2.1 Limits, Rates of Change, and Tangent Lines. Preliminary Questions 1. Average velocity is defined as a ratio of which two quantities?

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1 LIMITS. Limits, Rates of Change, and Tangent Lines Preinar Questions. Average velocit is defined as a ratio of which two quantities? Average velocit is defined as the ratio of distance traveled to time elapsed.. Average velocit is equal to the slope of a secant line through two points on a graph. Which graph? Average velocit is the slope of a secant line through two points on the graph of position as a function of time. 3. Can instantaneous velocit be defined as a ratio? If not, how is instantaneous velocit computed? Instantaneous velocit cannot be defined as a ratio. It is defined as the it of average velocit as time elapsed shrinks to zero. 4. What is the graphical interpretation of instantaneous velocit at a moment t = t 0? Instantaneous velocit at time t = t 0 is the slope of the line tangent to the graph of position as a function of time at t = t What is the graphical interpretation of the following statement: The average ROC approaches the instantaneous ROC as the interval [ 0, ] shrinks to 0? The slope of the secant line over the interval [ 0, ] approaches the slope of the tangent line at = The ROC of atmospheric temperature with respect to altitude is equal to the slope of the tangent line to a graph. Which graph? What are possible units for this rate? The rate of change of atmospheric temperature with respect to altitude is the slope of the line tangent to the graph of atmospheric temperature as a function of altitude. Possible units for this rate of change are F/ft or C/m. Eercises. A ball is dropped from a state of rest at time t = 0. The distance traveled after t seconds is s(t) = 6t ft. (a) How far does the ball travel during the time interval [,.5]? (b) Compute the average velocit over [,.5]. (c) Compute the average velocit over time intervals [,.0], [,.005], [,.00], [,.0000]. Use this to estimate the object s instantaneous velocit at t =. (a) Galileo s formula is s(t) = 6t. The ball thus travels s = s(.5) s() = 6(.5) 6() = 36 ft. (b) The average velocit over [,.5] is (c) s t = s(.5) s().5 = 36 = 7 ft/s. 0.5 The instantaneous velocit at t = is64ft/s. time interval [,.0] [,.005] [,.00] [,.0000] average velocit A wrench is released from a state of rest at time t = 0. Estimate the wrench s instantaneous velocit at t =, assuming that the distance traveled after t seconds is s(t) = 6t. To find the instantaneous velocit, we compute the average velocities: time interval [,.0] [,.005] [,.00] [,.0000] average velocit The instantaneous velocit is approimatel 3 ft/s. 3. Let v = 0 T as in Eample. Estimate the instantaneous ROC of v with respect to T when T = 300 K.

2 76 CHAPTER LIMITS T interval [300, 300.0] [300, ] average rate of change T interval [300, ] [300, ] average ROC The instantaneous rate of change is approimatel m/(s K). 4. Compute / for the interval [, 5],where = 4 9. What is the instantaneous ROC of with respect to at =? / = ((4(5) 9) (4() 9))/(5 ) = 4. Because the graph of = 4 9 is a line with slope 4, the average rate of change of calculated over an interval will be equal to 4; hence, the instantaneous rate of change at an will also be equal to 4. In Eercises 5 6, a stone is tossed in the air from ground level with an initial velocit of 5 m/s. Its height at time t is h(t) = 5t 4.9t m. 5. Compute the stone s average velocit over the time interval [0.5,.5] and indicate the corresponding secant line on a sketch of the graph of h(t). The average velocit is equal to The secant line is plotted with h(t) below. h(.5) h(0.5) = 0.3. h t 6. Compute the stone s average velocit over the time intervals [,.0], [,.00], [,.000] and [0.99, ], [0.999, ], [0.9999, ]. Use this to estimate the instantaneous velocit at t =. With h(t) = 5t 4.9t, the average velocit over the time interval [t, t ] is given b h t = h (t ) h (t ) t t. time interval [,.0] [,.00] [,.000] [0.99, ] [0.999, ] [0.9999, ] average velocit The instantaneous velocit at t = second is 5. m/s. 7. With an initial deposit of $00, the balance in a bank account after t ears is f (t) = 00(.08) t dollars. (a) What are the units of the ROC of f (t)? (b) Find the average ROC over [0, 0.5] and [0, ]. (c) Estimate the instantaneous rate of change at t = 0.5 b computing the average ROC over intervals to the left and right of t = 0.5. (a) The units of the rate of change of f (t) are dollars/ear or $/r. (b) The average rate of change of f (t) = 00(.08) t over the time interval [t, t ] is given b f t = f (t ) f (t ) t t. time interval [0,.5] [0, ] average rate of change

3 SECTION. Limits, Rates of Change, and Tangent Lines 77 (c) The rate of change at t = 0.5 is approimatel $8/r. time interval [.5,.5] [.5,.50] [.5,.500] average rate of change time interval [.49,.5] [.499,.5] [.4999,.5] average ROC The distance traveled b a particle at time t is s(t) = t 3 + t. Compute the average velocit over the time interval [, 4] and estimate the instantaneous velocit at t =. The average velocit over the time interval [, 4] is s(4) s() 4 = 68 3 =. To estimate the instantaneous velocit at t =, we eamine the following table. time interval [,.0] [,.00] [,.000] [.99, ] [.999, ] [.9999, ] average ROC The rate of change at t = is approimatel 4. In Eercises 9 6, estimate the instantaneous rate of change at the point indicated. 9. P() = 4 3; = interval [,.0] [,.00] [,.000] [.99, ] [.999, ] [.9999, ] average rate of change The rate of change at = is approimatel f (t) = 3t 5; t = 9 t interval [ 9, 8.99] [ 9, 8.999] [ 9, ] average ROC t interval [ 9.0, 9] [ 9.00, 9] [ 9.000, 9] average ROC The rate of change at t = 9is3, as the graph of = f (t) is a line with slope 3.. () = + ; = interval [,.0] [,.00] [,.000] [.99, ] [.999, ] [.9999, ] average ROC The rate of change at = is approimatel (t) = 3t + ; t = t interval [,.0] [,.00] [,.000] [0.99, ] [0.999, ] [0.9999, ] average ROC The rate of change at t = is approimatel 0.75.

