Solucao impares Calculo 6ed CAP.2

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(1)F. TX.10. 2. LIMITS AND DERIVATIVES. 2.1 The Tangent and Velocity Problems 1. (a) Using S (15> 250), we construct the following table:. w. T. slope = pS T. 5. (5> 694). 694250 515. = 444 = 44=4 10. 10. (10> 444). 444250 1015. = 194 = 38=8 5. 20. (20> 111). 111250 2015. = 139 5 = 27=8. 25. (25> 28). 28250 2515. 30. (30> 0). 0250 3015. (b) Using the values of w that correspond to the points closest to S (w = 10 and w = 20), we have 38=8 + (27=8) = 33=3 2. = 222 = 22=2 10 = 250 15 = 16=6. (c) From the graph, we can estimate the slope of the tangent line at S to be. 3. (a). {. 300 9. = 33=3.. T. pS T. (i). 0=5. (0=5> 0=333333). 0=333333. (ii). 0=9. (0=9> 0=473684). 0=263158. (iii). 0=99. (0=99> 0=497487). 0=251256. (iv). 0=999. (0=999> 0=499750). 0=250125. (v). 1=5. (1=5> 0=6). 0=2. (vi). 1=1. (1=1> 0=523810). 0=238095. (vii). 1=01. (1=01> 0=502488). 0=248756. (viii). 1=001. (1=001> 0=500250). 0=249875. (b) The slope appears to be 14 . (c) | . 1 2. = 14 ({ 1) or | = 14 { + 14 .. 5. (a) | = |(w) = 40w 16w2 . At w = 2, | = 40(2) 16(2)2 = 16. The average velocity between times 2 and 2 + k is. 40(2 + k) 16(2 + k)2 16 24k 16k2 |(2 + k) |(2) = = = 24 16k, if k 6= 0. (2 + k) 2 k k (ii) [2> 2=1]: k = 0=1, yave = 25=6 ft@s (i) [2> 2=5]: k = 0=5, yave = 32 ft@s. yave =. (iii) [2> 2=05]: k = 0=05, yave = 24=8 ft@s. (iv) [2> 2=01]: k = 0=01, yave = 24=16 ft@s. (b) The instantaneous velocity when w = 2 (k approaches 0) is 24 ft@s. 41.

(2) F. 42. ¤. CHAPTER 2. TX.10. LIMITS AND DERIVATIVES. 7. (a) (i) On the interval [1> 3], yave =. 10=7 1=4 9=3 v(3) v(1) = = = 4=65 m@s. 31 2 2. (ii) On the interval [2> 3], yave =. v(3) v(2) 10=7 5=1 = = 5=6 m@s. 32 1. (iii) On the interval [3> 5], yave =. 25=8 10=7 15=1 v(5) v(3) = = = 7=55 m@s. 53 2 2. (iv) On the interval [3> 4], yave =. 17=7 10=7 v(4) v(3) = = 7 m@s. 43 1. (b). Using the points (2> 4) and (5> 23) from the approximate tangent line, the instantaneous velocity at w = 3 is about. 23 4 6=3 m@s. 52. 9. (a) For the curve | = sin(10@{) and the point S (1> 0):. {. T. pS T. {. T. pS T. 2. (2> 0). 0. 0=5. (0=5> 0). 0. 1=5. (1=5> 0=8660). 1=7321. 0=6. (0=6> 0=8660). 2=1651. 1=4. (1=4> 0=4339). 1=0847. 0=7. (0=7> 0=7818). 2=6061. 1=3. (1=3> 0=8230). 2=7433. 0=8. (0=8> 1). 5. 1=2. (1=2> 0=8660). 4=3301. 0=9. (0=9> 0=3420). 1=1. (1=1> 0=2817). 3=4202. 2=8173. As { approaches 1, the slopes do not appear to be approaching any particular value. We see that problems with estimation are caused by the frequent. (b). oscillations of the graph. The tangent is so steep at S that we need to take {-values much closer to 1 in order to get accurate estimates of its slope.. (c) If we choose { = 1=001, then the point T is (1=001> 0=0314) and pS T 31=3794. If { = 0=999, then T is (0=999> 0=0314) and pS T = 31=4422. The average of these slopes is 31=4108. So we estimate that the slope of the tangent line at S is about 31=4..

(3) F. TX.10. SECTION 2.2 THE LIMIT OF A FUNCTION. ¤. 43. 2.2 The Limit of a Function 1. As { approaches 2, i ({) approaches 5. [Or, the values of i ({) can be made as close to 5 as we like by taking { sufciently. close to 2 (but { 6= 2).] Yes, the graph could have a hole at (2> 5) and be dened such that i (2) = 3. 3. (a) lim i ({) = means that the values of i ({) can be made arbitrarily large (as large as we please) by taking { {3. sufciently close to 3 (but not equal to 3). (b) lim i({) = means that the values of i ({) can be made arbitrarily large negative by taking { sufciently close to 4 {4+. through values larger than 4. 5. (a) i ({) approaches 2 as { approaches 1 from the left, so lim i ({) = 2. {1. (b) i ({) approaches 3 as { approaches 1 from the right, so lim i ({) = 3. {1+. (c) lim i ({) does not exist because the limits in part (a) and part (b) are not equal. {1. (d) i ({) approaches 4 as { approaches 5 from the left and from the right, so lim i ({) = 4. {5. (e) i(5) is not dened, so it doesn’t exist. 7. (a) lim j(w) = 1. (b) lim j(w) = 2. w0. w0+. (c) lim j(w) does not exist because the limits in part (a) and part (b) are not equal. w0. (d) lim j(w) = 2. (e) lim j(w) = 0. w2. w2+. (f ) lim j(w) does not exist because the limits in part (d) and part (e) are not equal. w2. (g) j(2) = 1. (h) lim j(w) = 3 w4. 9. (a) lim i ({) = . (b) lim i({) = . (d) lim i ({) = . (e) lim i ({) = . {7 {6. (c) lim i ({) = . {3. {0. {6+. (f ) The equations of the vertical asymptotes are { = 7, { = 3, { = 0, and { = 6. 11. (a) lim i ({) = 1 {0. (b) lim i({) = 0 {0+. (c) lim i ({) does not exist because the limits {0. in part (a) and part (b) are not equal. 13. lim i ({) = 2, {1. lim i ({) = 2, i (1) = 2. {1+. 15. lim i ({) = 4, {3+. lim i ({) = 2, lim i({) = 2,. {3. i (3) = 3, i (2) = 1. {2.

(4) F. 44. ¤. CHAPTER 2. TX.10. LIMITS AND DERIVATIVES. {2 2{ : {2. 17. For i ({) =. 19. For i ({) =. {2. {. i ({). 2=5. {. 0=714286. 1=9. {. i ({). i ({). 1. 0=718282. 1. 0=367879. 0=594885. 0=5. 0=426123. 2=1. 0=677419. 1=95. 0=661017. 2=05. 0=672131. 1=99. 0=665552. 0=1. 0=517092. 0=1. 0=483742. 2=01. 0=667774. 1=995. 0=666110. 0=05. 0=508439. 0=05. 0=491770. 2=005 2=001. 0=667221 0=666778. 1=999. 0=666556. 0=01. 0=501671. 0=01. 0=498337. {2. 21. For i ({) =. {. It appears that lim. { 2{ = 0=¯ 6 = 23 . {2 { 2. {0. {+42 : {. 23. For i ({) =. i ({). {. i ({). h{ 1 { = 0=5 = 12 . {2. {6 1 : {10 1. {. i ({). {. i ({). 1. 0=236068. 1. 0=267949. 0=5. 0=985337. 1=5. 0=183369. 0=5. 0=242641. 0=5. 0=258343. 0=9. 0=719397. 1=1. 0=484119. 0=1. 0=248457. 0=1. 0=251582. 0=95. 0=660186. 1=05. 0=540783. 0=05. 0=249224. 0=05. 0=250786. 0=99. 0=612018. 1=01. 0=588022. 0=01. 0=249844. 0=01. 0=250156. 0=999. 0=601200. 1=001. 0=598800. It appears that lim. {0. lim. {3+. {1. {. 0=5. It appears that lim. 27. lim. i ({). 0=655172. 2. 25.. h{ 1 { : {2. {+42 = 0=25 = 14 . {. It appears that lim. {1. {6 1 = 0=6 = 35 . {10 1. {+2 = since the numerator is negative and the denominator approaches 0 from the positive side as { 3+ . {+3. 2{ = since the numerator is positive and the denominator approaches 0 through positive values as { 1. ({ 1)2. 29. Let w = {2 9. Then as { 3+ , w 0+ , and lim ln({2 9) = lim ln w = by (3). {3+. 31.. lim { csc { = lim. {2 . {2 . w0+. { = since the numerator is positive and the denominator approaches 0 through negative sin {. values as { 2 . 33. (a) i ({) =. 1 . {3 1. {. From these calculations, it seems that lim i ({) = and lim i ({) = .. {1. {1+. i ({). {. i ({). 0=5. 1=14. 1=5. 0=42. 0=9. 3=69. 1=1. 3=02. 0=99. 33=7. 1=01. 33=0. 0=999. 333=7. 1=001. 333=0. 0=9999. 3333=7. 1=0001. 3333=0. 0=99999. 33,333=7. 1=00001. 33,333=3.

(5) F. TX.10. SECTION 2.2 THE LIMIT OF A FUNCTION. ¤. 45. (b) If { is slightly smaller than 1, then {3 1 will be a negative number close to 0, and the reciprocal of {3 1, that is, i ({), will be a negative number with large absolute value. So lim i({) = . {1. If { is slightly larger than 1, then {3 1 will be a small positive number, and its reciprocal, i ({), will be a large positive number. So lim i ({) = . {1+. (c) It appears from the graph of i that lim i ({) = and lim i ({) = .. {1. {1+. 35. (a) Let k({) = (1 + {)1@{ .. { 0=001 0=0001 0=00001 0=000001 0=000001 0=00001 0=0001 0=001. (b). k({) 2=71964 2=71842 2=71830 2=71828 2=71828 2=71827 2=71815 2=71692. It appears that lim (1 + {)1@{ 2=71828, which is approximately h. {0. In Section 3.6 we will see that the value of the limit is exactly h. 37. For i ({) = {2 (2{@1000):. (a). {. i ({). 1 0=8 0=6 0=4 0=2 0=1. 0=998000 0=638259 0=358484 0=158680 0=038851 0=008928. 0=05. 0=001465. It appears that lim i ({) = 0. {0. (b). {. i ({). 0=04 0=02 0=01 0=005 0=003. 0=000572 0=000614 0=000907 0=000978 0=000993. 0=001. 0=001000. It appears that lim i ({) = 0=001. {0.

(6) F. 46. ¤. CHAPTER 2. LIMITS AND DERIVATIVES. 39. No matter how many times we zoom in toward the origin, the graphs of i ({) = sin(@{) appear to consist of almost-vertical. lines. This indicates more and more frequent oscillations as { 0.. There appear to be vertical asymptotes of the curve | = tan(2 sin {) at { ±0=90. 41.. and { ±2=24. To nd the exact equations of these asymptotes, we note that the graph of the tangent function has vertical asymptotes at { = must have 2 sin { =. 2. + q, or equivalently, sin { =. 4. 2. + q. Thus, we. + 2 q. Since. 1 sin { 1, we must have sin { = ± 4 and so { = ± sin1. 4. (corresponding. to { ±0=90). Just as 150 is the reference angle for 30 , sin1 4 is the reference angle for sin1 4 . So { = ± sin1 4 are also equations of vertical asymptotes (corresponding to { ±2=24).. 2.3 Calculating Limits Using the Limit Laws 1. (a) lim [i ({) + 5j({)] = lim i ({) + lim [5j({)] {2. {2. {2. = lim i ({) + 5 lim j({) {2. {2. = 4 + 5(2) = 6 (b) lim [j({)]3 = {2. k. l3 lim j({). {2. [Limit Law 6]. = ( 2)3 = 8 (c) lim. {2. t s i ({) = lim i ({) {2. =. 4=2. [Limit Law 11]. [Limit Law 1] [Limit Law 3].

