lim f() = lim (0.8-0.08) = 0, " "!10!10 lim f() = lim 0 = 0.!10!10 Therefore, lim f() = 0.!10 lim g() = lim (0.8 - "!10!10 0.042-3) = 1, " lim g() = lim 1 = 1.!10!0 Therefore, lim g() = 1.!10 EXERCISE 3-2 Things to remember: 1. CONTINUITY A function f is CONTINUOUS AT THE POINT = c if: (a) lim f() eists;!c (b) f(c) eists; (c) lim f() = f(c)!c If one or more of the three conditions fails, then f is DISCONTINUOUS at = c. A function is CONTINUOUS ON THE OPEN INTERVAL (a, b) if it is continuous at each point on the interval. 2. ONE-SIDED CONTINUITY A function f is CONTINUOUS ON THE LEFT AT = c if lim f() = f(c); f is CONTINUOUS ON THE RIGHT AT = c!c if lim f() = f(c).!c The function f is continuous on the closed interval [a, b] if it is continuous on the open interval (a, b), and is continuous on the right at a and continuous on the left at b. 3. CONTINUITY PROPERTIES OF SOME SPECIFIC FUNCTIONS (a) A constant function, f() = k, is continuous for all. (b) For n a positive integer, f() = n is continuous for all. (c) A polynomial function P() = a n n a n-1 n-1... a 1 a 0 is continuous for all. EXERCISE 3-2 91
(d) A rational function R() = P() Q(), P and Q polynomial functions, is continuous for all ecept those numbers = c such that Q(c) = 0. (e) For n an odd positive integer, n > 1, wherever f is continuous. n f() is continuous (f) For n an even positive integer, n f() is continuous wherever f is continuous and non-negative. 4. SIGN PROPERTIES ON AN INTERVAL (a, b) If f is continuous or (a, b) and f() 0 for all in (a, b), then either f() > 0 for all in (a, b) or f() < 0 for all in (a, b). 5. CONSTRUCTING SIGN CHARTS Given a function f: Step 1. Find all partition numbers. That is: (A) Find all numbers where f is discontinuous. (Rational functions are discontinuous for values of that make a denominator 0.) (B) Find all numbers where f() = 0. (For a rational function, this occurs where the numerator is 0 and the denominator is not 0.) Step 2. Plot the numbers found in step 1 on a real number line, dividing the number line into intervals. Step 3. Select a test number in each open interval determined in step 2, and evaluate f() at each test number to determine whether f() is positive () or negative (-) in each interval. Step 4. Construct a sign chart using the real number line in step 2. This will show the sign of f() on each open interval. [Note: From the sign chart, it is easy to find the solution for the inequality f() < 0 or f() > 0.] 1. f is continuous at = 1, since lim f() = f(1) = 2!1 f() 3 2 1 3. f is discontinuous at = 1, since lim f() f(1)!1 f() 3 2 1 1 1 1 2 3 1 1 1 2 3 2 2 3 3 92 CHAPTER 3 LIMITS AND THE DERIVATIVE
5. lim f() = 2, lim f() = -2!1!1 implies lim f() does not eist;!1 f is discontinuous at = 1, since lim f() does not eist!1 f() 3 2 1 1 1 2 3 1 2 3 7. (A) lim!1 f() = 2 (B) lim f() = 1!1 (C) lim f() does not eist (D) f(1) = 1!1 (E) No, because lim f() does not eist.!1 9. (A) lim!"2 f() = 1 (B) lim f() = 1!"2 (C) lim f() = 1 (D) f(-2) = 3!"2 (E) No, because lim f() f(-2).!"2 11. (A) lim!"3 g() = 1 (B) lim g() = 1!"3 (C) lim g() = 1 (D) g(-3) = 3!"3 (E) No, because lim g() g(-3).!"3 13. (A) lim!2 g() = 2 (B) lim g() = -1!2 (C) lim g() does not eist (D) g(2) = 2!2 (E) No, because lim g() does not eist.!2 15. f() = 3-4 is a polynomial function. Therefore, f is continuous for all [3(c)]. EXERCISE 3-2 93
