 S0 sin on q q sin cos he two its in Eample 8 pla a crucial role in establishing fundamental properties of the trigonometric functions. he its reappear in Section.6. q q q 0 q 0 cos on q q FIGURE. (a) (b) Copright 04 Pearson Education, Inc.

25 or Use Onl in Pilot Program F Chapter Limits b. In this case, we let f = 0, g = - cos, and h = (Figure.b). Because f = h = 0, the Squeeze heorem implies that S0 S0 g = - cos = 0. B the it laws, it follows that S0 - cos = 0, or cos =. Related Eercises S0 S0 S0 S0 0.5 cos cos sin 0 FIGURE. QUICK CHECK 4 Suppose f satisfies f + for all values of 6 near zero. Find f, if possible. S0 EXAMPLE 9 Squeeze heorem for an important it a. Use a graphing utilit to confirm that cos sin, for 0 6. cos b. Use part (a) and the Squeeze heorem to prove that sin =. S0 SOLUION a. Figure. shows the graphs of = cos (lower curve), = sin (middle curve), and = (upper curve) on the interval -. hese graphs confirm the cos given inequalities. b. B part (a), cos sin, for near 0, ecept at 0. Furthermore, cos cos = =. he conditions of the Squeeze heorem are satisfied and S0 S0 cos we conclude that sin =. S0 his important it is used in Chapter 3 to discover derivative rules for trigonometric functions. Related Eercises SECION.3 EXERCISES Review Questions. How is f calculated if f is a polnomial function? Sa. How are f and f calculated if f is a polnomial Sa - Sa + function? 3. For what values of a does Sa r = ra if r is a rational function? 4. Assume g = 4 and f = g whenever 3. S3 Evaluate f, if possible. S Eplain wh = - 4. S3-3 S3 6. If S f = -8, find S 3 f 4 >3. p 7. Suppose p and q are polnomials. If = 0 and S0 q q0 =, find p0. 8. Suppose f = h = 5. Find g, where S S S f g h, for all. 9. Evaluate S Suppose f = b 4 if 3 + if 7 3. Compute f and f. S3 - S3 + Basic Skills 6. Limits of linear functions Evaluate the following its S S 7 4. Appling it laws Assume S S S p S6 S-5 f = 8, g = 3, S and S h =. Compute the following its and state the it laws used to justif our computations. 7. S 34f 4 8. c f S h d 9. 3 f - g f h4 S S. c f g d. S h c f S g - h d 3. 3h f g + 3 S S Copright 04 Pearson Education, Inc.

26 or Use Onl in Pilot Program F echniques for Computing Limits Evaluating its Evaluate the following its S S bs 3b 4b S One-sided its Let 6. ts- t + 5t ts3 3 t S 3. hs h + 4 f = b + if if Ú -. Compute the following its or state that the do not eist. a. S- - f 34. One-sided its Let b. f c. f S- + S- 0 if -5 f = c 5 - if if Ú 5. Compute the following its, or state that the do not eist. a. f b. c. f S-5 S-5 - S-5 + d. S5 - f e. S5 + f f. f S5 35. One-sided its a. Evaluate -. + S b. Eplain wh - does not eist. - S 36. One-sided its a. Evaluate S3 - b. Eplain wh S A does not eist. A Absolute value it Show that = 0 b first evaluating S0 S0 - and S0 +. Recall that = b if Ú 0 - if Absolute value it Show that Sa = a, for an real number. (Hint: Consider the cases a 6 0 and a Ú 0.) Other techniques Evaluate the following its, where a and b are fied real numbers S S b b 43. Sb - b S3-3 3t - 7t + 4. ts - t + b b S-b 4 + b S S9-9 - a 49. Sa - a, a h hs0 h 53. Slope of a tangent line 46. hs0 5 + h - 5 h 48. ts3 a4t - t - 3 b6 + t - t - a 50. Sa - a, a a 5. Sa - a a. Sketch a graph of = and carefull draw three secant lines connecting the points P0, and Q,, for = -3, -, and -. b. Find the slope of the line that joins P0, and Q,, for 0. c. Complete the table and make a conjecture about the value of -. S Slope of a tangent line a. Sketch a graph of = 3 and carefull draw four secant lines connecting the points P0, and Q, 3, for = -, -,, and. b. Find the slope of the line that joins P0, and Q, 3, for 0. c. Complete the table and make a conjecture about the value 3 - of. S Appling the Squeeze heorem a. Show that - sin, for 0. b. Illustrate the inequalities in part (a) with a graph. c. Use the Squeeze heorem to show that sin S0 = A cosine it b the Squeeze heorem It can be shown that - cos, for near 0. a. Illustrate these inequalities with a graph. b. Use these inequalities to find cos. S0 57. A sine it b the Squeeze heorem It can be shown that - 6 sin, for near 0. a. Illustrate these inequalities with a graph. sin b. Use these inequalities to find S0. Copright 04 Pearson Education, Inc.