(7) F. SECTION 2.3. (d) lim. {2. lim [3i ({)] 3i ({) {2 = j({) lim j({). CALCULATING LIMITS USING THE LIMIT LAWS. ¤. 47. [Limit Law 5]. {2. 3 lim i ({) =. {2. [Limit Law 3]. lim j({). {2. =. 3(4) = 6 2. (e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim. {2. j({) , does not exist because the k({). denominator approaches 0 while the numerator approaches a nonzero number. (f ) lim. {2. lim [j({) k({)] j({) k({) {2 = i ({) lim i ({). [Limit Law 5]. {2. lim j({) · lim k({). =. {2. {2. [Limit Law 4]. lim i ({). {2. =. 2 · 0 =0 4. 3. lim (3{4 + 2{2 { + 1) = lim 3{4 + lim 2{2 lim { + lim 1 {2. {2. {2. {2. [Limit Laws 1 and 2]. {2. = 3 lim {4 + 2 lim {2 lim { + lim 1. [3]. = 3(2)4 + 2(2)2 (2) + (1). [9, 8, and 7]. {2. {2. {2. {2. = 48 + 8 + 2 + 1 = 59 5. lim (1 + {8. 3 { ) (2 6{2 + {3 ) = lim (1 + 3 { ) · lim (2 6{2 + {3 ) {8 {8 = lim 1 + lim 3 { · lim 2 6 lim {2 + lim {3 {8 {8 {8 {8 {8 = 1 + 3 8 · 2 6 · 82 + 83 = (3)(130) = 390. 7. lim. {1. 1 + 3{ 1 + 4{2 + 3{4. 3. = %. 3 [6]. lim (1 + 3{). &3. {1. =. [5]. lim (1 + 4{2 + 3{4 ). % =. 1 + 3{ {1 1 + 4{2 + 3{4 lim. {1. &3. lim 1 + 3 lim {. {1. {1 {2 + 3. lim {4 3 3 3 1 4 1 1 + 3(1) = = = = 1 + 4(1)2 + 3(1)4 8 2 8 lim 1 + 4 lim. {1. 9. lim. {4. {1. t 16 {2 = lim (16 {2 ) {4. =. t. lim 16 lim {2. {4. {4. s = 16 (4)2 = 0. [2, 1, and 3]. {1. [11] [2] [7 and 9]. [7, 8, and 9]. [Limit Law 4] [1, 2, and 3] [7, 10, 9].

(8) F. 48. ¤. CHAPTER 2. TX.10. LIMITS AND DERIVATIVES. 11. lim. {2 + { 6 ({ + 3)({ 2) = lim = lim ({ + 3) = 2 + 3 = 5 {2 {2 {2 {2. 13. lim. {2 { + 6 does not exist since { 2 0 but {2 { + 6 8 as { 2. {2. 15. lim. w2 9 (w + 3)(w 3) w3 3 3 6 6 = lim = lim = = = 2w2 + 7w + 3 w3 (2w + 1)(w + 3) w3 2w + 1 2(3) + 1 5 5. {2. {2. w3. (4 + k)2 16 (16 + 8k + k2 ) 16 k(8 + k) 8k + k2 = lim = lim = lim = lim (8 + k) = 8 + 0 = 8 k0 k0 k0 k0 k k k k. 17. lim. k0. 19. By the formula for the sum of cubes, we have. lim. {2. 21. lim. w9. {+2 {+2 1 1 1 = lim = lim = = . {3 + 8 {2 ({ + 2)({2 2{ + 4) {2 {2 2{ + 4 4+4+4 12. 3+ w 3 w 9w = lim = lim 3 + w = 3 + 9 = 6 w9 3 w w9 3 w. {+23 {+23 {+2+3 ({ + 2) 9 = lim · = lim 23. lim {7 {7 {7 {7 { + 2 + 3 {7 ({ 7) { + 2 + 3 = lim. {7. 25.. {7 1 1 1 = lim = = {7 6 ({ 7) { + 2 + 3 {+2+3 9+3. 1 {+4 1 + 1 1 {+4 1 { = lim 4{ = lim = lim = = lim 4 {4 4 + { {4 4 + { {4 4{(4 + {) {4 4{ 4(4) 16. (4 { )(4 + { ) 4 { 16 { 27. lim = lim = lim {16 16{ {2 {16 (16{ {2 )(4 + { ) {16 {(16 {)(4 + { ) = lim. {16. 29. lim. w0. 1 1 w w 1+w. 1 1 1 1 = = = 16(8) 128 {(4 + { ) 16 4 + 16. = lim. w0. 1 1+w 1+ 1+w 1 1+w w = lim = lim w0 w0 w w 1+w w w+1 1+ 1+w 1+w 1+ 1+w. 1 1 1 = = = lim w0 2 1+w 1+ 1+w 1+0 1+ 1+0 (b). 31. (a). { 2 lim 3 1 + 3{ 1. {0. {. i ({). 0=001 0=0001 0=00001 0=000001 0=000001 0=00001 0=0001 0=001. 0=6661663 0=6666167 0=6666617 0=6666662 0=6666672 0=6666717 0=6667167 0=6671663. The limit appears to be. 2 . 3.

(9) F. TX.10 (c) lim. {0. SECTION 2.3. CALCULATING LIMITS USING THE LIMIT LAWS. ¤. 49. { 1 + 3{ + 1 { 1 + 3{ + 1 1 + 3{ + 1 { = lim · = lim {0 {0 (1 + 3{) 1 3{ 1 + 3{ 1 1 + 3{ + 1 1 lim 1 + 3{ + 1 3 {0 1 t lim (1 + 3{) + lim 1 = {0 {0 3 1 t lim 1 + 3 lim { + 1 = {0 {0 3. [Limit Law 3]. =. [1 and 11] [1, 3, and 7]. 1 1+3·0+1 3 2 1 = (1 + 1) = 3 3. [7 and 8]. =. 33. Let i({) = {2 , j({) = {2 cos 20{ and k({) = {2 . Then. 1 cos 20{ 1 {2 {2 cos 20{ {2. i ({) j({) k({).. So since lim i ({) = lim k({) = 0, by the Squeeze Theorem we have {0. {0. lim j({) = 0.. {0. . . 35. We have lim (4{ 9) = 4(4) 9 = 7 and lim {2 4{ + 7 = 42 4(4) + 7 = 7. Since 4{ 9 i ({) {2 4{ + 7 {4. {4. for { 0, lim i ({) = 7 by the Squeeze Theorem. {4. 37. 1 cos(2@{) 1. {4 {4 cos(2@{) {4 . Since lim {4 = 0 and lim {4 = 0, we have {0. lim {4 cos(2@{) = 0 by the Squeeze Theorem.. {0. {0. + 39. |{ 3| =. {3. if { 3 0. ({ 3). if { 3 ? 0. + =. {3. if { 3. 3{. if { ? 3. Thus, lim (2{ + |{ 3|) = lim (2{ + { 3) = lim (3{ 3) = 3(3) 3 = 6 and {3+. {3+. {3+. lim (2{ + |{ 3|) = lim (2{ + 3 {) = lim ({ + 3) = 3 + 3 = 6. Since the left and right limits are equal,. {3. {3. {3. lim (2{ + |{ 3|) = 6.. {3. . . . . . . 41. 2{3 {2 = {2 (2{ 1) = {2 · |2{ 1| = {2 |2{ 1|. + |2{ 1| =. 2{ 1. if 2{ 1 0. (2{ 1). if 2{ 1 ? 0. + =. 2{ 1. if { 0=5. (2{ 1). if { ? 0=5. So 2{3 {2 = {2 [(2{ 1)] for { ? 0=5. Thus,. lim. {0=5. 2{ 1 2{ 1 1 1 1 = lim = lim = 4. = = |2{3 {2 | {0=5 {2 [(2{ 1)] {0=5 {2 (0=5)2 0=25.

(10) F. 50. ¤. CHAPTER 2. TX.10. LIMITS AND DERIVATIVES. 43. Since |{| = { for { ? 0, we have lim. {0. 1 1 { |{|. . = lim. {0. 1 1 { {. = lim. {0. 2 , which does not exist since the {. denominator approaches 0 and the numerator does not. (b) (i) Since sgn { = 1 for { A 0, lim sgn { = lim 1 = 1.. 45. (a). {0+. {0+. (ii) Since sgn { = 1 for { ? 0, lim sgn { = lim 1 = 1. {0. {0. (iii) Since lim sgn { 6= lim sgn {, lim sgn { does not exist. {0. {0. {0+. (iv) Since |sgn {| = 1 for { 6= 0, lim |sgn {| = lim 1 = 1. {0. 47. (a) (i) lim I ({) = lim {1+. {1+. (ii) lim I ({) = lim {1. {1. {2 1 {2 1 = lim = lim ({ + 1) = 2 |{ 1| {1+ { 1 {1+. {0. (c). {2 1 {2 1 = lim = lim ({ + 1) = 2 |{ 1| {1 ({ 1) {1. (b) No, lim I ({) does not exist since lim I ({) 6= lim I ({). {1. {1+. 49. (a) (i) [[{]] = 2 for 2 { ? 1, so. (ii) [[{]] = 3 for 3 { ? 2, so. {1. lim [[{]] =. {2+. lim [[{]] =. {2. lim (2) = 2. {2+. lim (3) = 3.. {2. The right and left limits are different, so lim [[{]] does not exist. {2. (iii) [[{]] = 3 for 3 { ? 2, so. lim [[{]] =. {2=4. lim (3) = 3.. {2=4. (b) (i) [[{]] = q 1 for q 1 { ? q, so lim [[{]] = lim (q 1) = q 1. {q. {q. (ii) [[{]] = q for q { ? q + 1, so lim [[{]] = lim q = q. {q+. {q+. (c) lim [[{]] exists d is not an integer. {d. 51. The graph of i ({) = [[{]] + [[{]] is the same as the graph of j({) = 1 with holes at each integer, since i (d) = 0 for any. integer d. Thus, lim i ({) = 1 and lim i ({) = 1, so lim i ({) = 1. However, {2. {2+. {2. i (2) = [[2]] + [[2]] = 2 + (2) = 0, so lim i ({) 6= i (2). {2. 53. Since s({) is a polynomial, s({) = d0 + d1 { + d2 {2 + · · · + dq {q . Thus, by the Limit Laws,. lim s({) = lim d0 + d1 { + d2 {2 + · · · + dq {q = d0 + d1 lim { + d2 lim {2 + · · · + dq lim {q. {d. {d. {d. = d0 + d1 d + d2 d2 + · · · + dq dq = s(d) Thus, for any polynomial s, lim s({) = s(d). {d. {d. {d.

(11) F. TX.10. SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT. ¤. 51. i ({) 8 i({) 8 · ({ 1) = lim · lim ({ 1) = 10 · 0 = 0. 55. lim [i({) 8] = lim {1 {1 {1 {1 {1 {1 . Thus, lim i ({) = lim {[i ({) 8] + 8} = lim [i ({) 8] + lim 8 = 0 + 8 = 8. {1. {1. Note: The value of lim. {1. {1. {1. i ({) 8 i ({) 8 does not affect the answer since it’s multiplied by 0. What’s important is that lim {1 { 1 {1. exists. 57. Observe that 0 i ({) {2 for all {, and lim 0 = 0 = lim {2 . So, by the Squeeze Theorem, lim i ({) = 0. {0. {0. {0. 59. Let i({) = K({) and j({) = 1 K({), where K is the Heaviside function dened in Exercise 1.3.57.. Thus, either i or j is 0 for any value of {. Then lim i ({) and lim j({) do not exist, but lim [i ({)j({)] = lim 0 = 0. {0. {0. {0. {0. 61. Since the denominator approaches 0 as { 2, the limit will exist only if the numerator also approaches. 0 as { 2. In order for this to happen, we need lim. {2. 2 3{ + d{ + d + 3 = 0. 3(2)2 + d(2) + d + 3 = 0 12 2d + d + 3 = 0 d = 15. With d = 15, the limit becomes lim. {2. 3(2 + 3) 3({ + 2)({ + 3) 3({ + 3) 3 3{2 + 15{ + 18 = lim = lim = = = 1. {2 ({ 1)({ + 2) {2 { 1 {2 + { 2 2 1 3. 2.4 The Precise Definition of a Limit . . 1. On the left side of { = 2, we need |{ 2| ? 10 2 = 7. 4 . 7. On the right side, we need |{ 2| ? 10 2 = 43 . For both of 3. these conditions to be satised at once, we need the more restrictive of the two to hold, that is, |{ 2| ? 47 . So we can choose = 47 , or any smaller positive number. 3. The leftmost question mark is the solution of. { = 1=6 and the rightmost, { = 2=4. So the values are 1=62 = 2=56 and. 2=42 = 5=76. On the left side, we need |{ 4| ? |2=56 4| = 1=44. On the right side, we need |{ 4| ? |5=76 4| = 1=76. To satisfy both conditions, we need the more restrictive condition to hold — namely, |{ 4| ? 1=44. Thus, we can choose = 1=44, or any smaller positive number. 5.. From the graph, we nd that tan { = 0=8 when { 0=675, so 4. 1 0=675 1 . when { 0=876, so. 4. 4. 0=675 0=1106. Also, tan { = 1=2. + 2 0=876 2 = 0=876 . 4. 0=0906.. Thus, we choose = 0=0906 (or any smaller positive number) since this is the smaller of 1 and 2 ..