17. g() = 3 is a rational function and the denominator 2 is 0 at 2 = -2. Thus, g is continuous for all ecept = -2 [3(d)]. 1 19. m() = is a rational function and the denominator (! 1)( 4) ( - 1)( 4) is 0 at = 1 or = -4. Thus, m is continuous for all ecept = 1, = -4 [3(d)]. 21. F() = 2 2 9 is a rational function and the denominator 2 9 0 for all. Thus, F is continuous for all. 23. M() =! 1 4 2! 9 is a rational function and the denominator 42-9 = 0 at = 3 2, - 3 2. Thus, M is continuous for all ecept = ± 3 2. 25. f() = " 2 if is an integer $ 1 if is not an integer (A) The graph of f is: (B) lim f() = 1 (C) f(2) = 2!2 (D) f is not continuous at = 2 since lim f() f(2).!2 (E) f is discontinuous at = n for all integers n. 27. 2 - - 12 < 0 Let f() = 2 - - 12 = ( - 4)( 3). Then f is continuous for all and f(-3) = f(4) = 0. Thus, = -3 and = 4 are partition numbers. - - - - - - f()!4 8() -4-3 0 4 5 0!12 (!) 5 8() Thus, 2 - - 12 < 0 for: -3 < < 4 (inequality notation) (-3, 4) (interval notation) 94 CHAPTER 3 LIMITS AND THE DERIVATIVE
29. 2 21 > 10 or 2-10 21 > 0 Let f() = 2-10 21 = ( - 7)( - 3). Then f is continuous for all and f(3) = f(7) = 0. Thus, = 3 and = 7 are partition numbers. - - - 0 3 4 7 8 Thus, 2-10 21 > 0 for: < 3 or > 7 (inequality notation) (-, 3) (7, ) (interval notation) f() 0 21() 4!3(!) 8 5() 31. 3 < 4 or 3-4 < 0 Let f() = 3-4 = ( 2-4) = ( - 2)( 2). Then f is continuous for all and f(-2) = f(0) = f(2) = 0. Thus, = -2, = 0 and = 2 are partition numbers. - - - - - f() -3-2 -1 0 1 2 3!3!15(!)!1 3 () 1!3(!) 3 15() Thus, 3 < 4 for: - < < -2 or 0 < < 2 (inequality notation) (-, -2) (0, 2) (interval notation) 33. 2 5 > 0! 3 Let f() = 2 5 ( 5) =. Then f is discontinuous at = 3 and! 3! 3 f(0) = f(-5) = 0. Thus, = -5, = 0, and = 3 are partition numbers. - - - - - - f()!6! 2-6 -5-1 0 1 3 4 (!) 3!1 1() 1!3(!) 4 36() Thus, 2 5! 3 > 0 for: -5 < < 0 or > 3 (inequality notation) (-5, 0) (3, ) (interval notation) EXERCISE 3-2 95
35. (A) f() > 0 on (-4, -2) (0, 2) (4, ) (B) f() < 0 on (-, -4) (-2, 0) (2, 4) 37. f() = 4-6 2 3 5 Partition numbers: 1-2.5308, 2-0.7198 (A) f() > 0 on (-, -2.5308) (-0.7198, ) (B) f() < 0 on (-2.5308, -0.7198) 3 6! 3 39. f() = 2! 1 Partition numbers: 1-2.1451, 2 = -1, 3-0.5240, 4 = 1, 5 2.6691 (A) f() > 0 on (-, -2.1451) (-1, -0.5240) (1, 2.6691) (B) f() < 0 on (-2.1451, -1) (-0.5240, 1) (2.6691, ) 41. f() = - 6 is continuous for all since it is a polynomial function. Therefore, g() = " 6 is continuous for all such that - 6 0, that is, for all in [6, ) [see 3(f)]. 43. f() = 5 - is continuous for all since it is a polynomial function. Therefore, F() = 3 5 " is continuous for all, that is, for all in (-, ). 45. f() = 2-9 is continuous for all since it is a polynomial function. Therefore, g() = 2-9 = ( - 3)( 3) 0. - - - - - -4-3 0 3 4 2 " 9 is continuous on (-, -3] [3, ). 2 " 9 is continuous for all such that f() 0!9!4 7 4 7 47. f() = 2 1 is continuous for all since it is a polynomial function. Also 2 1 1 > 0 for all. Therefore, 2 1 is continuous for all, that is, for all in (-, ). 96 CHAPTER 3 LIMITS AND THE DERIVATIVE