4 78 CHAPTER LIMITS 3. f () = e ; = 0 interval [ 0.0, 0] [ 0.00, 0] [ 0.000, 0] [0, 0.0] [0, 0.00] [0, 0.000] average ROC The rate of change at = 0 is approimatel f () = e ; = e interval [e 0.0, e] [e 0.00, e] [e 0.000, e] [e, e + 0.0] [e, e ] [e, e ] average ROC The rate of change at = e is approimatel f () = ln ; = 3 interval [.99, 3] [.999, 3] [.9999, 3] [3, 3.0] [3, 3.00] [3, 3.000] average ROC The rate of change at = 3 is approimatel f () = tan ; = 4 π interval [ π4 0.0, 4 π ] [ π4 0.00, 4 π ] [ π , 4 π ] [ π4, 4 π + 0.0] [ π 4, 4 π ] [ π 4, 4 π ] average ROC The rate of change at = 4 π is approimatel The atmospheric temperature T (in F) above a certain point on earth is T = h,whereh is the altitude in feet (valid for h 37,000). What are the average and instantaneous rates of change of T with respect to h? Whare the the same? Sketch the graph of T for h 37,000. The average and instantaneous rates of change of T with respect to h are both F/ft. The rates of change are the same because T is a linear function of h with slope Temp ( F) ,000 0,000 30,000 Altitude (ft) 8. The height (in feet) at time t (in seconds) of a small weight oscillating at the end of a spring is h(t) = 0.5cos(8t). (a) Calculate the weight s average velocit over the time intervals [0, ] and [3, 5]. (b) Estimate its instantaneous velocit at t = 3. (a) The average velocit over the time interval [t, t ] is given b h t (b) = h (t ) h (t ) t t. time interval [0, ] [3, 5] average velocit ft/s 0.78 ft/s time interval [3, 3.0] [3, 3.00] [3, 3.000] [.99, 3] [.999, 3] [.9999, 3] average velocit

5 SECTION. Limits, Rates of Change, and Tangent Lines 79 The instantaneous velocit at t = 3 seconds is approimatel 3.6 ft/s. 9. The number P(t) of E. coli cells at time t (hours) in a petri dish is plotted in Figure 9. (a) Calculate the average ROC of P(t) over the time interval [, 3] and draw the corresponding secant line. (b) Estimate the slope m of the line in Figure 9. What does m represent? P(t) 0,000 8,000 6,000 4,000,000,000 3 t (hours) FIGURE 9 Number of E. coli cells at time t. (a) Looking at the graph, we can estimate P() = 000 and P(3) = Assuming these values of P(t), the average rate of change is The secant line is here: P(3) P() 3 = 6000 = 3000 cells/hour. P(t) 0,000 8,000 6,000 4,000,000,000 3 t (hours) (b) The line in Figure 9 goes through two points with approimate coordinates (, 000) and (.5, 4000). This line has approimate slope m = = cells/hour. m is close to the slope of the line tangent to the graph of P(t) at t =, and so m represents the instantaneous rate of change of P(t) at t = hour. 0. Calculate + for =, 0, 00,,000. Does change more rapidl when is large or small? Interpret our answer in terms of tangent lines to the graph of = changes less rapidl as gets larger. This can be seen in the tangent lines to the graph of, which become less steep as gets larger.. Assume that the period T (in seconds) of a pendulum (the time required for a complete back-and-forth ccle) is T = 3 L,whereL is the pendulum s length (in meters).

6 80 CHAPTER LIMITS (a) What are the units for the ROC of T with respect to L? Eplain what this rate measures. (b) Which quantities are represented b the slopes of lines A and B in Figure 0? (c) Estimate the instantaneous ROC of T with respect to L when L = 3 m. B A Period (s) 3 Length (m) FIGURE 0 The period T is the time required for a pendulum to swing back and forth. (a) The units for the rate of change of T with respect to L are seconds per meter. This rate measures the sensitivit of the period of the pendulum to a change in the length of the pendulum. (b) The slope of the line B represents the average rate of change in T from L = mtol = 3 m. The slope of the line A represents the instantaneous rate of change of T at L = 3m. (c) time interval [3, 3.0] [3, 3.00] [3, 3.000] [.99, 3] [.999, 3] [.9999, 3] average velocit The instantaneous rate of change at L = m is approimatel s/m.. The graphs in Figure represent the positions of moving particles as functions of time. (a) Do the instantaneous velocities at times t, t, t 3 in (A) form an increasing or decreasing sequence? (b) Is the particle speeding up or slowing down in (A)? (c) Is the particle speeding up or slowing down in (B)? Distance Distance t (A)t t 3 Time FIGURE (B) Time (a) As the value of the independent variable increases, we note that the slope of the tangent lines decreases. Since Figure (A) displas position as a function of time, the slope of each tangent line is equal to the velocit of the particle; consequentl, the velocit of the particle is decreasing. (b) Based on the solution to part (a), the velocit of the particle is decreasing; hence, the particle is slowing down. (c) If we were to draw several lines tangent to the graph in Figure (B), we would find that the slopes would be increasing. Accordingl, the velocit of the particle associated with Figure (B) is increasing, and the particle is speeding up. 3. The graphs in Figure represent the positions s of moving particles as functions of time t. Match each graph with one of the following statements: (a) Speeding up (b) Speeding up and then slowing down (c) Slowing down (d) Slowing down and then speeding up

7 SECTION. Limits, Rates of Change, and Tangent Lines 8 s s (A) t (B) t s s (C) t FIGURE (D) t When a particle is speeding up over a time interval, its graph is bent upward over that interval. When a particle is slowing down, its graph is bent downward over that interval. Accordingl, In graph (A), the particle is (c) slowing down. In graph (B), the particle is (b) speeding up and then slowing down. In graph (C), the particle is (d) slowing down and then speeding up. In graph (D), the particle is (a) speeding up. 4. An epidemiologist finds that the percentage N(t) of susceptible children who were infected on da t during the first three weeks of a measles outbreak is given, to a reasonable approimation, b the formula A graph of N(t) appears in Figure 3. 00t N(t) = t 3 + 5t 00t % Infected Time (das) FIGURE 3 Graph of N(t). (a) Draw the secant line whose slope is the average rate of increase in infected children over the intervals between das 4 and 6 and between das and 4. Then compute these average rates (in units of percent per da). (b) Estimate the ROC of N(t) on da. (a) % Infected Time (das) The average rate of change of N(t) over the interval between da 4 and da 6 is given b N t = N(6) N(4) 6 4 = 3.776%/da. Similarl, we calculate the average rate of change of N(t) over the interval between da and da 4 as N t = N(4) N() 4 = %/da. (b)