(12) F. 52. ¤. CHAPTER 2 LIMITS AND DERIVATIVES. . . . 7. For % = 1, the denition of a limit requires that we nd such that 4 + { 3{3 2 ? 1. 1 ? 4 + { 3{3 ? 3. whenever 0 ? |{ 1| ? . If we plot the graphs of | = 1, | = 4 + { 3{3 and | = 3 on the same screen, we see that we need 0=86 { 1=11. So since |1 0=86| = 0=14 and |1 1=11| = 0=11, we choose = 0=11 (or any smaller positive number). For % = 0=1, we must nd such that 4 + { 3{3 2 ? 0=1 1=9 ? 4 + { 3{3 ? 2=1 whenever 0 ? |{ 1| ? . From the graph, we see that we need 0=988 { 1=012. So since |1 0=988| = 0=012 and |1 1=012| = 0=012, we choose = 0=012 (or any smaller positive number) for the inequality to hold.. From the graph, we nd that | = tan2 { = 1000 when { 1=539 and. 9. (a). { 1=602 for { near. . 2. Thus, we get 1=602 . 2. 0=031 for. P = 1000.. From the graph, we nd that | = tan2 { = 10,000 when { 1=561 and. (b). { 1=581 for { near. . 2. Thus, we get 1=581 . 2. 0=010 for. P = 10,000.. 11. (a) D = u2 and D = 1000 cm2. u2 = 1000 u2 =. 1000 . u=. t. 1000 . (u A 0). 17=8412 cm.. (b) |D 1000| 5 5 u2 1000 5 1000 5 u2 1000 + 5 t t t t t t 995 1000 u 1005 17=7966 u 17=8858. 995 0=04466 and 1005 1000 0=04455. So if the machinist gets the radius within 0=0445 cm of 17=8412, the area will be within 5 cm2 of 1000. (c) { is the radius, i ({) is the area, d is the target radius given in part (a), O is the target area (1000), % is the tolerance in the area (5), and is the tolerance in the radius given in part (b). 13. (a) |4{ 8| = 4 |{ 2| ? 0=1. |{ 2| ?. 0=1 0=1 , so = = 0=025. 4 4. (b) |4{ 8| = 4 |{ 2| ? 0=01 |{ 2| ?. 0=01 0=01 , so = = 0=0025. 4 4.

(13) F. TX.10. SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT. ¤. 53. 15. Given % A 0, we need A 0 such that if 0 ? |{ 1| ? , then. |(2{ + 3) 5| ? %. But |(2{ + 3) 5| ? %. |2{ 2| ? % 2 |{ 1| ? % |{ 1| ? %@2. So if we choose = %@2, then 0 ? |{ 1| ? . . |(2{ + 3) 5| ? %. Thus, lim (2{ + 3) = 5 by the denition of a limit. {1. 17. Given % A 0, we need A 0 such that if 0 ? |{ (3)| ? , then. |(1 4{) 13| ? %. But |(1 4{) 13| ? %. |4{ 12| ? % |4| |{ + 3| ? % |{ (3)| ? %@4. So if we choose = %@4, then 0 ? |{ (3)| ? . . |(1 4{) 13| ? %. Thus, lim (1 4{) = 13 by the denition of {3. x. a limit. 3 ? %. 5 5. {. 19. Given % A 0, we need A 0 such that if 0 ? |{ 3| ? , then . So choose = 5%. Then 0 ? |{ 3| ? of a limit, lim. {3. |{ 3| ? 5% . 1 5. |{ 3| ? % |{ 3| ? 5%.. { 3 |{ 3| ? % ? %. By the denition 5 5 5. { 3 = . 5 5 2 { + { 6. 21. Given % A 0, we need A 0 such that if 0 ? |{ 2| ? , then . ({ + 3)({ 2) ?% 5 {2 Then 0 ? |{ 2| ? . {2. 5 ? %. |{ + 3 5| ? % [{ 6= 2] |{ 2| ? %. So choose = %.. ({ + 3)({ 2) |{ 2| ? % |{ + 3 5| ? % 5 ? % [{ 6= 2] {2. 2 2 { + { 6 ? %. By the denition of a limit, lim { + { 6 = 5. 5 {2 {2 {2 23. Given % A 0, we need A 0 such that if 0 ? |{ d| ? , then |{ d| ? %. So = % will work.. . . 25. Given % A 0, we need A 0 such that if 0 ? |{ 0| ? , then {2 0 ? %. Then 0 ? |{ 0| ? . {2 ? %. |{| ?. %. Take = %.. {2 0 ? %. Thus, lim {2 = 0 by the denition of a limit. {0. . . . . 27. Given % A 0, we need A 0 such that if 0 ? |{ 0| ? , then |{| 0 ? %. But |{| = |{|. So this is true if we pick = %.. Thus, lim |{| = 0 by the denition of a limit. {0.

(14) F. 54. ¤. TX.10. CHAPTER 2 LIMITS AND DERIVATIVES. . . . {2 4{ + 4 ? %. . |{ 2| ? % ({ 2)2 ? %. Thus, . 29. Given % A 0, we need A 0 such that if 0 ? |{ 2| ? , then {2 4{ + 5 1 ? %. ({ 2)2 ? %. So take = %. Then 0 ? |{ 2| ? lim {2 4{ + 5 = 1 by the denition of a limit. {2. . . . 31. Given % A 0, we need A 0 such that if 0 ? |{ (2)| ? , then {2 1 3 ? % or upon simplifying we need. 2 { 4 ? % whenever 0 ? |{ + 2| ? . Notice that if |{ + 2| ? 1, then 1 ? { + 2 ? 1 5 ? { 2 ? 3 |{ 2| ? 5. So take = min {%@5> 1}. Then 0 ? |{ + 2| ? |{ 2| ? 5 and |{ + 2| ? %@5, so 2 { 1 3 = |({ + 2)({ 2)| = |{ + 2| |{ 2| ? (%@5)(5) = %. Thus, by the denition of a limit, lim ({2 1) = 3. {2.

(15) . If 0 ? |{ 3| ? , then |{ 3| ? 2 2 ? { 3 ? 2 4 ? { + 3 ? 8 |{ + 3| ? 8. Also |{ 3| ? 8% , so {2 9 = |{ + 3| |{ 3| ? 8 · 8% = %. Thus, lim {2 = 9.. 33. Given % A 0, we let = min 2>. % 8. {3. 35. (a) The points of intersection in the graph are ({1 > 2=6) and ({2 > 3=4). with {1 0=891 and {2 1=093. Thus, we can take to be the smaller of 1 {1 and {2 1. So = {2 1 0=093.. (b) Solving {3 + { + 1 = 3 + % gives us two nonreal complex roots and one real root, which is 2@3 216 + 108% + 12 336 + 324% + 81%2 12 {(%) = 1@3 . Thus, = {(%) 1. 2 6 216 + 108% + 12 336 + 324% + 81% (c) If % = 0=4, then {(%) 1=093 272 342 and = {(%) 1 0=093, which agrees with our answer in part (a). 37. 1. Guessing a value for . Given % A 0, we must nd A 0 such that | { d| ? % whenever 0 ? |{ d| ? . But. |{ d| ? % (from the hint). Now if we can nd a positive constant F such that { + d A F then | { d| = {+ d |{ d| |{ d| ? ? %, and we take |{ d| ? F%. We can nd this number by restricting { to lie in some interval F {+ d t { + d A 12 d + d, and so centered at d. If |{ d| ? 12 d, then 12 d ? { d ? 12 d 12 d ? { ? 32 d F=. t. 1 2d. = min. q. +. t 1 d is a suitable choice for the constant. So |{ d| ? d %. This suggests that we let 2d +. 1 d> 2. t. 1 d 2. +. 2. Showing that works |{ d| ? 12 d . r d % . Given % A 0, we let = min. q. 1 d> 2. t. 1 d 2. +. r d % . If 0 ? |{ d| ? , then. t t 1 { + d A 12 d + d (as in part 1). Also |{ d| ? d + d %, so 2. s d@2 + d % |{ d| = %. Therefore, lim { = d by the denition of a limit. ? s | { d| = {d {+ d d@2 + d.

(16) F. TX.10 39. Suppose that lim i ({) = O. Given % = {0. 1 2,. SECTION 2.5 CONTINUITY. there exists A 0 such that 0 ? |{| ? . ¤. 55. |i({) O| ? 12 . Take any rational. number u with 0 ? |u| ? . Then i (u) = 0, so |0 O| ? 12 , so O |O| ? 12 . Now take any irrational number v with 0 ? |v| ? . Then i (v) = 1, so |1 O| ? 12 . Hence, 1 O ? 12 , so O A 12 . This contradicts O ? 12 , so lim i({) does not {0. exist. 41.. 1 1 A 10,000 ({ + 3)4 ? ({ + 3)4 10,000. 1. |{ + 3| ? 4 10,000. |{ (3)| ?. 1 10. 43. Given P ? 0 we need A 0 so that ln { ? P whenever 0 ? { ? ; that is, { = hln { ? hP whenever 0 ? { ? . This. suggests that we take = hP . If 0 ? { ? hP , then ln { ? ln hP = P . By the denition of a limit, lim ln { = . {0+. 2.5 Continuity 1. From Denition 1, lim i ({) = i (4). {4. 3. (a) The following are the numbers at which i is discontinuous and the type of discontinuity at that number: 4 (removable),. 2 ( jump), 2 ( jump), 4 (innite). (b) i is continuous from the left at 2 since. lim i ({) = i (2). i is continuous from the right at 2 and 4 since. {2. lim i({) = i (2) and lim i ({) = i (4). It is continuous from neither side at 4 since i (4) is undened.. {2+. {4+. 5. The graph of | = i ({) must have a discontinuity at { = 3 and must show that lim i({) = i (3). {3. (b) There are discontinuities at times w = 1, 2, 3, and 4. A person. 7. (a). parking in the lot would want to keep in mind that the charge will jump at the beginning of each hour.. 9. Since i and j are continuous functions,. lim [2i ({) j({)] = 2 lim i ({) lim j({). {3. {3. {3. = 2i(3) j(3). [by Limit Laws 2 and 3] [by continuity of i and j at { = 3]. = 2 · 5 j(3) = 10 j(3) Since it is given that lim [2i ({) j({)] = 4, we have 10 j(3) = 4, so j(3) = 6. {3.

(17) F. 56. ¤. CHAPTER 2. 11. lim i ({) = lim {1. TX.10. LIMITS AND DERIVATIVES. {1. 4 { + 2{3 =. . lim { + 2 lim {3. {1. 4. {1. 4 = 1 + 2(1)3 = (3)4 = 81 = i(1).. By the denition of continuity, i is continuous at d = 1. 13. For d A 2, we have. lim (2{ + 3) 2{ + 3 {d = {d { 2 lim ({ 2). lim i ({) = lim. {d. [Limit Law 5]. {d. 2 lim { + lim 3 =. {d. {d. lim { lim 2. {d. =. [1, 2, and 3]. {d. 2d + 3 d2. [7 and 8]. = i (d) Thus, i is continuous at { = d for every d in (2> ); that is, i is continuous on (2> ). 15. i ({) = ln |{ 2| is discontinuous at 2 since i(2) = ln 0 is not dened.. + 17. i ({) =. h{. if { ? 0. 2. if { 0. {. The left-hand limit of i at d = 0 is lim i ({) = lim h{ = 1. The {0. {0. right-hand limit of i at d = 0 is lim i ({) = lim {2 = 0. Since these {0+. {0+. limits are not equal, lim i ({) does not exist and i is discontinuous at 0= {0. ; cos { A ? 19. i ({) = 0 A = 1 {2. if { ? 0 if { = 0 if { A 0. lim i ({) = 1, but i (0) = 0 6= 1, so i is discontinuous at 0.. {0. { is a rational function. So by Theorem 5 (or Theorem 7), I is continuous at every number in its domain, {2 + 5{ + 6.