49. The graph of f is shown at the right. This function is discontinuous at = 1. [ lim f() does not eist.]!1 51. The graph of f is: 53. The graph of f is: This function is continuous & for all. lim f() = f(2) = 3. $ % '( "2 This function is discontinuous at = 0. $ ' lim f() = 0 f(0) = 1. %& () "0 55. f is discontinuous at = 2: f is not defined at 2, lim f() = 0 and lim 4.!2!2 57. f is discontinuous at = -1 and = 1 because f is not defined at these points. However, lim f() = 2 and lim f() = 2.!"1!1 59. (A) Yes; g is continuous on (-1, 2). (B) Since lim g() = -1 = g(-1), g is continuous from the right!"1 at = -1. (C) Since lim g() = 2 = g(2), g is continuous from the left!2 at = 2. (D) Yes; g is continuous on the closed interval [-1, 2]. 61. (A) Since lim f() = f(0) = 0, f is continuous from the right!0 at = 0. (B) Since lim f() = -1 f(0) = 0, f is not continuous from the left!0 at = 0. (C) f is continuous on the open interval (0, 1). (D) f is not continuous on the closed interval [0, 1] since lim f() = 0 f(1) = 1, i.e., f is not continuous from the left!1 at = 1. (E) f is continuous on the half-closed interval [0, 1). EXERCISE 3-2 97
f() 10 f() 10 63. intercepts: = -5, 2 10 5 5 5 5 10 65. intercepts: = -6, -1, 4 10 5 5 5 5 10 10 2 67. f() = 0 for all. This does not contradict Theorem 2 1! because f is not continuous on (-1, 3); f is discontinuous at = 1. 69. The following sketches illustrate that either condition is possible. Theorem 2 implies that one of these two conditions must occur. f() 5 f() 5 10 5 10 5 10 5 5 71. (A) P() = 0.39 if 0 < " 1 % 0.63 if 1 < " 2 $ 0.87 if 2 < " 3 1.11 if 3 < " 4 &% 1.35 if 4 < " 5 (B) (C) P is continuous at = 4.5 since P is a continuous function on (4, 5]; P() = 1.35 for 4 < 5. P is not continuous at = 4 since P() = 1.11 lim P() = 1.35. lim!4!4 73. (A) Q is defined for all real numbers whereas P is defined only for > 0. (B) For each positive integer n, P(n) = 0.37 (n - 1)(0.23) while Q(n) = 0.37 n(0.23); for each positive number, not an integer P() = Q(). 75. (A) S() = 5.00 0.63 if 0 50; S(50) = 36.50; S() = 36.50 0.45( - 50) = 14 0.45 if > 50 " Therefore, S() = 5.00 0.63 if 0 % % 50 $ 14.00 0.45 if > 50 98 CHAPTER 3 LIMITS AND THE DERIVATIVE
(B) S() $60 $40 $20 (C) S() is continuous at = 50; lim S() = lim S()!50!50 = lim S()!50 = S(50) = 36.5. 50 100 77. (A) E(s) = % ' 1000,0 " s " 10,000 & 1000 0.05(s 10,000),10,000 < s < 20,000 (' 1500 0.05(s 10,000),s $ 20,000 The graph of E is: (B) From the graph, (C) From the graph, lim E(s) = $1000 and E(10,000) = $1000. s!10,000 lim E(s) does not eist. E(20,000) = $2000. s!20,000 (D) E is continuous at 10,000; E is not continuous at 20,000. 79. (A) From the graph, N is discontinuous at t = t 2, t = t 3, t = t 4, t = t 6, and t = t 7. (B) From the graph, lim t!t 5 N(t) = 7 and N(t 5 ) = 7. (C) From the graph, lim t!t 3 N(t) does not eist; N(t 3 ) = 4. EXERCISE 3-3 Things to remember: 1. VERTICAL ASYMPTOTES The vertical line = a is a VERTICAL ASYMPTOTE for the graph of y = f() if f() or f() - as a or a - That is, if f() either increases or decreases without bound as approaches a from the right or from the left. EXERCISE 3-3 99