8 8 CHAPTER LIMITS time interval [,.5] [,.] [,.0] [,.00] average ROC time interval [.5, ] [.8, ] [.99, ] [.999, ] average ROC The instantaneous rate of change of N(t) on da is 0.986%/da. 5. The fraction of a cit s population infected b a flu virus is plotted as a function of time (in weeks) in Figure 4. (a) Which quantities are represented b the slopes of lines A and B? Estimate these slopes. (b) Is the flu spreading more rapidl at t =,, or 3? (c) Is the flu spreading more rapidl at t = 4, 5, or 6? Fraction infected B 0. A FIGURE 4 Weeks (a) The slope of line A is the average rate of change over the interval [4, 6], whereas the slope of the line B is the instantaneous rate of change at t = 6. Thus, the slope of the line A ( )/ = 0.045/week, whereas the slope of the line B ( )/6 = 0.07/week. (b) Among times t =,, 3, the flu is spreading most rapidl at t = 3 since the slope is greatest at that instant; hence, the rate of change is greatest at that instant. (c) Among times t = 4, 5, 6, the flu is spreading most rapidl at t = 4 since the slope is greatest at that instant; hence, the rate of change is greatest at that instant. 6. The fungus fusarium eosporium infects a field of fla plants through the roots and causes the plants to wilt. Eventuall, the entire field is infected. The percentage f (t) of infected plants as a function of time t (in das) since planting is shown in Figure 5. (a) What are the units of the rate of change of f (t) with respect to t? What does this rate measure? (b) Use the graph to rank (from smallest to largest) the average infection rates over the intervals [0, ], [0, 3], and [40, 5]. (c) Use the following table to compute the average rates of infection over the intervals [30, 40], [40, 50], [30, 50]: Das Percent infected (d) Draw the tangent line at t = 40 and estimate its slope. Choose an two points on the tangent line for the computation. Percent infected Das after planting FIGURE 5 (a) The units of the rate of change of f (t) with respect to t are percent /da or %/d. This rate measures how quickl the population of fla plants is becoming infected. (b) From smallest to largest, the average rates of infection are those over the intervals [40, 5], [0, ], [0, 3]. Thisis because the slopes of the secant lines over these intervals are arranged from smallest to largest.

9 SECTION. Limits, Rates of Change, and Tangent Lines 83 (c) The average rates of infection over the intervals [30, 40], [40, 50], [30, 50] are.9,.5,.7 %/d, respectivel. (d) The tangent line sketched in the graph below appears to pass through the points (0, 80) and (40, 9). The estimate of the instantaneous rate of infection at t = 40 das is therefore = 0 = 0.55%/d Let v = 0 T as in Eample. Is the ROC of v with respect to T greater at low temperatures or high temperatures? Eplain in terms of the graph. (m/s) T (K) As the graph progresses to the right, the graph bends progressivel downward, meaning that the slope of the tangent lines becomes smaller. This means that the ROC of v with respect to T is lower at high temperatures. 8. If an object moving in a straight line (but with changing velocit) covers s feet in t seconds, then its average velocit is v 0 = s/ t ft/s. Show that it would cover the same distance if it traveled at constant velocit v 0 over the same time interval of t seconds. This is a justification for calling s/ t the average velocit. An object moving in a straight line from time t = t 0 to time t = t with average velocit v 0 satisfies: The distance travelled b the particle in t seconds is s/ t = (s s 0 )/(t t 0 ) = v 0. s s 0 = v 0 (t t 0 ). An object moving at constant velocit v 0 is at distance s(t) = s 0 + v 0 t. Between time t = t 0 and time t = t,ittravels distance: The two distances traveled are the same. s(t ) s(t 0 ) = (s 0 + v 0 t ) (s 0 + v 0 t 0 ) = v 0 (t t 0 ). 9. Sketch the graph of f () = ( ) over [0, ]. Refer to the graph and, without making an computations, find: (a) The average ROC over [0, ] (b) The (instantaneous) ROC at = (c) The values of at which the ROC is positive (a) f (0) = f (), so there is no change between = 0and =. The average ROC is zero.

10 84 CHAPTER LIMITS (b) The tangent line to the graph of f () is horizontal at = ; the instantaneous ROC is zero at this point. (c) The ROC is positive at all points where the graph is rising, because the slope of the tangent line is positive at these points. This is so for all between = 0and = Which graph in Figure 6 has the following propert: For all, the average ROC over [0, ] is greater than the instantaneous ROC at? Eplain. (A) FIGURE 6 (B) (a) The average rate of change over [0, ] is greater than the instantaneous rate of change at : (B). (b) The average rate of change over [0, ] is less than the instantaneous rate of change at : (A) The graph in (B) bends downward, so the slope of the secant line through (0, 0) and (, f ()) is larger than the slope of the tangent line at (, f ()). On the other hand, the graph in (A) bends upward, so the slope of the tangent line at (, f ()) is larger than the slope of the secant line through (0, 0) and (, f ()). Further Insights and Challenges 3. The height of a projectile fired in the air verticall with initial velocit 64 ft/sish(t) = 64t 6t ft. (a) Compute h(). Show that h(t) h() can be factored with (t ) as a factor. (b) Using part (a), show that the average velocit over the interval [, t] is 6(t 3). (c) Use this formula to find the average velocit over several intervals [, t] with t close to. Then estimate the instantaneous velocit at time t =. (a) With h(t) = 64t 6t,wehaveh() = 48 ft, so h(t) h() = 6t + 64t 48. Taking out the common factor of 6 and factoring the remaining quadratic, we get (b) The average velocit over the interval [, t] is h(t) h() = 6(t 4t + 3) = 6(t )(t 3). h(t) h() t = 6(t )(t 3) t = 6(t 3). (c) t average velocit over [, t] The instantaneous velocit is approimatel 3 ft/s. Plugging t = second into the formula in (b) ields 6( 3) = 3 ft/s eactl. 3. Let Q(t) = t. As in the previous eercise, find a formula for the average ROC of Q over the interval [, t] and use it to estimate the instantaneous ROC at t =. Repeat for the interval [, t] and estimate the ROC at t =. The average ROC is Q(t) Q() t = t t. Appling the difference of squares formula gives that the average ROC is ((t + )(t ))/(t ) = (t + ) for t =. As t gets closer to, this gets closer to + =. The instantaneous ROC is. For t 0 =, the average ROC is Q(t) Q() t = t 4 t, which simplifies to t + fort =. As t approaches, the average ROC approaches 4. The instantaneous ROC is therefore 4.