(18) { | {2 + 5{ + 6 6= 0 = {{ | ({ + 3)({ + 2) 6= 0} = {{ | { 6= 3, 2} or (> 3) (3> 2) (2> ).. 21. I ({) =. 23. By Theorem 5, the polynomials {2 and 2{ 1 are continuous on (> ). By Theorem 7, the root function. continuous on [0> ). By Theorem 9, the composite function 2{ 1 is continuous on its domain, [ 12 > ). By part 1 of Theorem 4, the sum U({) = {2 + 2{ 1 is continuous on [ 12 > ).. { is.

(19) F. TX.10. SECTION 2.5 CONTINUITY. ¤. 57. 25. By Theorem 7, the exponential function h5w and the trigonometric function cos 2w are continuous on (> ).. By part 4 of Theorem 4, O(w) = h5w cos 2w is continuous on (> ). 27. By Theorem 5, the polynomial w4 1 is continuous on (> ). By Theorem 7, ln { is continuous on its domain, (0> ).. By Theorem 9, ln w4 1 is continuous on its domain, which is.

(20)

(21) w | w4 1 A 0 = w | w4 A 1 = {w | |w| A 1} = (> 1) (1> ) 29. The function | =. 1 is discontinuous at { = 0 because the 1 + h1@{. left- and right-hand limits at { = 0 are different.. 31. Because we are dealing with root functions, 5 +. { is continuous on [0> ), { + 5 is continuous on [5> ), so the. 5+ { is continuous on [0> ). Since i is continuous at { = 4, lim i ({) = i (4) = 73 . quotient i ({) = {4 5+{ 33. Because {2 { is continuous on R, the composite function i ({) = h{. 2 {. is continuous on R, so. lim i ({) = i (1) = h1 1 = h0 = 1.. {1. + 35. i ({) =. {2 {. if { ? 1 if { 1. By Theorem 5, since i ({) equals the polynomial {2 on (> 1), i is continuous on (> 1). By Theorem 7, since i ({) equals the root function { on (1> )> i is continuous on (1> ). At { = 1, lim i({) = lim {2 = 1 and {1. {1. lim i ({) = lim { = 1. Thus, lim i({) exists and equals 1. Also, i (1) = 1 = 1. Thus, i is continuous at { = 1.. {1+. {1. {1+. We conclude that i is continuous on (> ). ; A 1 + {2 A ? 37. i ({) = 2 { A A = ({ 2)2. if { 0 if 0 ? { 2 if { A 2. i is continuous on (> 0), (0> 2), and (2> ) since it is a polynomial on each of these intervals. Now lim i ({) = lim (1 + {2 ) = 1 and lim i ({) = lim (2 {) = 2> so i is {0. {0. {0+. {0+. discontinuous at 0. Since i (0) = 1, i is continuous from the left at 0= Also, lim i ({) = lim (2 {) = 0> {2. {2. lim i ({) = lim ({ 2)2 = 0, and i (2) = 0, so i is continuous at 2= The only number at which i is discontinuous is 0.. {2+. {2+.

(22) F. 58. ¤. CHAPTER 2. TX.10. LIMITS AND DERIVATIVES. ; {+2 A A ? 39. i ({) = h{ A A = 2{. if { ? 0 if 0 { 1 if { A 1. i is continuous on (> 0) and (1> ) since on each of these intervals it is a polynomial; it is continuous on (0> 1) since it is an exponential. Now lim i ({) = lim ({ + 2) = 2 and lim i ({) = lim h{ = 1, so i is discontinuous at 0. Since i (0) = 1, i is {0. {0. {0+. {0+. continuous from the right at 0. Also lim i ({) = lim h{ = h and lim i ({) = lim (2 {) = 1, so i is discontinuous {1. {1. {1+. {1+. at 1. Since i (1) = h, i is continuous from the left at 1. + 41. i ({) =. f{2 + 2{ 3. { f{. if { ? 2 if { 2. i is continuous on (> 2) and (2> ). Now lim i ({) = lim {2. lim i ({) = lim. {2+. {2+. {2. 2 f{ + 2{ = 4f + 4 and. 3 { f{ = 8 2f. So i is continuous 4f + 4 = 8 2f 6f = 4 f = 23 . Thus, for i. to be continuous on (> ), f = 23 . 43. (a) i ({) =. ({2 + 1)({2 1) ({2 + 1)({ + 1)({ 1) {4 1 = = = ({2 + 1)({ + 1) [or {3 + {2 + { + 1] {1 {1 {1. for { 6= 1. The discontinuity is removable and j({) = {3 + {2 + { + 1 agrees with i for { 6= 1 and is continuous on R. (b) i ({) =. {({2 { 2) {({ 2)({ + 1) {3 {2 2{ = = = {({ + 1) [or {2 + {] for { 6= 2. The discontinuity {2 {2 {2. is removable and j({) = {2 + { agrees with i for { 6= 2 and is continuous on R. (c) lim i ({) = lim [[sin {]] = lim 0 = 0 and lim i ({) = lim [[sin {]] = lim (1) = 1, so lim i ({) does not { . { . { . { +. { +. { +. {. exist. The discontinuity at { = is a jump discontinuity. 45. i ({) = {2 + 10 sin { is continuous on the interval [31> 32], i (31) 957, and i(32) 1030. Since 957 ? 1000 ? 1030,. there is a number c in (31> 32) such that i (f) = 1000 by the Intermediate Value Theorem. Note: There is also a number c in (32> 31) such that i (f) = 1000= 47. i ({) = {4 + { 3 is continuous on the interval [1> 2]> i (1) = 1, and i(2) = 15. Since 1 ? 0 ? 15, there is a number f. in (1> 2) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation {4 + { 3 = 0 in the interval (1> 2)= 49. i ({) = cos { { is continuous on the interval [0> 1], i (0) = 1, and i(1) = cos 1 1 0=46. Since 0=46 ? 0 ? 1, there. is a number f in (0> 1) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos { { = 0, or cos { = {, in the interval (0> 1)..

(23) F. TX.10. SECTION 2.5 CONTINUITY. ¤. 59. 51. (a) i ({) = cos { {3 is continuous on the interval [0> 1], i (0) = 1 A 0, and i (1) = cos 1 1 0=46 ? 0. Since. 1 A 0 A 0=46, there is a number f in (0> 1) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos { {3 = 0, or cos { = {3 , in the interval (0> 1). (b) i (0=86) 0=016 A 0 and i (0=87) 0=014 ? 0, so there is a root between 0=86 and 0=87, that is, in the interval (0=86> 0=87). 53. (a) Let i ({) = 100h{@100 0=01{2 = Then i (0) = 100 A 0 and. i (100) = 100h1 100 63=2 ? 0. So by the Intermediate Value Theorem, there is a number f in (0> 100) such that i (f) = 0. This implies that 100hf@100 = 0=01f2 . (b) Using the intersect feature of the graphing device, we nd that the root of the equation is { = 70=347, correct to three decimal places. 55. () If i is continuous at d, then by Theorem 8 with j(k) = d + k, we have. lim i (d + k) = i lim (d + k) = i (d).. k0. k0. (

(24) ) Let % A 0. Since lim i(d + k) = i (d), there exists A 0 such that 0 ? |k| ? k0. . |i (d + k) i (d)| ? %. So if 0 ? |{ d| ? , then |i ({) i (d)| = |i (d + ({ d)) i (d)| ? %. Thus, lim i ({) = i (d) and so i is continuous at d. {d. 57. As in the previous exercise, we must show that lim cos(d + k) = cos d to prove that the cosine function is continuous. k0. lim cos(d + k) = lim (cos d cos k sin d sin k) = lim (cos d cos k) lim (sin d sin k). k0. k0. . = + 59. i ({) =. k0. k0. lim cos d lim cos k lim sin d lim sin k = (cos d)(1) (sin d)(0) = cos d. k0. 0 if { is rational 1 if { is irrational. k0. k0. k0. is continuous nowhere. For, given any number d and any A 0, the interval (d > d + ). contains both innitely many rational and innitely many irrational numbers. Since i (d) = 0 or 1, there are innitely many numbers { with 0 ? |{ d| ? and |i ({) i (d)| = 1. Thus, lim i({) 6= i (d). [In fact, lim i ({) does not even exist.] {d. 61. If there is such a number, it satises the equation {3 + 1 = {. {d. {3 { + 1 = 0. Let the left-hand side of this equation be. called i ({). Now i (2) = 5 ? 0, and i (1) = 1 A 0. Note also that i ({) is a polynomial, and thus continuous. So by the Intermediate Value Theorem, there is a number f between 2 and 1 such that i(f) = 0, so that f = f3 + 1. 63. i ({) = {4 sin(1@{) is continuous on (> 0) (0> ) since it is the product of a polynomial and a composite of a. trigonometric function and a rational function. Now since 1 sin(1@{) 1, we have {4 {4 sin(1@{) {4 . Because.

(25) F. 60. ¤. CHAPTER 2 LIMITS AND DERIVATIVES. TX.10. lim ({4 ) = 0 and lim {4 = 0, the Squeeze Theorem gives us lim ({4 sin(1@{)) = 0, which equals i(0). Thus, i is. {0. {0. {0. continuous at 0 and, hence, on (> ). 65. Dene x(w) to be the monk’s distance from the monastery, as a function of time, on the rst day, and dene g(w) to be his. distance from the monastery, as a function of time, on the second day. Let G be the distance from the monastery to the top of the mountain. From the given information we know that x(0) = 0, x(12) = G, g(0) = G and g(12) = 0. Now consider the function x g, which is clearly continuous. We calculate that (x g)(0) = G and (x g)(12) = G. So by the Intermediate Value Theorem, there must be some time w0 between 0 and 12 such that (x g)(w0 ) = 0 x(w0 ) = g(w0 ). So at time w0 after 7:00 AM, the monk will be at the same place on both days.. 2.6 Limits at Infinity; Horizontal Asymptotes 1. (a) As { becomes large, the values of i ({) approach 5.. (b) As { becomes large negative, the values of i ({) approach 3. 3. (a) lim i ({) = {2. (d) lim i ({) = 1 {. 5. i (0) = 0,. i(1) = 1,. lim i ({) = 0,. {. (b). lim i ({) = . {1. (e) lim i ({) = 2 {. 7. lim i ({) = , {2. lim i ({) = 0,. {. i is odd. lim i ({) = . {0. (c). lim i ({) = . {1+. (f ) Vertical: { = 1, { = 2; Horizontal: | = 1, | = 2 lim i({) = ,. {. lim i ({) = ,. {0+. 9. i (0) = 3,. lim i ({) = 4,. {0. lim i ({) = 2,. {0+. lim i ({) = ,. {. lim i ({) = ,. {4+. lim i ({) = ,. {4. lim i ({) = 3. {. 11. If i ({) = {2@2{ , then a calculator gives i(0) = 0, i (1) = 0=5, i(2) = 1, i (3) = 1=125, i (4) = 1, i (5) = 0=78125,. i (6) = 0=5625, i (7) = 0=3828125, i (8) = 0=25, i (9) = 0=158203125, i(10) = 0=09765625, i (20) 0=00038147, i (50) 2=2204 × 1012 , i (100) 7=8886 × 1027 . It appears that lim {2@2{ = 0. {.