2. LOCATING VERTICAL ASYMPTOTES A polynomial function has no vertical asymptotes. If f() = n()/d() is a rational function, d(c) = 0 and n(c) 0, then the line = c is a vertical asymptote of the graph of f. 3. HORIZONTAL ASYMPTOTES The horizontal line y = b is a HORIZONTAL ASYMPTOTE for the graph of y = f() if lim "$ f() = b or lim f() = b. " 4. LIMITS AT INFINITY FOR POWER FUNCTIONS If p is a positive real number and k is a nonzero constant, then k (a) lim "$ p = 0 (b) lim (c) lim k p = ± (d) lim "$ " " k p = 0 k p = ± provided that p is defined for negative values of. The limits in (c) and (d) will be either - or, depending on k and p. 5. LIMITS AT INFINITY FOR POLYNOMIAL FUNCTIONS If then lim and p() = a n n a n-1 n-1 a 1 a 0, a n 0, n 1, " p() = lim a " n n = ± lim p() = lim a "$ "$ n n = ± Each limit will be either - or, depending on a n and n. 6. LIMITS AT INFINITY AND HORIZONTAL ASYMPTOTES FOR RATIONAL FUNCTIONS If f() = a m m a m "1 m "1 a 1 a 0 b n n b n "1 n "1 b 1 b 0, a m 0, b n 0 a then lim f() = lim m m " " b n n " a and lim f() = lim m m "$ " b n n (a) If m < n, then lim f() = lim f() = 0 and the line "$ y = 0 (the -ais) is a horizontal asymptote for f(). 100 CHAPTER 3 LIMITS AND THE DERIVATIVE
(b) If m = n, then lim f() = lim f() = a m " "$ b n asymptote for f(). is a horizontal (c) If m > n, then each limit will be or -, depending on m, n, a m, and b n, and f() does not have a horizontal 1. lim f() = -2 " asymptote. 5. lim f() does not eist "2 9. f() = " 5 (A) lim f() = - "5 (C) lim "5 11. f() = 2 " 4 ( " 4) 2 (A) lim "4 f() = f() does not eist. (B) lim f() = "5 (B) lim f() = "4 13. f() = 2 " 2 ( " 1)( 2) = ( " 1) " 1 (A) lim f() = lim ( 2) = 3 "1 "1 (B) lim f() = lim ( 2) = 3 "1 "1 (C) lim f() = lim ( 2) = 3 "1 "1 15. f() = 2 " 3 2 2 (A) lim f() = - "2 (C) lim "2 f() does not eist. (B) lim f() = "2 3. lim f() = - "2 7. lim "2 f() = 0 (C) lim f() = "4 = 2, provided 1 17. p() = 4 5 3 4 1 (A) lim p() = lim 4 5 = " " (B) lim p() = lim 4 5 = - "$ "$ 19. p() = 2 5 2 6 11 (A) lim p() = lim -2 6 = - " " (B) lim "$ p() = lim "$ -2 6 = - 21. f() = lim "3 1 ; f is discontinuous at = -3. 3 f() = -, lim "3 f() = ; = -3 is a vertical asymptote. EXERCISE 3-3 101
23. h() = 2 4 2 " 4 = 2 4 ; h is discontinuous at = -2, = 2. ( " 2)( 2) At = -2: h() =, lim h() = - ; = -2 is a vertical asymptote. lim "2 "2 At = 2: lim h() = -, lim h() = ; = 2 is a vertical asymptote. "2 "2 25. F() = 2 4 2 " 4. Since 2 4 0 for all, F is continuous for all ; there are no vertical asymptotes. 27. H() = 2 " 2 " 3 2 " 4 3 ( " 3)( 1) = ; H is discontinuous at = 1, = 3. ( " 3)( " 1) At = 1: lim H() = -, lim H() = ; = 1 is a vertical asymptote. "1 "1 At = 3: ( " 3)( 1) Since ( " 3)( " 1) = 1 provided 3, " 1 1& lim H() = lim % ( = 4 = 2; H does not have a vertical asymptote "3 "3 $ " 1' 2 at = 3. 8 " 16 29. T() = 4 " 8 3 16 2 = 8( " 2) 2 ( 2 " 8 " 16) T is discontinuous at = 0, = 4. = 8( " 2) 2 ( " 4) 2 At = 0: lim T() =, lim T() =, = 0 is a vertical asymptote. "0 "0 At = 4: lim T() =, lim T() = ; = 4 is a vertical asymptote. "4 "4 31. lim 4 7 " 5 " 9 = lim " 4 5 = 4 5 5 2 11 33. lim " 7 " 2 5 2 = lim " 7 = since m = 2 > n = 1 7 4 " 14 2 35. lim " 6 5 3 7 4 = lim " 6 5 = 0 since m = 4 < n = 5 102 CHAPTER 3 LIMITS AND THE DERIVATIVE