11 SECTION. Limits: A Numerical and Graphical Approach Show that the average ROC of f () = 3 over [, ] is equal to + +. Use this to estimate the instantaneous ROC of f () at =. The average ROC is f () f () = 3. Factoring the numerator as the difference of cubes means the average rate of change is ( )( + + ) = + + (for all = ). The closer gets to, the closer the average ROC gets to + + = 3. The instantaneous ROC is Find a formula for the average ROC of f () = 3 over [, ] and use it to estimate the instantaneous ROC at =. The average ROC is f () f () = 3 8. Appling the difference of cubes formula to the numerator, we find that the average ROC is ( + + 4)( ) = for =. The closer gets to, the closer the average ROC gets to + () + 4 =. 35. Let T = 3 L as in Eercise. The numbers in the second column of Table 4 are increasing and those in the last column are decreasing. Eplain wh in terms of the graph of T as a function of L. Also, eplain graphicall wh the instantaneous ROC at L = 3 lies between and TABLE 4 Interval Average Rates of Change of T with Respect to L Average ROC Interval Average ROC [3, 3.] [.8, 3] [3, 3.] [.9, 3] [3, 3.00] [.999, 3] [3, ] [.9995, 3] Since the average ROC is increasing on the intervals [3, L] as L get close to 3, we know that the slopes of the secant lines between points on the graph over these intervals are increasing. The more rows we add with smaller intervals, the greater the average ROC. This means that the instantaneous ROC is probabl greater than all of the numbers in this column. Likewise, since the average ROC is decreasing on the intervals [L, 3] as L gets closer to 3, we know that the slopes of the secant lines between points over these intervals are decreasing. This means that the instantaneous ROC is probabl less than all the numbers in this column. The tangent slope is somewhere between the greatest value in the first column and the least value in the second column. Hence, it is between.4399 and The first column underestimates the instantaneous ROC b secant slopes; this estimate improves as L decreases toward L = 3. The second column overestimates the instantaneous ROC b secant slopes; this estimate improves as L increases toward L = 3.. Limits: A Numerical and Graphical Approach Preinar Questions. What is the it of f () = as π? π =.. What is the it of g(t) = t as t π? t π t = π. 3. Can f () approach a it as c if f (c) is undefined? If so, give an eample.

12 86 CHAPTER LIMITS Yes. The it of a function f as c does not depend on what happens at = c, onl on the behavior of f as c. As an eample, consider the function The function is clearl not defined at = but f () =. 4. Is 0 equal to 0 or 0? 0 f () = = ( + ) =. 0 0 = What does the following table suggest about f () and f ()? f () The values in the table suggest that f () = and + f () = Is it possible to tell if f () eists b onl eamining values f () for close to but greater than 5? Eplain. 5 No. B eamining values of f () for close to but greater than 5, we can determine whether the one-sided it 5+ f () eists. To determine whether 5 f () eists, we must eamine value of f () on both sides of = If ou know in advance that f () eists, can ou determine its value just knowing the values of f () for all 5 > 5? Yes. If 5 f () eists, then both one-sided its must eist and be equal. 8. Which of the following pieces of information is sufficient to determine whether f () eists? Eplain. 5 (a) The values of f () for all (b) The values of f () for in [4.5, 5.5] (c) The values of f () for all in [4.5, 5.5] other than = 5 (d) The values of f () for all 5 (e) f (5) To determine whether 5 f () eists, we must know the values of f () for values of near 5, both smaller than and larger than 5. Thus, the information in (a), (b) or (c) would be sufficient. The information in (d) does not include values of f for < 5, so this information is not sufficient to determine whether the it eists. The it does not depend at all on the value f (5), so the information in (e) is also not sufficient to determine whether the it eists. Eercises In Eercises 6, fill in the tables and guess the value of the it.. f (), where f () = 3. f () f () f ()

13 SECTION. Limits: A Numerical and Graphical Approach 87 The it as is 3.. h(t),whereh(t) = cos t t 0 t. Note that h(t) is even, that is, h(t) = h( t). t ±0.00 ±0.000 ± ± h(t) The it as t 0is. 3. f (),where f () = + 6. t ±.00 ±.000 h(t) t ± ±.0000 h(t) f () f () The it as is 3 5. sin θ θ 4. f (θ), where f (θ) = θ 0 θ f () θ ±0.00 ±0.000 ± ± f (θ) The it as θ 0is f (),where f () = e 0. θ ±.00 ±.000 f (θ) θ ± ±.0000 f (θ) ±0.5 ±0. ±0.05 ±0.0 f () f () f ()

14 88 CHAPTER LIMITS The it as 0is. 6. f (),where f () = ln f () f () The it as 0+ is Determine f () for the function f () shown in Figure FIGURE 9 The graph suggests that f ().5 as Do either of the two oscillating functions in Figure 0 appear to approach a it as 0? (A) FIGURE 0 (B) (A) does not appear to approach a it as 0; the values of the function oscillate wildl as 0. The values of the function graphed in (B) seem to settle to 0 as 0, so the it seems to eist. In Eercises 9 0, evaluate the it. 9. As, f () =. You can see this, for eample, on the graph of f () = The graph of f () = 3 is a horizontal line. f () = 3 for all values of, so the it is also equal to In Eercises 0, verif each it using the it definition. For eample, in Eercise, show that 6 can be made as small as desired b taking close to = 6 6 = 3. 6 can be made arbitraril small b making close enough to 3, thus making 3 small.. 4 = f () 4 = 4 4 =0 for all values of so f () 4 is alread smaller than an positive number as