(26) F. TX.10 SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 13. lim. {. 3{2 { + 4 (3{2 { + 4)@{2 = lim 2 { 2{ + 5{ 8 (2{2 + 5{ 8)@{2. ¤. [divide both the numerator and denominator by {2 (the highest power of { thatappears in the denominator)]. lim (3 1@{ + 4@{2 ). =. {. [Limit Law 5]. lim (2 + 5@{ 8@{2 ). {. lim 3 lim (1@{) + lim (4@{2 ). =. {. {. {. {. {. {. [Limit Laws 1 and 2]. lim 2 + lim (5@{) lim (8@{2 ) 3 lim (1@{) + 4 lim (1@{2 ). =. {. {. 15. lim. {. 17.. lim. [Limit Laws 7 and 3]. 2 + 5 lim (1@{) 8 lim (1@{2 ) {. {. =. 3 0 + 4(0) 2 + 5(0) 8(0). =. 3 2. [Theorem 5 of Section 2.5]. lim (1@{) lim (1@{) 1 1@{ 0 0 { { = lim = = = = =0 2{ + 3 { (2{ + 3)@{ lim (2 + 3@{) lim 2 + 3 lim (1@{) 2 + 3(0) 2. {. {. {. (1 { {2 )@{2 1 { {2 = lim = 2 { (2{2 7)@{2 2{ 7. {. lim (1@{2 1@{ 1). {. lim (2 7@{2 ). {. lim (1@{2 ) lim (1@{) lim 1. =. {. {. {. lim 2 7 lim (1@{2 ). {. {. 61. =. 001 1 = 2 7(0) 2. 19. Divide both the numerator and denominator by {3 (the highest power of { that occurs in the denominator).. 5 {3 + 5{ 5 lim 1 + 2 1 + { { { + 5{ {3 {2 lim = lim = lim = 4 1 { 2{3 {2 + 4 { 2{3 {2 + 4 { 4 1 2 + 3 + lim 2 { { { { {3 {3 1 lim 1 + 5 lim 2 1 1 + 5(0) { { { = = = 1 1 2 0 + 4(0) 2 + 4 lim 3 lim 2 lim { { { { { 3. 21. First, multiply the factors in the denominator. Then divide both the numerator and denominator by x4 .. 5 4x4 + 5 4+ 4 4 4x4 + 5 4x4 + 5 x x = lim = lim lim = lim 5 2 x (x2 2)(2x2 1) x 2x4 5x2 + 2 x 2x4 5x2 + 2 x 2 2 + 4 x x x4 5 1 lim 4 + 4 lim 4 + 5 lim 4 x x 4 4 + 5(0) x x x = = =2 = = 1 1 2 5(0) + 2(0) 2 2 5 lim 2 5 lim 2 + 2 lim 4 lim 2 2 + 4 x x x x x x x x.

(27) F. 62. ¤. TX.10. CHAPTER 2 LIMITS AND DERIVATIVES. 9{6 { 9{6 { @{3 = lim = 3 { { ({3 + 1)@{3 { +1. 23. lim. s (9{6 {)@{6 lim. {. s lim 9 1@{5. {. =. lim 1 + lim (1@{3 ) { { = 90 =3. 1@{3 ). lim (1 + { t lim 9 lim (1@{5 ) {. =. [since {3 =. {6 for { A 0]. {. 1+0. 2 9{2 + { 3{ 9{2 + { + 3{ 9{2 + { (3{)2 = lim { { 9{2 + { + 3{ 9{2 + { + 3{ 2 2 9{ + { 9{ 1@{ { = lim · = lim { { 9{2 + { + 3{ 9{2 + { + 3{ 1@{ {@{ 1 1 1 1 = lim s = = lim s = = { 3+3 6 9+3 9{2 @{2 + {@{2 + 3{@{ { 9 + 1@{ + 3. 25. lim 9{2 + { 3{ = lim {. 27. lim. {. {2 + d{ {2 + e{ = lim. {. {2 + d{ {2 + e{ {2 + d{ + {2 + e{ {2 + d{ + {2 + e{. ({2 + d{) ({2 + e{) [(d e){]@{ = lim = lim 2 2 2 { { { + d{ + { + e{ { + d{ + {2 + e{ @ {2 de de de s = = = lim s { 2 1 + 0 + 1 + 0 1 + d@{ + 1 + e@{ 29. lim. {. { + {3 + {5 ({ + {3 + {5 )@{4 = lim { (1 {2 + {4 )@{4 1 {2 + {4 = lim. {. [divide by the highest power of { in the denominator]. 1@{3 + 1@{ + { = 1@{4 1@{2 + 1. because (1@{3 + 1@{ + {) and (1@{4 1@{2 + 1) 1 as { . 31.. lim ({4 + {5 ) = lim {5 ( {1 + 1) [factor out the largest power of {] = because {5 and 1@{ + 1 1. {. {. as { . Or: lim {4 + {5 = lim {4 (1 + {) = . {. 33. lim. {. {. 1 h{ (1 h{ )@h{ 1@h{ 1 01 1 = = = lim = lim { (1 + 2h{ )@h{ { 1@h{ + 2 1 + 2h{ 0+2 2. 35. Since 1 cos { 1 and h2{ A 0, we have h2{ h2{ cos { h2{ . We know that lim (h2{ ) = 0 and {. lim h2{ = 0, so by the Squeeze Theorem, lim (h2{ cos {) = 0.. {. {. 37. (a). (b). From the graph of i ({) =. {2 + { + 1 + {, we estimate. the value of lim i ({) to be 0=5. {. {. i ({). 10,000. 0=4999625. 100,000. 0=4999962. 1,000,000. 0=4999996. From the table, we estimate the limit to be 0=5..

(28) F. TX.10 SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 2 2 { + { + 1 {2 { +{+1{ (c) lim {2 + { + 1 + { = lim {2 + { + 1 + { = lim { { { {2 + { + 1 { {2 + { + 1 { ({ + 1)(1@{) 1 + (1@{) s = lim = lim { {2 + { + 1 { (1@{) { 1 + (1@{) + (1@{2 ) 1 1 1+0 = = 2 1+0+01 Note that for { ? 0, we have {2 = |{| = {, so when we divide the radical by {, with { ? 0, we get s 1 2 1 2 { + { + 1 = { + { + 1 = 1 + (1@{) + (1@{2 ). { {2 1 1 2{ + 1 1 lim 2 + lim 2 + lim 2+ { { 2{ + 1 { { { { { = = lim = = lim 39. lim 2 2 { { 2 { { 2 { 2 1 lim 1 lim lim 1 { { { { { { { =. 2+0 = 2, so | = 2 is a horizontal asymptote. 10. The denominator { 2 is zero when { = 2 and the numerator is not zero, so we investigate | = i ({) =. 2{ + 1 as { approaches 2. {2. lim i ({) = because as. {2. { 2 the numerator is positive and the denominator approaches 0 through negative values. Similarly, lim i ({) = . Thus, { = 2 is a vertical asymptote. {2+. The graph conrms our work.. 1 1 2{2 + { 1 1 1 lim 2 + 2 2+ 2 { { 2{2 + { 1 {2 { { = { = lim 41. lim = lim 2 2 1 { {2 + { 2 { { + { 2 { 2 1 1+ 2 lim 1 + { { { { {2 {2 1 1 lim 2 lim 2 + lim 2+00 { { { { { = 2, so | = 2 is a horizontal asymptote. = = 1 1 1 + 0 2(0) 2 lim 2 lim 1 + lim { { { { { | = i ({) =. (2{ 1)({ + 1) 2{2 + { 1 = , so lim i({) = , {2 + { 2 ({ + 2)({ 1) {2. lim i ({) = , lim i ({) = , and lim i ({) = . Thus, { = 2. {2+. {1. {1+. and { = 1 are vertical asymptotes. The graph conrms our work. 43. | = i ({) =. {2. {({2 1) {({ + 1)({ 1) {({ + 1) {3 { = = = = j({) for { 6= 1. 6{ + 5 ({ 1)({ 5) ({ 1)({ 5) {5. The graph of j is the same as the graph of i with the exception of a hole in the graph of i at { = 1. By long division, j({) =. {2 + { 30 ={+6+ . {5 {5. As { ±, j({) ±, so there is no horizontal asymptote. The denominator of j is zero when { = 5. lim j({) = and lim j({) = , so { = 5 is a {5. {5+. vertical asymptote. The graph conrms our work.. ¤. 63.

(29) F. 64. ¤. CHAPTER 2 LIMITS AND DERIVATIVES. TX.10. 45. From the graph, it appears | = 1 is a horizontal asymptote.. 3{3 + 500{2 3{3 + 500{2 3 + (500@{) {3 = lim 3 = lim lim { {3 + 500{2 + 100{ + 2000 { { + 500{2 + 100{ + 2000 { 1 + (500@{) + (100@{2 ) + (2000@{3 ) {3 3+0 = 3, so | = 3 is a horizontal asymptote. = 1+0+0+0 The discrepancy can be explained by the choice of the viewing window. Try [100,000> 100,000] by [1> 4] to get a graph that lends credibility to our calculation that | = 3 is a horizontal asymptote.. 47. Let’s look for a rational function.. (1). lim i({) = 0 degree of numerator ? degree of denominator. {±. (2) lim i ({) = there is a factor of {2 in the denominator (not just {, since that would produce a sign {0. change at { = 0), and the function is negative near { = 0. (3) lim i ({) = and lim i({) = vertical asymptote at { = 3; there is a factor of ({ 3) in the {3. {3+. denominator. (4) i (2) = 0 2 is an {-intercept; there is at least one factor of ({ 2) in the numerator. Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us i ({) =. 2{ as one possibility. {2 ({ 3). 49. | = i ({) = {4 {6 = {4 (1 {2 ) = {4 (1 + {)(1 {). The |-intercept is. i (0) = 0. The {-intercepts are 0, 1, and 1 [found by solving i ({) = 0 for {]. Since {4 A 0 for { 6= 0, i doesn’t change sign at { = 0. The function does change sign at { = 1 and { = 1. As { ±, i ({) = {4 (1 {2 ) approaches because {4 and (1 {2 ) . 51. | = i ({) = (3 {)(1 + {)2 (1 {)4 . The |-intercept is i (0) = 3(1)2 (1)4 = 3.. The {-intercepts are 3, 1, and 1. There is a sign change at 3, but not at 1 and 1. When { is large positive, 3 { is negative and the other factors are positive, so lim i ({) = . When { is large negative, 3 { is positive, so. {. lim i({) = .. {. sin { 1 1 for { A 0. As { , 1@{ 0 and 1@{ 0, so by the Squeeze { { { sin { = 0. Theorem, (sin {)@{ 0. Thus, lim { {. 53. (a) Since 1 sin { 1 for all {> .

(30) F. TX.10 SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES (b) From part (a), the horizontal asymptote is | = 0. The function | = (sin {)@{ crosses the horizontal asymptote whenever sin { = 0; that is, at { = q for every integer q. Thus, the graph crosses the asymptote an innite number of times.. 55. Divide the numerator and the denominator by the highest power of { in T({).. (a) If deg S ? deg T, then the numerator 0 but the denominator doesn’t. So lim [S ({)@T({)] = 0. {. (b) If deg S A deg T, then the numerator ± but the denominator doesn’t, so lim [S ({)@T({)] = ± {. (depending on the ratio of the leading coefcients of S and T). 5 { 5 1@ { 5 = · = lim s = 5 and 57. lim { { { 1 1@ { 10 1 (1@{) lim. {. 10h{ 21 1@h{ 10 (21@h{ ) 10 0 10h{ 21 5 { = = 5. Since , · = lim ? i ({) ? { 2h{ 1@h{ 2 2 2h{ {1. we have lim i({) = 5 by the Squeeze Theorem. {. . 59. (a) lim y(w) = lim y 1 hjw@y w. . . w. = y (1 0) = y . (b) We graph y(w) = 1 h9=8w and y(w) = 0=99y , or in this case, y(w) = 0=99. Using an intersect feature or zooming in on the point of intersection, we nd that w 0=47 s.. 61. Let j({) =. 3{2 + 1 and i ({) = |j({) 1=5|. Note that 2{2 + { + 1. lim j({) =. {. 3 2. and lim i({) = 0. We are interested in nding the {. {-value at which i ({) ? 0=05. From the graph, we nd that { 14=804, so we choose Q = 15 (or any larger number). 4{2 + 1. 63. For % = 0=5, we need to nd Q such that . (2) ? 0=5. {+1 4{2 + 1 2=5 ? ? 1=5 whenever { Q. We graph the three parts of this {+1 inequality on the same screen, and see that the inequality holds for { 6= So we choose Q = 6 (or any smaller number). 4{2 + 1 For % = 0=1, we need 2=1 ? ? 1=9 whenever { Q= From the {+1 graph, it seems that this inequality holds for { 22. So we choose Q = 22 (or any smaller number).. ¤. 65.