15 SECTION. Limits: A Numerical and Graphical Approach (4 + 3) = (4 + 3) = 4 8 =4. Therefore, if ou make small enough, ou can make (4 + 3) as small as desired. 4. (5 7) = 8 3 As 3, note that (5 7) 8 = 5 5 = 5 3. If ou make 3 small enough, ou can make (5 7) 8 as small as desired. 5. ( ) = 8 9 We have ( 8) = ( )( 9) = 9. If ou make 9 small enough, ou can make 9 = ( 8) as small as desired. 6. ( ) = 5 As 5, we have ( ) = 0 = ( )( + 5) = ( 5). B making close enough to 5 sothat ( 5) is sufficientl small, ou can make ( 5) = ( ) as small as desired. 7. = 0 0 As 0, we have 0 = To simplif things, suppose that <, so that = <. Bmaking sufficientl small, so that = is even smaller, ou can make 0 as small as desired. 8. ( + 4) = = =. If ou make <, <, sothatmaking 0 small enough can make as small as desired. 9. ( + + 3) = 3 0 As 0, we have = + = +.If <, + can be no bigger than 3, so + < 3. Therefore, b making 0 = sufficientl small, ou can make = + as small as desired. 0. ( 3 + 9) = 9 0 ( 3 + 9) 9 = 3.Ifwemake <, then 3 <. Therefore, b making 0 = sufficientl small, we can make ( 3 + 9) 9 as small as desired. In Eercises 36, estimate the it numericall or state that the it does not eist.. The it as is f () The it as 3is f ()

16 90 CHAPTER LIMITS f () The it as is f () The it as 3is3.75. sin f () The it as 0is. sin f () The it as 0is5. sin f () The it does not eist. As 0, f () ; similarl, as 0+, f (). cos θ 8. θ 0 θ The it as 0is0. 9. h 0 cos h f () h ±0. ±0.0 ±0.00 ±0.000 f (h) The it does not eist since cos (/h) oscillates infinitel often as h h 0 sin h cos h

17 SECTION. Limits: A Numerical and Graphical Approach 9 The it as 0is0. h 3. h 0 h h f (h) h f (h) The it as 0 is approimatel (The eact answer is ln.) e f () The it as 0 is approimatel (The eact answer is ln.) sec f () The it as + is approimatel.44. (The eact answer is.) The it as 0is tan sin f () The it as 0 does not eist. tan sin f () f () The it as 0 is approimatel.00. (The eact answer is.)

18 9 CHAPTER LIMITS 37. Determine + f () and f () for the function shown in Figure. FIGURE The left-hand it is f () =, whereas the right-hand it is f () =. Accordingl, the + two-sided it does not eist. 38. Determine the one-sided its at c =,, 4, 5 of the function g(t) shown in Figure and state whether the it eists at these points FIGURE At c =, the left-hand it is g(t) =.5, whereas the right-hand it is g(t) = 0.7. Accordingl, t t + the two-sided it does not eist at c =. At c =, the left-hand it is g(t) =, whereas the right-hand it is g(t) = 3. Accordingl, the t t + two-sided it does not eist at c =. At c = 4, the left-hand it is g(t) =, whereas the right-hand it is g(t) =. Accordingl, the t 4 t 4+ two-sided it eists at c = 4 and equals. At c = 5, the left hand it is g(t).3, and the right-hand it is g(t).3. Accordingl, the t 5 t 5+ two-sided it eists at c = 5 and is equal to roughl.3. Note that g(5).3 as well. 39. The greatest integer function is defined b [] =n, wheren is the unique integer such that n < n +. See Figure 3. (a) For which values of c does [] eist? What about c c+ []? (b) For which values of c does [] eist? c 3 FIGURE 3 Graph of =[]. (a) The one-sided its eist for all real values of c. (b) For each integer value of c, the one-sided its differ. In particular, [] = c, whereas c c+ [] = c.(for noninteger values of c, the one-sided its both equal [c].) The it [] eists when c c [] = c+ [], namel for noninteger values of c: n < c < n +, where n is an integer.

19 SECTION. Limits: A Numerical and Graphical Approach Draw a graph of f () = and use it to determine the one-sided its + f () and f (). The graph of f () is a step function. The left-hand it is f () =, whereas the right-hand it is f () =. Accordingl, the two-sided it does not eist In Eercises 4 43, determine the one-sided its numericall. sin 4. 0± f () The left-hand it is f () =, whereas the right-hand it is f () = ± /...5. f () The left-hand it is f () =, whereas the right-hand it is f () = sin( ) 43. 0± f () The left-hand it is f () =, whereas the right-hand it is f () = Determine the one- or two-sided infinite its in Figure 4. (A) FIGURE 4 4 (B) In (A), the left-hand it is +, whereas the right-hand it is. Thatis,as, the function takes on arbitraril large positive values. Similarl, as +, the function takes on arbitraril large negative values. In (B), the it is + ; thatis,as 4, the function takes on arbitraril large positive values.

20 94 CHAPTER LIMITS 45. Determine the one-sided its of f () at c = andc = 4, for the function shown in Figure 5. For c =, we have f () = and f () =. + For c = 4, we have f () = and f () = FIGURE Determine the infinite one- and two-sided its in Figure FIGURE 6 f () = f () = + f () = 3 f () = 5 In Eercises 47 50, draw the graph of a function with the given its. 47. f () =, f () = 0, f () = f () =, f () = 0, f () =

21 SECTION. Limits: A Numerical and Graphical Approach f () = f () = 3, f () =, f () = = f (4) f () =, f () = 3, f () = In Eercises 5 56, graph the function and use the graph to estimate the value of the it. sin 3θ 5. θ 0 sin θ = sin 3 sin The it as θ 0is = The it as 0is. cos 53. 0

22 96 CHAPTER LIMITS = cos The it as 0 is approimatel (The eact answer is ln.) sin 4θ 54. θ 0 cos θ = sin 4q cosq The it as θ 0is 3. cos 3θ cos 4θ 55. θ 0 θ = cos 3 cos The it as θ 0is3.5. cos 3θ cos 5θ 56. θ 0 θ 8.0 = cos 3 cos The it as θ 0is8. Further Insights and Challenges 57. Light waves of frequenc λ passing through a slit of width a produce a Fraunhofer diffraction pattern of light and dark fringes (Figure 7). The intensit as a function of the angle θ is given b ( ) sin(r sin θ) I (θ) = I m R sin θ where R = πa/λ and I m is a constant. Show that the intensit function is not defined at θ = 0. Then check numericall that I (θ) approaches I m as θ 0 for an two values of R (e.g., choose two integer values).

23 SECTION. Limits: A Numerical and Graphical Approach 97 a Incident light waves Slit Viewing screen Intensit pattern FIGURE 7 Fraunhofer diffraction pattern. If ou plug in θ = 0, ou get a division b zero in the epression ( ) sin R sin θ ; R sin θ thus, I (0) is undefined. If R =, a table of values as θ 0 follows: The it as θ 0is I m = I m. If R = 3, the table becomes: θ I (θ) I m I m I m I m θ I (θ) I m I m I m I m Again, the it as θ 0isI m = I m. sin nθ 58. Investigate numericall for several values of n and then guess the value in general. θ 0 θ For n = 3, we have θ sin nθ θ The it as θ 0is3. For n = 5, we have θ sin nθ θ The it as θ 0is 5. We surmise that, in general, θ 0 sin nθ θ = n. b 59. Show numericall that for b = 3, 5 appears to equal ln 3, ln 5, where ln is the natural logarithm. Then 0 make a conjecture (guess) for the value in general and test our conjecture for two additional values of b We have ln

24 98 CHAPTER LIMITS We have ln We conjecture that 0 b = ln b for an positive number b. Here are two additional test cases ( ) We have ln We have ln n 60. Investigate m for (m, n) equal to (, ), (, ), (, 3), and(3, ). Then guess the value of the it in general and check our guess for at least three additional pairs The it as is The it as is The it as is The it as is 3. n For general m and n, we have m = n m

25 SECTION. Limits: A Numerical and Graphical Approach 99 The it as is The it as is The it as is sin(sin ) 6. Find b eperimentation the positive integers k such that 0 k eists. sin(sin ) For k =, we have f () = = f () For k =, we have 0 f () = 0 sin(sin ) =. For k = 3, the it does not eist f () f () Indeed, as 0, f () = sin(sin ) 3, whereas as 0+, f () = sin(sin ) 3. sin(sin ) For k = 4, we have f () = =. For k = 5, the it does not eist f () f () Indeed, as 0, f () = sin(sin ) 5, whereas as 0+, f () = sin(sin ) 5. sin(sin ) For k = 6, we have f () = =. SUMMARY f ()

26 00 CHAPTER LIMITS For k =, the it is 0. For k =, the it is. For odd k >, the it does not eist. For even k >, the it is. 6. Sketch a graph of f () = 8 with a graphing calculator. Observe that f (3) is not defined. 3 (a) Zoom in on the graph to estimate L = f (). 3 (b) Observe that the graph of f () is increasing. Eplain how this implies that Use this to determine L to three decimal places. (a) f (.99999) L f (3.0000) = = 3 (b) It is clear that the graph of f rises as we move to the right. Mathematicall, we ma epress this observation as: whenever u <v, f (u) < f (v). Because it follows that < 3 = 3 f () <3.0000, f (.99999) <L = f () < f (3.0000). 3 With f (.99999) and f (3.0000) , the above inequalit becomes < L < ; hence, to three decimal places, L = The function f () = / / / is defined for = 0. + / (a) Investigate f () and f () numericall (b) Produce a graph of f on a graphing utilit and describe its behavior near = 0. (a) f () (b) As 0, f (), whereas as 0+, f () Show that f () = sin is equal to the slope of a secant line through the origin and the point (, sin ) on the graph of = sin (Figure 8). Use this to give a geometric interpretation of f (). 0

27 SECTION.3 Basic Limit Laws 0 (, sin ) sin FIGURE 8 Graph of = sin. The slope of a secant line through the points (0, 0) and (, sin ) is given b sin 0 = sin.since 0 f () is the slope of the secant line, the it f () is equal to the slope of the tangent line to = sin at = Basic Limit Laws Preinar Questions. State the Sum Law and Quotient Law. Suppose c f () and c g() both eist. The Sum Law states that ( f () + g()) = f () + c c c g(). Provided c g() = 0, the Quotient Law states that f () c g() = c f () c g().. Which of the following is a verbal version of the Product Law? (a) The product of two functions has a it. (b) The it of the product is the product of the its. (c) The product of a it is a product of functions. (d) A it produces a product of functions. The verbal version of the Product Law is (b): The it of the product is the product of the its. 3. Which of the following statements are incorrect (k and c are constants)? (a) c k = c (c) c = c (b) k = k c (d) = c Statements (a) and (d) are incorrect. Because k is constant, statement (a) should read c k = k. Statement (d) should be c = c. 4. Which of the following statements are incorrect? (a) The Product Law does not hold if the it of one of the functions is zero. (b) The Quotient Law does not hold if the it of the denominator is zero. (c) The Quotient Law does not hold if the it of the numerator is zero. Statements (a) and (c) are incorrect. The Product Law remains valid when the it of one or both of the functions is zero, and the Quotient Law remains valid when the it of the numerator is zero. Eercises In Eercises, evaluate the its using the Limit Laws and the following two facts, where c and k are constants:. 9 = = 3. 3 c = c, c k = k