(31) F. 66. ¤. CHAPTER 2. TX.10. LIMITS AND DERIVATIVES. 65. (a) 1@{2 ? 0=0001. {2 A 1@0=0001 = 10 000 { A 100 ({ A 0). (b) If % A 0 is given, then 1@{2 ? % {2 A 1@% { A 1@ %. Let Q = 1@ %. 1 1 1 1 2 0 = 2 ? %, so lim 2 = 0. Then { A Q { A { { { { % 67. For { ? 0, |1@{ 0| = 1@{. If % A 0 is given, then 1@{ ? %. Take Q = 1@%. Then { ? Q. { ? 1@%.. { ? 1@% |(1@{) 0| = 1@{ ? %, so lim (1@{) = 0. {. 69. Given P A 0, we need Q A 0 such that { A Q. h{ A P . Now h{ A P. { A ln P, so take. Q = max(1> ln P ). (This ensures that Q A 0.) Then { A Q = max(1> ln P) h{ A max(h> P) P, so lim h{ = . {. 71. Suppose that lim i ({) = O. Then for every % A 0 there is a corresponding positive number Q such that |i ({) O| ? % {. whenever { A Q. If w = 1@{, then { A Q. 0 ? 1@{ ? 1@Q. 0 ? w ? 1@Q. Thus, for every % A 0 there is a. corresponding A 0 (namely 1@Q) such that |i(1@w) O| ? % whenever 0 ? w ? . This proves that lim i (1@w) = O = lim i({).. w0+. {. Now suppose that lim i ({) = O. Then for every % A 0 there is a corresponding negative number Q such that {. |i ({) O| ? % whenever { ? Q. If w = 1@{, then { ? Q. 1@Q ? 1@{ ? 0 1@Q ? w ? 0. Thus, for every. % A 0 there is a corresponding A 0 (namely 1@Q) such that |i (1@w) O| ? % whenever ? w ? 0. This proves that lim i (1@w) = O = lim i ({).. w0. {. 2.7 Derivatives and Rates of Change 1. (a) This is just the slope of the line through two points: pS T =. i ({) i (3) | = . { {3. (b) This is the limit of the slope of the secant line S T as T approaches S : p = lim. {3. i({) i (3) . {3. 3. (a) (i) Using Denition 1 with i ({) = 4{ {2 and S (1> 3),. i ({) i (d) (4{ {2 ) 3 ({2 4{ + 3) ({ 1)({ 3) = lim = lim = lim {d {1 {1 {1 {d {1 {1 {1. p = lim. = lim (3 {) = 3 1 = 2 {1. (ii) Using Equation 2 with i ({) = 4{ {2 and S (1> 3),. 4(1 + k) (1 + k)2 3 i (d + k) i(d) i (1 + k) i (1) = lim = lim k0 k0 k0 k k k. p = lim = lim. k0. 4 + 4k 1 2k k2 3 k2 + 2k k(k + 2) = lim = lim = lim (k + 2) = 2 k0 k0 k0 k k k. (b) An equation of the tangent line is | i(d) = i 0 (d)({ d) | i (1) = i 0 (1)({ 1) | 3 = 2({ 1), or | = 2{ + 1..

(32) F. TX.10. SECTION 2.7 DERIVATIVES AND RATES OF CHANGE. ¤. 67. The graph of | = 2{ + 1 is tangent to the graph of | = 4{ {2 at the. (c). point (1> 3). Now zoom in toward the point (1> 3) until the parabola and the tangent line are indistiguishable.. 5. Using (1) with i ({) =. {1 and S (3> 2), {2 {1 { 1 2({ 2) 2 i ({) i (d) { 2 {2 p = lim = lim = lim {d {3 {3 {d {3 {3 = lim. {3. 3{ 1 1 = lim = = 1 ({ 2)({ 3) {3 { 2 1. Tangent line: | 2 = 1({ 3) | 2 = { + 3. | = { + 5. { 1 ( { 1)( { + 1) {1 1 1 = lim = lim = lim = . {1 {1 ({ 1)( { + 1) {1 ({ 1)( { + 1) {1 {1 2 {+1. 7. Using (1), p = lim. Tangent line: | 1 = 12 ({ 1). | = 12 { +. 1 2. 9. (a) Using (2) with | = i ({) = 3 + 4{2 2{3 ,. i (d + k) i (d) 3 + 4(d + k)2 2(d + k)3 (3 + 4d2 2d3 ) = lim k0 k0 k k. p = lim. 3 + 4(d2 + 2dk + k2 ) 2(d3 + 3d2 k + 3dk2 + k3 ) 3 4d2 + 2d3 k0 k. = lim. 3 + 4d2 + 8dk + 4k2 2d3 6d2 k 6dk2 2k3 3 4d2 + 2d3 k0 k. = lim. 8dk + 4k2 6d2 k 6dk2 2k3 k(8d + 4k 6d2 6dk 2k2 ) = lim k0 k0 k k. = lim. = lim (8d + 4k 6d2 6dk 2k2 ) = 8d 6d2 k0. (b) At (1> 5): p = 8(1) 6(1)2 = 2, so an equation of the tangent line. (c). is | 5 = 2({ 1) | = 2{ + 3. At (2> 3): p = 8(2) 6(2)2 = 8, so an equation of the tangent line is | 3 = 8({ 2) | = 8{ + 19.. 11. (a) The particle is moving to the right when v is increasing; that is, on the intervals (0> 1) and (4> 6). The particle is moving to. the left when v is decreasing; that is, on the interval (2> 3). The particle is standing still when v is constant; that is, on the intervals (1> 2) and (3> 4)..

(33) F. 68. ¤. CHAPTER 2. LIMITS AND DERIVATIVES. TX.10. (b) The velocity of the particle is equal to the slope of the tangent line of the graph. Note that there is no slope at the corner points on the graph. On the interval (0> 1)> the slope is. 30 = 3. On the interval (2> 3), the slope is 10. 31 13 = 2. On the interval (4> 6), the slope is = 1. 32 64 13. Let v(w) = 40w 16w2 .. 8 2w2 5w + 2 40w 16w2 16 v(w) v(2) 16w2 + 40w 16 = lim = lim = lim w2 w2 w2 w2 w2 w2 w2 w2. y(2) = lim. = lim. w2. 8(w 2)(2w 1) = 8 lim (2w 1) = 8(3) = 24 w2 w2. Thus, the instantaneous velocity when w = 2 is 24 ft@s. 1 d2 (d + k)2 1 2 2 v(d + k) v(d) (d + k) d d2 (d + k)2 d2 (d2 + 2dk + k2 ) = lim = lim = lim 15. y(d) = lim k0 k0 k0 k0 k k k kd2 (d + k)2 (2dk + k2 ) k(2d + k) (2d + k) 2d 2 = lim = lim 2 = 2 2 = 3 m@s k0 kd2 (d + k)2 k0 kd2 (d + k)2 k0 d (d + k)2 d ·d d. = lim. So y (1) =. 2 2 1 2 2 m@s. = 2 m@s, y(2) = 3 = m@s, and y(3) = 3 = 13 2 4 3 27. 17. j 0 (0) is the only negative value. The slope at { = 4 is smaller than the slope at { = 2 and both are smaller than the slope at. { = 2. Thus, j 0 (0) ? 0 ? j 0 (4) ? j0 (2) ? j 0 (2). 19. We begin by drawing a curve through the origin with a. slope of 3 to satisfy i (0) = 0 and i 0 (0) = 3. Since i 0 (1) = 0, we will round off our gure so that there is a horizontal tangent directly over { = 1. Last, we make sure that the curve has a slope of 1 as we pass over { = 2. Two of the many possibilities are shown. 21. Using Denition 2 with i ({) = 3{2 5{ and the point (2> 2), we have. 3(2 + k)2 5(2 + k) 2 i(2 + k) i(2) (12 + 12k + 3k2 10 5k) 2 = lim = lim i (2) = lim k0 k0 k0 k k k 3k2 + 7k = lim = lim (3k + 7) = 7 k0 k0 k 0. So an equation of the tangent line at (2> 2) is | 2 = 7({ 2) or | = 7{ 12..

(34) F. TX.10. SECTION 2.7 DERIVATIVES AND RATES OF CHANGE. 23. (a) Using Denition 2 with I ({) = 5{@(1 + {2 ) and the point (2> 2), we have. ¤. 69. (b). 5(2 + k) 2 I (2 + k) I (2) 1 + (2 + k)2 I (2) = lim = lim k0 k0 k k 0. = lim. 5k + 10 5k + 10 2(k2 + 4k + 5) 2 + 4k + 5 k2 + 4k + 5 = lim k0 k k. k2. k0. = lim. k0. 2k2 3k k(2k 3) 2k 3 3 = lim = lim = k(k2 + 4k + 5) k0 k(k2 + 4k + 5) k0 k2 + 4k + 5 5. So an equation of the tangent line at (2> 2) is | 2 = 35 ({ 2) or | = 35 { +. 16 . 5. 25. Use Denition 2 with i ({) = 3 2{ + 4{2 .. i (d + k) i (d) [3 2(d + k) + 4(d + k)2 ] (3 2d + 4d2 ) = lim k0 k0 k k (3 2d 2k + 4d2 + 8dk + 4k2 ) (3 2d + 4d2 ) = lim k0 k 2 2k + 8dk + 4k k(2 + 8d + 4k) = lim = lim (2 + 8d + 4k) = 2 + 8d = lim k0 k0 k0 k k. i 0 (d) = lim. 27. Use Denition 2 with i (w) = (2w + 1)@(w + 3).. 2(d + k) + 1 2d + 1 i (d + k) i (d) (d + k) + 3 d+3 (2d + 2k + 1)(d + 3) (2d + 1)(d + k + 3) i (d) = lim = lim = lim k0 k0 k0 k k k(d + k + 3)(d + 3) 0. (2d2 + 6d + 2dk + 6k + d + 3) (2d2 + 2dk + 6d + d + k + 3) k0 k(d + k + 3)(d + 3) 5k 5 5 = lim = = lim k0 k(d + k + 3)(d + 3) k0 (d + k + 3)(d + 3) (d + 3)2 = lim. . 29. Use Denition 2 with i ({) = 1@ { + 2.. 1 1 d+2 d+k+2 s d+2 (d + k) + 2 i (d + k) i (d) d+k+2 d+2 = lim = lim i 0 (d) = lim k0 k0 k0 k k k d+2 d+k+2 d+2+ d+k+2 (d + 2) (d + k + 2) = lim · = lim k0 k0 k d + k + 2 k d+k+2 d+2 d+2+ d+k+2 d+2 d+2+ d+k+2 = lim. k0. k 1 = lim k0 k d+k+2 d+2 d+2+ d+k+2 d+k+2 d+2 d+2+ d+k+2. 1 1 = 2 = 2(d + 2)3@2 2 d+2 d+2 Note that the answers to Exercises 31 – 36 are not unique.. (1 + k)10 1 = i 0 (1), where i ({) = {10 and d = 1. k0 k (1 + k)10 1 = i 0 (0), where i ({) = (1 + {)10 and d = 0. Or: By Denition 2, lim k0 k. 31. By Denition 2, lim.