28 0 CHAPTER LIMITS 4 = = (3 + 4) We appl the Laws for Sums, Products, and Constants: (3 + 4) = = = 3( 3) + 4 = = 4 = 4( 3) = ( + 4) 3 8. ( + 4) 3 = 3() = 33. (Here we used the result from Eercise 7.) 9. t 4 (3t 4) ( + 4) = 3 ( ( + 4) = = =. 3 )( ) ( + 4) 3 (3t 4) = 3 t 4 = =. t 4 t 4 t (3 + ) ( ) ( 3 + = 5. (4 + )( ) 5 ) = ( 5)3 + ( 5) = 35. ( (4 + )( ) = / 4 = ( 4 )( ) + / / / / ( ) )( ( ) ) + = 3 0 = 0.. ( ) 3. ( + )( + ) We appl the Sum Law and Product Law: ( ) = = ( ) 4 ( ) 3 = 3 + 4( ) = 3( 4 ) ( 3 ) 4 = =

29 SECTION.3 Basic Limit Laws 03 We appl the Product Law and Sum Law: ( )( ( + )( + ) = = ( + ) )( ( + = ( + )( + ) = 4 )( ) ( + ) ) + 4. ( + )(3 9) ( ) ( ) ( ( ) ) ( + ) 3 9 = = ( + ) ( ) 3() 9 = 3 3 = 9. t 5. t 9 t + t 9 3t 4 6. t 4 t t t 0. 5 t t + = 3t 4 t 4 t + t 9 t t + = t 9 t 9 3 t 4 = t 4 t 4 t + = t 4 t = = ( 3 = = = = 4 =. ) We appl the definition of t, and then the Quotient Law. = t t t = = t t t =. t ( ) 3 + 4( ) = 4 = 5. We appl the definition of =, and then the Quotient and Product Laws (since = ).. 3 ( ) 5 = 5 = 5 ( 5 ) = 5 = 5.

30 04 CHAPTER LIMITS We appl the Sum, Product, and Quotient Laws. The Product Law is applied to the eponentiations = and 3 =. ( ) ( ) 3 ( ) = = z + z. z z + = ( ) 3 3 We appl the Quotient and Sum Laws: z + z z z + = z (z + z) (z + ) = z = z z + z z z + = z z 3. Use the Quotient Law to prove that if c f () eists and is nonzero, then c f () = c f () ( ) = / + + =. Since c f () is nonzero, we can appl the Quotient Law: ( ) = c f () ( ) c ( ) = c f () c f (). 4. Assume that 6 f () = 4 and compute: (a) 6 f () (b) 6 f () (c) 6 f() (a) Appl the Product Law: ( )( ) f 6 () = f () f () = (4)(4) = (b) Since 6 f () = 0, we ma appl the Quotient Law: (c) Appl the Product Law: 6 f () = f () = 4. 6 ( )( ) f() = f () = 6(4) = In Eercises 5 8, evaluate the it assuming that f () = 3 and g() = f ()g() 4 f ()g() = f () g() = 3 = ( f () + 3g()) 4 ( f () + 3g()) = f () + 3 g() = = = 9.

31 g() 7. 4 SECTION.3 Basic Limit Laws 05 Since 4 = 0, we ma appl the Quotient Law, then appling the Product Law (from = ): f () g() 9 g() 4 = 4 g() = 4 ) = 6. 4 f () + f () + 4 3g() 9 = g() 9 = = 4 6 = sin 9. Can the Quotient Law be applied to evaluate? Eplain. 0 sin Theit Quotient Lawcannot be applied to evaluate since = 0. This violates a condition 0 0 of the Quotient Law. Accordingl, the rule cannot be emploed. 30. Show that the Product Law cannot be used to evaluate the ( π/) tan. π/ The it Product Law cannot be applied to evaluate ( π/) tan since tan does not π/ π/ eist (for eample, as π/, tan ). This violates a hpothesis of the Product Law. Accordingl, the rule cannot be emploed. 3. Give an eample where ( f () + g()) eists but neither f () nor g() eists Let f () = / and g() = /. Then ( f () + g()) = 0 = 0 However, f () = / and g() = / do not eist a 3. Assume that the it L a = eists and that a = foralla > 0. Prove that L ab = L a + L b for 0 0 a, b > 0. Hint: (ab) = a (b ) + (a ). Verif numericall that L = L 3 + L 4. Let a, b > 0. Then (ab) a (b ) + (a ) L ab = = 0 0 = a b a = L b + L a = L a + L b. From the table below, we estimate that, to three decimal places, L 3 =.099, L 4 =.386 and L =.485. Thus, L =.485 = = L 3 + L (3 )/ (4 )/ ( )/ ( 33. Use the Limit Laws and the result = c to show that n = c n for all whole numbers n. If ou are c c familiar with induction, give a formal proof b induction.

32 06 CHAPTER LIMITS Correct answers can var. An eample is given: Let P[n] be the proposition : n = c n, and proceed b induction. c P[] is true, as = c. Suppose that P[n] is true, so that n = c n. We must prove P[n + ], thatis,that c c c n+ = c n+. Appling the Product Law and P[n], we see: ( ) c n+ = n = c c n)( = c n c = c n+. c Therefore, P[n] true implies that P[n + ] true. B induction, the it is equal to c n for all n. 34. Etend Eercise 33 to negative integers. Let m be an arbitrar negative integer. m = nfor some positive integer n. Appling the Quotient Law and the results of Eercise 33, we obtain: c m = n = c c n = c = c n c n = c n = c m Further Insights and Challenges 35. Show that if both f () g() and g() eist and g() is nonzero, then f () eists. Hint: Write c c c c f () = ( f () g())/g() and appl the Quotient Law. Given that c f ()g() = L and c g() = M = 0 both eist, observe that f () = c c f ()g() g() also eists. 36. Show that if tg(t) =, then g(t) eists and equals 4. t 3 t 3 = c f ()g() c g() = We are given that t 3 tg(t) =. Since t 3 t = 3 = 0, we ma appl the Quotient Law: h(t) 37. Prove that if = 5, then h(t) = 5. t 3 t t 3 tg(t) tg(t) t 3 g(t) = = t 3 t 3 t t = 3 = 4. t 3 h(t) Given that = 5, observe that t = 3. Now use the Product Law: t 3 t t 3 h(t) = t h(t) = t 3 t 3 t ( t 3 t )( t 3 L M ) h(t) = 3 5 = 5. t f () 38. Assuming that =, which of the following statements is necessaril true? Wh? 0 (a) f (0) = 0 (b) f () = 0 0 f () (a) Given that =, it is not necessaril true that f (0) = 0. A countereample is provided b f () = { 0, = 0 5, = 0. f () (b) Given that =, it is necessaril true that f () = 0. For note that = 0, whence f () = f () ( )( ) = 0 0 f () = 0 =

33 f () 39. Prove that if f () = L = 0 and g() = 0, then does not eist. c c c g() f () Suppose that eists. Then c g() L = c f () = c g() f () g() = c g() c SECTION.3 Basic Limit Laws 07 f () g() = 0 c f () g() = 0. f () But, we were given that L = 0, so we have arrived at a contradiction. Thus, does not eist. c g() 40. Suppose that g(h) = L. h 0 (a) Eplain wh g(ah) = L for an constant a = 0. h 0 (b) If we assume instead that g(h) = L, is it still necessaril true that g(ah) = L? h h (c) Illustrate the conclusions ou reach in (a) and (b) with the function f () =. (a) As h 0, ah 0 as well; hence, if we make the change of variable w = ah,then g(ah) = g(w) = L. h 0 w 0 (b) No. As h, ah a, so we should not epect g(ah) = g(h). h h (c) Let g() =.Then On the other hand, g(h) = 0 and g(ah) = h 0 h 0 h 0 (ah) = 0. g(h) = while g(ah) = h h h 0 (ah) = a, which is equal to the previous it if and onl if a =±. 4. Show that if h 0 g(h) = f (h) h. f (h) h Let g(h) = f (h) h. Then, h 0 using the result from Eercise 40. = L, then h 0 f (ah) h f (ah) h ( = a h 0 = al for all a = 0. Hint: Appl the result of Eercise 40 to ) f (ah) = a g(ah) = al, ah h 0 4. In Section., we mentioned that the number e is characterized b the propert h 0 show that for b > 0, Hint: ApplEercise4to f (h) = eh h Let f (h) = e h. Then b h = ln b h 0 h and a = ln b. e h h =. Assuming this, and, b Eercise 4, h 0 f (h) h =, b h e h ln b = = h 0 h h 0 h h 0 f (h ln b) h = ln b = ln b. 43. There is a Limit Law for composite functions but it is not stated in the tet. Which of the following is the correct statement? Give an intuitive eplanation.

34 08 CHAPTER LIMITS (a) f (g()) = f () c c (b) f (g()) = f (), wherel = c L c g() (c) f (g()) = g(),wherel = c L c f () Use the correct version to evaluate sin(g()), where g() = π 6 (b) is the correct law. Let t = g(). We alread know that, as c, t L. f (g()) = f (t), soitis reasonable to assume that f (g()) = f (t). Since the variable is a dumm variable, we can freel change the c t L name back to, so that the it is f (), as stated in (b). L B (b), sin(g()) = sin() = sin(π/6) = π/6..4 Limits and Continuit Preinar Questions. Which propert of f () = 3 allows us to conclude that 3 = 8? We can conclude that 3 = 8 because the function 3 is continuous at =.. What can be said about f (3) if f is continuous and f () = 3? If f is continuous and 3 f () =,then f (3) =. 3. Suppose that f () <0if is positive and f () >if is negative. Can f be continuous at = 0? Since f () <0when is positive and f () >when is negative, it follows that f () 0 and f () Thus, 0 f () does not eist, so f cannot be continuous at = Is it possible to determine f (7) if f () = 3forall < 7and f is right-continuous at = 7? No. To determine f (7), we need to combine either knowledge of the values of f () for < 7 with left-continuit or knowledge of the values of f () for > 7 with right-continuit. 5. Are the following true or false? If false, state a correct version. (a) f () is continuous at = a if the left- and right-hand its of f () as a eist and are equal. (b) f () is continuous at = a if the left- and right-hand its of f () as a eist and equal f (a). (c) If the left- and right-hand its of f () as a eist, then f has a removable discontinuit at = a. (d) If f () and g() are continuous at = a, then f () + g() is continuous at = a. (e) If f () and g() are continuous at = a, then f ()/g() is continuous at = a. (a) False. The correct statement is f () is continuous at = a if the left- and right-hand its of f () as a eist and equal f (a). (b) True. (c) False. The correct statement is If the left- and right-hand its of f () as a are equal but not equal to f (a), then f has a removable discontinuit at = a. (d) True. (e) False. The correct statement is If f () and g() are continuous at = a and g(a) = 0, then f ()/g() is continuous at = a.

35 SECTION.4 Limits and Continuit 09 Eercises. Find the points of discontinuit of the function shown in Figure 4 and state whether it is left- or right-continuous (or neither) at these points FIGURE 4 The function f is discontinuous at = ; it is left-continuous there. The function f is discontinuous at = 3; it is neither left-continuous nor right-continuous there. The function f is discontinuous at = 5; it is left-continuous there. In Eercises 4, refer to the function f () in Figure FIGURE 5. Find the points of discontinuit of f () and state whether f () is left- or right-continuous (or neither) at these points. The function f is discontinuous at = ; it is left-continuous there. The function f is discontinuous at = 3; it is neither left-continuous nor right-continuous there. The function f is discontinuous at = 5; it is right-continuous there. 3. At which point c does f () have a removable discontinuit? What value should be assigned to f (c) to make f continuous at = c? Because 3 f () eists, the function f has a removable discontinuit at = 3. Assigning f (3) = 4.5 makes f continuous at = Find the point c at which f () has a jump discontinuit but is left-continuous. What value should be assigned to f (c ) to make f right-continuous at = c? The function f has a jump discontinuit at =, but is left-continuous there. Assigning f () = 3 makes f right-continuous at = (but no longer left-continuous). 5. (a) For the function shown in Figure 6, determine the one-sided its at the points of discontinuit. (b) Which of these discontinuities is removable and how should f be redefined to make it continuous at this point? 6 4 FIGURE 6