(35) F. 70. ¤. CHAPTER 2. TX.10. LIMITS AND DERIVATIVES {. 33. By Equation 3, lim. {5. 2 32 = i 0 (5), where i({) = 2{ and d = 5. {5. cos( + k) + 1 = i 0 (), where i ({) = cos { and d = . k cos( + k) + 1 Or: By Denition 2, lim = i 0 (0), where i ({) = cos( + {) and d = 0. k0 k. 35. By Denition 2, lim. k0. i (5 + k) i (5) [100 + 50(5 + k) 4=9(5 + k)2 ] [100 + 50(5) 4=9(5)2 ] = lim k0 k k (100 + 250 + 50k 4=9k2 49k 122=5) (100 + 250 122=5) 4=9k2 + k = lim = lim k0 k0 k k k(4=9k + 1) = lim (4=9k + 1) = 1 m@s = lim k0 k0 k. 37. y(5) = i 0 (5) = lim. k0. The speed when w = 5 is |1| = 1 m@s. 39. The sketch shows the graph for a room temperature of 72 and a refrigerator. temperature of 38 . The initial rate of change is greater in magnitude than the rate of change after an hour.. 41. (a) (i) [2000> 2002]:. 77 55 22 S (2002) S (2000) = = = 11 percent@year 2002 2000 2 2. (ii) [2000> 2001]:. 68 55 S (2001) S (2000) = = 13 percent@year 2001 2000 1. (iii) [1999> 2000]:. 55 39 S (2000) S (1999) = = 16 percent@year 2000 1999 1. (b) Using the values from (ii) and (iii), we have. 13 + 16 = 14=5 percent@year. 2. (c) Estimating D as (1999> 40) and E as (2001> 70), the slope at 2000 is 70 40 30 = = 15 percent@year. 2001 1999 2 F(105) F(100) 6601=25 6500 F = = = $20=25@unit. { 105 100 5 F F(101) F(100) 6520=05 6500 (ii) = = = $20=05@unit. { 101 100 1 5000 + 10(100 + k) + 0=05(100 + k)2 6500 20k + 0=05k2 F(100 + k) F(100) = = (b) k k k = 20 + 0=05k, k 6= 0. 43. (a) (i). So the instantaneous rate of change is lim. k0. F(100 + k) F(100) = lim (20 + 0=05k) = $20@unit. k0 k. 45. (a) i 0 ({) is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are. dollars per ounce..

(36) F. TX.10. SECTION 2.8. THE DERIVATIVE AS A FUNCTION. ¤. 71. (b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is $17@ounce. So the cost of producing the 800th (or 801st) ounce is about $17. (c) In the short term, the values of i 0 ({) will decrease because more efcient use is made of start-up costs as { increases. But eventually i 0 ({) might increase due to large-scale operations. 47. W 0(10) is the rate at which the temperature is changing at 10:00 AM . To estimate the value of W 0(10), we will average the. difference quotients obtained using the times w = 8 and w = 12. Let D = E=. 72 81 W (8) W (10) = = 4=5 and 8 10 2. W (w) W (10) 88 81 D+E 4=5 + 3=5 W (12) W (10) = = 3=5. Then W 0(10) = lim = = 4 F@h. w10 12 10 2 w 10 2 2. 49. (a) V 0 (W ) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mg@L)@ C.. (b) For W = 16 C, it appears that the tangent line to the curve goes through the points (0> 14) and (32> 6). So 8 6 14 = = 0=25 (mg@L)@ C. This means that as the temperature increases past 16 C, the oxygen 32 0 32. V 0 (16) . solubility is decreasing at a rate of 0=25 (mg@L)@ C. 51. Since i ({) = { sin(1@{) when { 6= 0 and i (0) = 0, we have. i 0 (0) = lim. k0. i (0 + k) i (0) k sin(1@k) 0 = lim = lim sin(1@k). This limit does not exist since sin(1@k) takes the k0 k0 k k. values 1 and 1 on any interval containing 0. (Compare with Example 4 in Section 2.2.). 2.8 The Derivative as a Function 1. It appears that i is an odd function, so i 0 will be an even. function—that is, i 0 (d) = i 0 (d). (a) i 0 (3) 1=5. (b) i 0 (2) 1. (c) i 0 (1) 0. (d) i 0 (0) 4. (e) i 0 (1) 0. (f ) i 0 (2) 1. (g) i 0 (3) 1=5. 3. (a)0 = II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then. negative again. The actual function values in graph II follow the same pattern. (b)0 = IV, since from left to right, the slopes of the tangents to graph (b) start out at a xed positive quantity, then suddenly become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents. (c)0 = I, since the slopes of the tangents to graph (c) are negative for { ? 0 and positive for { A 0, as are the function values of graph I. (d)0 = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then positive, then 0, then negative again, and the function values in graph III follow the same pattern..

(37) F. 72. ¤. CHAPTER 2. TX.10. LIMITS AND DERIVATIVES. Hints for Exercises 4 –11: First plot {-intercepts on the graph of i 0 for any horizontal tangents on the graph of i . Look for any corners on the graph of i — there will be a discontinuity on the graph of i 0 . On any interval where i has a tangent with positive (or negative) slope, the graph of i 0 will be positive (or negative). If the graph of the function is linear, the graph of i 0 will be a horizontal line. 5.. 7.. 9.. 11.. 13. It appears that there are horizontal tangents on the graph of P for w = 1963. and w = 1971. Thus, there are zeros for those values of w on the graph of P 0 . The derivative is negative for the years 1963 to 1971.. 15.. The slope at 0 appears to be 1 and the slope at 1 appears to be 2=7. As { decreases, the slope gets closer to 0. Since the graphs are so similar, we might guess that i 0 ({) = h{ ..

(38) F. TX.10 17. (a) By zooming in, we estimate that i 0 (0) = 0, i 0. 1 2. SECTION 2.8. THE DERIVATIVE AS A FUNCTION. = 1, i 0 (1) = 2,. and i 0 (2) = 4.. (b) By symmetry, i 0 ({) = i 0 ({). So i 0 12 = 1, i 0 (1) = 2, and i 0 (2) = 4. (c) It appears that i 0 ({) is twice the value of {, so we guess that i 0 ({) = 2{. i({ + k) i ({) ({ + k)2 {2 = lim k0 k0 k k 2 { + 2k{ + k2 {2 k(2{ + k) 2k{ + k2 = lim = lim = lim (2{ + k) = 2{ = lim k0 k0 k0 k0 k k k. (d) i 0 ({) = lim. i ({ + k) i({) = lim k0 k0 k. 19. i 0 ({) = lim. = lim. k0. 1 k 2. k. 1 k0 2. = lim. =. 1. 2 ({. + k) . 1 3. . k. . 1. 2{. . 1 3. = lim. k0. 1 { 2. + 12 k 13 12 { + k. 1 3. 1 2. Domain of i = domain of i 0 = R. 5(w + k) 9(w + k)2 (5w 9w2 ) i (w + k) i(w) = lim 21. i (w) = lim k0 k0 k k 0. 5w + 5k 9(w2 + 2wk + k2 ) 5w + 9w2 5w + 5k 9w2 18wk 9k2 5w + 9w2 = lim k0 k0 k k. = lim. k(5 18w 9k) 5k 18wk 9k2 = lim = lim (5 18w 9k) = 5 18w k0 k0 k0 k k. = lim. Domain of i = domain of i 0 = R. ({ + k)3 3({ + k) + 5 ({3 3{ + 5) i ({ + k) i ({) = lim k0 k0 k k 3 3 2 2 3 { + 3{ k + 3{k + k 3{ 3k + 5 { 3{ + 5 3{2 k + 3{k2 + k3 3k = lim = lim k0 k0 k k 2 2 k 3{ + 3{k + k 3 = lim 3{2 + 3{k + k2 3 = 3{2 3 = lim k0 k0 k. 23. i 0({) = lim. Domain of i = domain of i 0 = R. j({ + k) j({) = lim 25. j ({) = lim k0 k0 k 0. = lim. k0. %s & s 1 + 2({ + k) 1 + 2{ 1 + 2({ + k) + 1 + 2{ s k 1 + 2({ + k) + 1 + 2{. 1 (1 + 2{ + 2k) (1 + 2{) 2 2 ks l = lim = = k0 2 1 + 2{ 1 + 2{ 1 + 2{ + 2k + 1 + 2{ k 1 + 2({ + k) + 1 + 2{. Domain of j = 12 > , domain of j0 = 12 > .. ¤. 73.

(39) F. 74. ¤. CHAPTER 2. TX.10. LIMITS AND DERIVATIVES. 4(w + k) 4(w + k)(w + 1) 4w(w + k + 1) 4w J(w + k) J(w) (w + k) + 1 w + 1 (w + k + 1)(w + 1) = lim = lim 27. J0(w) = lim k0 k0 k0 k k k 2 2 4w + 4kw + 4w + 4k 4w + 4kw + 4w 4k = lim = lim k0 k0 k(w + k + 1)(w + 1) k(w + k + 1)(w + 1) = lim. k0. 4 4 = (w + k + 1)(w + 1) (w + 1)2. Domain of J = domain of J0 = (> 1) (1> ). 4 { + 4{3 k + 6{2 k2 + 4{k3 + k4 {4 i({ + k) i ({) ({ + k)4 {4 = lim = lim 29. i ({) = lim k0 k0 k0 k k k 3 2 2 3 4 4{ k + 6{ k + 4{k + k = lim 4{3 + 6{2 k + 4{k2 + k3 = 4{3 = lim k0 k0 k 0. Domain of i = domain of i 0 = R. i ({ + k) i ({) [({ + k)4 + 2({ + k)] ({4 + 2{) = lim k0 k k. 31. (a) i 0 ({) = lim. k0. = lim. {4 + 4{3 k + 6{2 k2 + 4{k3 + k4 + 2{ + 2k {4 2{ k. = lim. 4{3 k + 6{2 k2 + 4{k3 + k4 + 2k k(4{3 + 6{2 k + 4{k2 + k3 + 2) = lim k0 k k. k0. k0. = lim (4{3 + 6{2 k + 4{k2 + k3 + 2) = 4{3 + 2 k0. (b) Notice that i 0 ({) = 0 when i has a horizontal tangent, i 0 ({) is positive when the tangents have positive slope, and i 0 ({) is negative when the tangents have negative slope.. 33. (a) X 0 (w) is the rate at which the unemployment rate is changing with respect to time. Its units are percent per year.. (b) To nd X 0 (w), we use lim. k0. For 1993: X 0 (1993) . X (w + k) X(w) X(w + k) X (w) for small values of k. k k. 6=1 6=9 X (1994) X (1993) = = 0=80 1994 1993 1. For 1994: We estimate X 0 (1994) by using k = 1 and k = 1, and then average the two results to obtain a nal estimate. k = 1 X 0 (1994) k = 1 X 0 (1994) . 6=9 6=1 X (1993) X (1994) = = 0=80; 1993 1994 1. X (1995) X (1994) 5=6 6=1 = = 0=50. 1995 1994 1. So we estimate that X 0 (1994) 12 [(0=80) + (0=50)] = 0=65. w 0. X (w). 1993. 1994. 1995. 1996. 1997. 1998. 1999. 2000. 2001. 2002. 0=80. 0=65. 0=35. 0=35. 0=45. 0=35. 0=25. 0=25. 0=90. 1=10.

(40) F. TX.10. SECTION 2.8. THE DERIVATIVE AS A FUNCTION. ¤. 75. 35. i is not differentiable at { = 4, because the graph has a corner there, and at { = 0, because there is a discontinuity there. 37. i is not differentiable at { = 1, because the graph has a vertical tangent there, and at { = 4, because the graph has a corner. there. 39. As we zoom in toward (1> 0), the curve appears more and more like a. straight line, so i ({) = { +. s |{| is differentiable at { = 1. But no. matter how much we zoom in toward the origin, the curve doesn’t straighten out—we can’t eliminate the sharp point (a cusp). So i is not differentiable at { = 0. 41. d = i , e = i 0 , f = i 00 . We can see this because where d has a horizontal tangent, e = 0, and where e has a horizontal tangent,. f = 0. We can immediately see that f can be neither i nor i 0 , since at the points where f has a horizontal tangent, neither d nor e is equal to 0. 43. We can immediately see that d is the graph of the acceleration function, since at the points where d has a horizontal tangent,. neither f nor e is equal to 0. Next, we note that d = 0 at the point where e has a horizontal tangent, so e must be the graph of the velocity function, and hence, e0 = d. We conclude that f is the graph of the position function. 1 + 4({ + k) ({ + k)2 (1 + 4{ {2 ) i ({ + k) i ({) = lim 45. i ({) = lim k0 k0 k k 2 2 (1 + 4{ + 4k { 2{k k ) (1 + 4{ {2 ) 4k 2{k k2 = lim = lim = lim (4 2{ k) = 4 2{ k0 k0 k0 k k 0. i 00 ({) = lim. k0. i 0 ({ + k) i 0 ({) [4 2({ + k)] (4 2{) 2k = lim = lim = lim (2) = 2 k0 k0 k k0 k k. We see from the graph that our answers are reasonable because the graph of i 0 is that of a linear function and the graph of i 00 is that of a constant function..

(41) F. 76. ¤. CHAPTER 2. LIMITS AND DERIVATIVES. 2({ + k)2 ({ + k)3 (2{2 {3 ) i ({ + k) i ({) = lim k0 k0 k k k(4{ + 2k 3{2 3{k k2 ) = lim (4{ + 2k 3{2 3{k k2 ) = 4{ 3{2 = lim k0 k0 k. 47. i 0 ({) = lim. 4({ + k) 3({ + k)2 (4{ 3{2 ) i 0 ({ + k) i 0 ({) k(4 6{ 3k) = lim = lim i ({) = lim k0 k0 k0 k k k = lim (4 6{ 3k) = 4 6{ 00. k0. i 000 ({) = lim. k0. i (4) ({) = lim. i 00 ({ + k) i 00 ({) [4 6({ + k)] (4 6{) 6k = lim = lim = lim (6) = 6 k0 k0 k k0 k k. k0. i 000 ({ + k) i 000 ({) 6 (6) 0 = lim = lim = lim (0) = 0 k0 k0 k k0 k k. The graphs are consistent with the geometric interpretations of the derivatives because i 0 has zeros where i has a local minimum and a local maximum, i 00 has a zero where i 0 has a local maximum, and i 000 is a constant function equal to the slope of i 00 . 49. (a) Note that we have factored { d as the difference of two cubes in the third step.. i 0(d) = lim. {d. i ({) i (d) {1@3 d1@3 {1@3 d1@3 = lim = lim 1@3 {d {d ({ {d {d d1@3 )({2@3 + {1@3 d1@3 + d2@3 ). 1 1 = 2@3 or 13 d2@3 {2@3 + {1@3 d1@3 + d2@3 3d 3 i(0 + k) i(0) k0 1 0 = lim = lim 2@3 . This function increases without bound, so the limit does not (b) i (0) = lim k0 k0 k0 k k k = lim. {d. exist, and therefore i 0(0) does not exist. (c) lim |i 0 ({)| = lim {0. {0. 51. i ({) = |{ 6| =. 1 = and i is continuous at { = 0 (root function), so i has a vertical tangent at { = 0. 3{2@3. + {6. ({ 6) if { 6 ? 0. So the right-hand limit is lim. {6+. is lim. {6. if { 6 6. =. + { 6 if { 6 6 { if { ? 6. i ({) i (6) |{ 6| 0 {6 = lim = lim = lim 1 = 1, and the left-hand limit {6 {6 {6+ {6+ { 6 {6+. i({) i(6) |{ 6| 0 6{ = lim = lim = lim (1) = 1. Since these limits are not equal, {6 {6 {6 {6 { 6 {6. i ({) i (6) does not exist and i is not differentiable at 6. {6 + 1 if { A 6 0 0 However, a formula for i is i ({) = 1 if { ? 6. i 0 (6) = lim. {6. Another way of writing the formula is i 0 ({) =. {6 . |{ 6|.

(42) F. TX.10 53. (a) i ({) = { |{| =. + 2 {. if { 0 2. THE DERIVATIVE AS A FUNCTION. ¤. (b) Since i ({) = {2 for { 0, we have i 0 ({) = 2{ for { A 0.. if { ? 0. {. SECTION 2.8. [See Exercise 2.8.17(d).] Similarly, since i({) = {2 for { ? 0, we have i 0 ({) = 2{ for { ? 0. At { = 0, we have i 0 (0) = lim. {0. i ({) i (0) { |{| = lim = lim |{| = 0= {0 { {0 {0. So i is differentiable at 0. Thus, i is differentiable for all {. +. 2{ if { 0. 0. (c) From part (b), we have i ({) =. 2{ if { ? 0. , = 2 |{|.. 55. (a) If i is even, then. i 0 ({) = lim. k0. = lim. k0. i ({ + k) i ({) i [({ k)] i ({) = lim k0 k k i ({ k) i({) i ({ k) i ({) = lim k0 k k. = lim. {0. [let { = k]. i ({ + {) i ({) = i 0 ({) {. Therefore, i 0 is odd. (b) If i is odd, then i 0 ({) = lim. k0. = lim. k0. = lim. i ({ + k) i ({) i [({ k)] i ({) = lim k0 k k i({ k) + i ({) i ({ k) i({) = lim k0 k k. {0. [let { = k]. i ({ + {) i ({) = i 0 ({) {. Therefore, i 0 is even. 57.. In the right triangle in the diagram, let | be the side opposite angle ! and { the side adjacent angle !. Then the slope of the tangent line c is p = |@{ = tan !. Note that 0 ? ! ?. . 2. We know (see Exercise 17). that the derivative of i({) = {2 is i 0 ({) = 2{. So the slope of the tangent to the curve at the point (1> 1) is 2. Thus, ! is the angle between 0 and tangent is 2; that is, ! = tan1 2 63 .. 2. whose. 77.

(43) F. 78. ¤. TX.10. CHAPTER 2 LIMITS AND DERIVATIVES. 2 Review 1. (a) lim i ({) = O: See Denition 2.2.1 and Figures 1 and 2 in Section 2.2. {d. (b) lim i ({) = O: See the paragraph after Denition 2.2.2 and Figure 9(b) in Section 2.2. {d+. (c) lim i({) = O: See Denition 2.2.2 and Figure 9(a) in Section 2.2. {d. (d) lim i ({) = : See Denition 2.2.4 and Figure 12 in Section 2.2. {d. (e) lim i ({) = O: See Denition 2.6.1 and Figure 2 in Section 2.6. {. 2. In general, the limit of a function fails to exist when the function does not approach a xed number. For each of the following. functions, the limit fails to exist at { = 2.. The left- and right-hand. There is an. There are an innite. limits are not equal.. innite discontinuity.. number of oscillations.. 3. (a) – (g) See the statements of Limit Laws 1– 6 and 11 in Section 2.3. 4. See Theorem 3 in Section 2.3. 5. (a) See Denition 2.2.6 and Figures 12–14 in Section 2.2.. (b) See Denition 2.6.3 and Figures 3 and 4 in Section 2.6. 6. (a) | = {4 : No asymptote. (c) | = tan {: Vertical asymptotes { =. (b) | = sin {: No asymptote 2. + q, q an integer. (d) | = tan1 {: Horizontal asymptotes | = ± 2. (e) | = h{ : Horizontal asymptote | = 0 lim h{ = 0. (f ) | = ln {: Vertical asymptote { = 0 lim ln { = . (g) | = 1@{: Vertical asymptote { = 0,. (h) | =. {. {0+. {: No asymptote. horizontal asymptote | = 0 7. (a) A function i is continuous at a number d if i({) approaches i (d) as { approaches d; that is, lim i ({) = i (d). {d. (b) A function i is continuous on the interval (> ) if i is continuous at every real number d. The graph of such a function has no breaks and every vertical line crosses it..

(44) F. TX.10. CHAPTER 2 REVIEW. ¤. 8. See Theorem 2.5.10. 9. See Denition 2.7.1. 10. See the paragraph containing Formula 3 in Section 2.7. 11. (a) The average rate of change of | with respect to { over the interval [{1 > {2 ] is. (b) The instantaneous rate of change of | with respect to { at { = {1 is lim. {2 {1. i ({2 ) i({1 ) . {2 {1. i ({2 ) i ({1 ) . {2 {1. 12. See Denition 2.7.2. The pages following the denition discuss interpretations of i 0 (d) as the slope of a tangent line to the. graph of i at { = d and as an instantaneous rate of change of i ({) with respect to { when { = d. 13. See the paragraphs before and after Example 6 in Section 2.8. 14. (a) A function i is differentiable at a number d if its derivative i 0 exists. (c). at { = d; that is, if i 0 (d) exists. (b) See Theorem 2.8.4. This theorem also tells us that if i is not continuous at d, then i is not differentiable at d. 15. See the discussion and Figure 7 on page 159.. 1. False.. Limit Law 2 applies only if the individual limits exist (these don’t).. 3. True.. Limit Law 5 applies.. 5. False.. Consider lim. {5. {({ 5) sin({ 5) or lim . The rst limit exists and is equal to 5. By Example 3 in Section 2.2, {5 {5 {5. we know that the latter limit exists (and it is equal to 1). 7. True.. A polynomial is continuous everywhere, so lim s({) exists and is equal to s(e).. 9. True.. See Figure 8 in Section 2.6.. {e. + 1@({ 1) if { 6= 1. 11. False.. Consider i ({) =. 13. True.. Use Theorem 2.5.8 with d = 2, e = 5, and j({) = 4{2 11. Note that i (4) = 3 is not needed.. 2. if { = 1. 15. True, by the denition of a limit with % = 1. 17. False.. 19. False.. See the note after Theorem 4 in Section 2.8. g 2| is the second derivative while g{2 2 g| g 2| then = 0, but = 1. g{2 g{. . g| g{. 2. is the rst derivative squared. For example, if | = {,. 79.

(45) F. 80. ¤. TX.10. CHAPTER 2 LIMITS AND DERIVATIVES. 1. (a) (i) lim i ({) = 3. (ii). {2+. lim i ({) = 0. {3+. (iii) lim i ({) does not exist since the left and right limits are not equal. (The left limit is 2.) {3. (iv) lim i ({) = 2 {4. (v) lim i ({) = . (vi) lim i ({) = . (vii) lim i ({) = 4. (viii) lim i ({) = 1. {0. {2. {. {. (b) The equations of the horizontal asymptotes are | = 1 and | = 4. (c) The equations of the vertical asymptotes are { = 0 and { = 2. (d) i is discontinuous at { = 3, 0, 2, and 4. The discontinuities are jump, innite, innite, and removable, respectively. 3. Since the exponential function is continuous, lim h{ {1. 5. lim. {3. 7. lim. k0. 3 {. = h11 = h0 = 1.. {2 9 ({ + 3)({ 3) {3 3 3 6 3 = lim = lim = = = {2 + 2{ 3 {3 ({ + 3)({ 1) {3 { 1 3 1 4 2. 3 k 3k2 + 3k 1 + 1 (k 1)3 + 1 k3 3k2 + 3k = lim = lim = lim k2 3k + 3 = 3 k0 k0 k0 k k k. Another solution: Factor the numerator as a sum of two cubes and then simplify. [(k 1) + 1] (k 1)2 1(k 1) + 12 (k 1)3 + 1 (k 1)3 + 13 = lim = lim lim k0 k0 k0 k k k 2 = lim (k 1) k + 2 = 1 0 + 2 = 3 k0. u u 4 = since (u 9) 0 as u 9 and A 0 for u 6= 9. u9 (u 9)4 (u 9)4. 9. lim. 11. lim. x1 x3. 2(2) (x2 + 1)(x2 1) (x2 + 1)(x + 1)(x 1) (x2 + 1)(x + 1) 4 x4 1 = lim = lim = lim = = 2 2 x1 x1 + 5x 6x x1 x(x + 5x 6) x(x + 6)(x 1) x(x + 6) 1(7) 7. 13. Since { is positive,. {2 = |{| = {. Thus, s 1 9@{2 {2 9 {2 9@ {2 10 1 lim = lim = lim = = { 2{ 6 { (2{ 6)@{ { 2 6@{ 20 2. 15. Let w = sin {. Then as { , sin { 0+ , so w 0+ . Thus, lim ln(sin {) = lim ln w = . { . 17. lim. {. w0+. 2 { + 4{ + 1 { {2 + 4{ + 1 + { ({2 + 4{ + 1) {2 · = lim { { 1 {2 + 4{ + 1 + { {2 + 4{ + 1 + { l k (4{ + 1)@{ = lim divide by { = {2 for { A 0 { ( {2 + 4{ + 1 + {)@{. {2 + 4{ + 1 { = lim. 4 + 1@{ 4+0 4 = lim s = = =2 { 2 1+0+0+1 1 + 4@{ + 1@{2 + 1.