Solution manual for calculus early transcendentals 11th edition by anton

Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at rch 26, arch 26, 2009 2009 06:39 06:39 ”ISM ”ISM LT LT chapter chapter 1” 1” Sheet Sheet number number 11 Page Page number number 26 26 black black CHAPTER CHAPTER 11 Limits Limits and and Continuity Continuity Limits and Continuity EXERCISE EXERCISE SET SET 1.1 1.1 (b) Exercise (a) (a) 33 Set 1.1 (b) 33 (c) (c) 33 (d) (d) 33 (a) 30 2 (a) (a) (b) (b) (b) 00 (c) (c) (c) 00 (a) 3 (a) (a) −1 −1 (b) (b) (b) 33 (c) (d) (c) (c) does does not not exist exist 4 5 6 7 (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) −1 22 00 11 −∞ −∞ −∞ (b) (c) does not exist (b) (b) 00 (b) (c) does not exist (b) (c) (b) 00 (c) 00 (b) (c) (d) (b) (c) (b) (c) (b) (c) (d) (b) (c) (b) −∞ −∞ (c) −∞ −∞ (b) −∞ (c) −∞ (a) +∞ (a) +∞ +∞ (a) (b) +∞ (b) +∞ +∞ (b) 10 (a) +∞ 10 (a) (a) does +∞not exist (d) (c) (c) does does not not exist exist (d) (d) (d) 33 (d) (d) 22 (d) (d) 00 (d) (d) 11 (d) (d) undef (d) can undef (d) not be found from graph (c) does not −2 does−∞ not exist exist (f ) (d) −2x = 0, x = (d) (c) (e) x(d) = −2, (c) (d) (e)exist +∞ (c)−1does does not not exist (c) 2 (d) (f ) 322 (d) (g) x = −2, x = The The limit limit appears appears to to be be 2 1.986 -0.1 1.986 -0.1 (ii) (ii) (ii) (b) 11 (b) (b)−∞ (d) (d) 11 −0.1 −0.01 −0.001 0.001 0.01 0.1 −0.1 −0.01 −0.001 0.001 0.01 0.1 −0.1 −0.01 −0.001 0.001 0.01 0.1 1.9866933 1.9998667 1.9999987 1.9999987 1.9998667 1.9866933 1.9866933 1.9998667 1.9999987 1.9999987 1.9998667 1.9866933 1.9866933 1.9998667 1.9999987 1.9999987 1.9998667 1.9866933 (ii) (ii) 12 (i) 12 (i) 12 (i) (c) +∞ (c) +∞ +∞ (c) (b) −∞ −∞ (b) +∞ (b) (c) (a) 11 (a) +∞ (a) 11 (i) 11 (i) 11 (i) (d) (d) (d) 00 0.1 0.1 The limit appears to be −0.5 −0.05 −0.005 0.005 0.05 0.5 −0.5 −0.05 −0.005 0.005 0.05 0.5 −0.489669752 −0.499895842 −0.499998958 −0.499998958 −0.499895842 −0.489669752 −0.489669752 −0.499895842 −0.499998958 −0.499998958 −0.499895842 −0.489669752 (ii) Full 13 file at 13 (a) (a) -0.4896698 -0.4896698 The The limit limit appears appears to to be be −1/2 −1/2 -0.5 -0.5 -0.5 -0.5 22 0.1429 1.5 1.5 0.2105 0.5 0.5 1.1 1.1 0.3021 1.01 1.01 0.3300 The limit appears to be −1/2 1.001 1.001 0.3330 00 1.0000 0.5 0.5 0.5714 0.9 0.9 0.3690 0.99 0.99 0.3367 0.999 0.999 0.3337 Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at 13 (a) Chapter 0.1429 1.5 0.2105 1.1 0.3021 1.01 0.3300 1.001 0.3330 1.0000 0.5 0.5714 0.9 0.99 0.3690 0.3367 (b) 0.4286 1.5 1.0526 The limit is 1/3 1.1 1.01 6.344 66.33 1.001 666.3 1.0001 6666.3 50 (c) 0 −1 0.5 −1.7143 0.9 −7.0111 The limit is +∞ 0.99 0.999 −67.001 −667.0 0.9999 −6667.0 The limit is −∞ -50 14 (a) −0.25 0.5359 −0.1 0.5132 −0.001 −0.0001 0.5001 0.5000 0.0001 0.5000 0.001 0.4999 0.1 0.4881 0.6 -0.25 (b) Full file at 0.25 8.4721 0.25 0.1 20.488 0.001 2000.5 0.0001 20001 The limit is 1/2 0.25 0.4721 0.999 0.3337 Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at Exercise Set 1.1 100 0.25 (c) −0.25 −7.4641 The limit is +∞ −0.1 −0.001 −19.487 −1999.5 -0.25 −0.0001 −20000 -100 15 (a) −0.25 2.7266 −0.1 2.9552 The limit is −∞ −0.001 −0.0001 3.0000 3.0000 0.0001 3.0000 0.001 3.0000 0.1 2.9552 0.25 2.7266 -0.25 (b) 0.25 −0.5 1.7552 −0.9 6.2161 −0.99 54.87 The limit is −0.999 541.1 −1.5 −0.1415 −1.1 −1.01 −1.001 −4.536 −53.19 −539.5 60 -1.5 -60 16 (a) 1.5574 −0.5 1.0926 −0.9 1.0033 The limit does not exist −0.99 1.0000 −0.999 1.0000 −1.5 1.0926 1.5 -1.5 Full file at The limit is −1.1 1.0033 −1.01 1.0000 −1.001 1.0000 Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at Chapter (b) −0.25 1.9794 −0.1 2.4132 −0.001 2.5000 −0.0001 2.5000 0.0001 2.5000 0.001 2.5000 0.1 2.4132 0.25 1.9794 2.5 -0.25 0.25 The limit is 5/2 17 False; define f (x) = x for x = a and f (a) = a + Then limx→a f (x) = a = f (a) = a + 18 True; by 1.1.3 19 False; define f (x) = for x < and f (x) = x + for x ≥ Then the left and right limits exist but are unequal 20 False; define f (x) = 1/x for x > and f (0) = 27 msec = x2 − = x − which gets close to −2 as x gets close to −1, thus y − = −2(x + 1) or y = −2x − x+1 28 msec = x2 = x which gets close to as x gets close to 0, thus y = x 29 msec = x4 − = x3 + x2 + x + which gets close to as x gets close to 1, thus y − = 4(x − 1) or y = 4x − x−1 30 msec = x4 − = x3 −x2 +x−1 which gets close to −4 as x gets close to −1, thus y −1 = −4(x+1) or y = −4x−3 x+1 31 (a) The length of the rod while at rest (b) The limit is zero The length of the rod approaches zero as its speed approaches c 32 (a) The mass of the object while at rest (b) The limiting mass as the velocity approaches the speed of light; the mass is unbounded 3.5 33 (a) –1 2.5 The limit appears to be 3.5 (b) Full file at – 0.001 2.5 0.001 The limit appears to be Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at Exercise Set 1.2 3.5 (c) – 0.000001 0.000001 2.5 The limit does not exist Exercise Set 1.2 (a) By Theorem 1.2.2, this limit is + · (−4) = −6 (b) By Theorem 1.2.2, this limit is − · (−4) + = 13 (c) By Theorem 1.2.2, this limit is · (−4) = −8 (d) By Theorem 1.2.2, this limit is (−4)2 = 16 (e) By Theorem 1.2.2, this limit is √ (f ) By Theorem 1.2.2, this limit is =− (−4) + = 2 (a) By Theorem 1.2.2, this limit is + = (b) The limit doesn’t exist because lim f doesn’t exist and lim g does (c) By Theorem 1.2.2, this limit is −2 + = (d) By Theorem 1.2.2, this limit is + = (e) By Theorem 1.2.2, this limit is 0/(1 + 0) = (f ) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t (g) The limit doesn’t exist because (h) By Theorem 1.2.2, this limit is f (x) is not defined for < x < √ = By Theorem 1.2.3, this limit is · · = By Theorem 1.2.3, this limit is 33 − · 32 + · = 27 By Theorem 1.2.4, this limit is (32 − · 3)/(3 + 1) = 3/4 By Theorem 1.2.4, this limit is (6 · − 9)/(03 − 12 · + 3) = −3 After simplification, x4 − = x3 + x2 + x + 1, and the limit is 13 + 12 + + = x−1 After simplification, t3 + = t2 − 2t + 4, and the limit is (−2)2 − · (−2) + = 12 t+2 After simplification, x2 + 6x + x+5 = , and the limit is (−1 + 5)/(−1 − 4) = −4/5 x − 3x − x−4 Full file at Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at Chapter 10 After simplification, x2 − 4x + x−2 = , and the limit is (2 − 2)/(2 + 3) = x2 + x − x+3 11 After simplification, 2x2 + x − = 2x − 1, and the limit is · (−1) − = −3 x+1 12 After simplification, 3x2 − x − 3x + = , and the limit is (3 · + 2)/(2 · + 3) = 2x + x − 2x + 13 After simplification, t3 + 3t2 − 12t + t2 + 5t − = , and the limit is (22 + · − 2)/(22 + · 2) = 3/2 t3 − 4t t2 + 2t 14 After simplification, t3 + t2 − 5t + t+3 = , and the limit is (1 + 3)/(1 + 2) = 4/3 t − 3t + t+2 15 The limit is +∞ 16 The limit is −∞ 17 The limit does not exist 18 The limit is +∞ 19 The limit is −∞ 20 The limit does not exist 21 The limit is +∞ 22 The limit is −∞ 23 The limit does not exist 24 The limit is −∞ 25 The limit is +∞ 26 The limit does not exist 27 The limit is +∞ 28 The limit is +∞ √ √ x−9 29 After simplification, √ = x + 3, and the limit is + = x−3 30 After simplification, 31 (a) √ 4−y √ √ = + y, and the limit is + = 2− y (b) 32 (a) does not exist (c) (b) 33 True, by Theorem 1.2.2 x2 = x→0 x 34 False; e.g lim Full file at (c) Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at Exercise Set 1.2 35 False; e.g f (x) = 2x, g(x) = x, so lim f (x) = lim g(x) = 0, but lim f (x)/g(x) = x→0 x→0 x→0 36 True, by Theorem 1.2.4 37 After simplification, 38 After simplification, √ √ x+4−2 =√ , and the limit is 1/4 x x+4+2 x2 + − x =√ , and the limit is x x +4+2 39 (a) After simplification, x3 − = x2 + x + 1, and the limit is x−1 y x (b) 40 (a) After simplification, x2 − = x − 3, and the limit is −6, so we need that k = −6 x+3 (b) On its domain (all real numbers), f (x) = x − 41 (a) Theorem 1.2.2 doesn’t apply; moreover one cannot subtract infinities (b) lim+ x→0 1 − x x = lim+ x→0 x−1 x2 = −∞ 42 (a) Theorem 1.2.2 assumes that L1 and L2 are real numbers, not infinities It is in general not true that ”∞·0 = ” (b) x2 − = = for x = 0, so that lim x→0 x x + 2x x(x2 + 2x) x+2 − x x2 + 2x = a x+1−a − = and for this to have a limit it is necessary that lim (x + − a) = 0, i.e x→1 x−1 x −1 x2 − 1 x+1−2 x−1 1 a = For this value, − = = = and lim = x→1 x + x − x2 − x2 − x −1 x+1 43 For x = 1, 44 (a) For small x, 1/x2 is much bigger than ±1/x (b) 1 x+1 + = Since the numerator has limit and x2 tends to zero from the right, the limit is +∞ x x x2 45 The left and/or right limits could be plus or minus infinity; or the limit could exist, or equal any preassigned real number For example, let q(x) = x − x0 and let p(x) = a(x − x0 )n where n takes on the values 0, 1, 46 If on the contrary lim g(x) did exist then by Theorem 1.2.2 so would lim [f (x) + g(x)], and that would be a x→a x→a contradiction 47 Clearly, g(x) = [f (x) + g(x)] − f (x) By Theorem 1.2.2, lim [f (x) + g(x)] − lim f (x) = lim [f (x) + g(x) − f (x)] = x→a lim g(x) x→a Full file at x→a x→a Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at Chapter 48 By Theorem 1.2.2, lim f (x) = x→a f (x) x→a g(x) lim lim g(x) = x→a f (x) x→a g(x) lim f (x) exists x→a g(x) · = 0, since lim Exercise Set 1.3 (a) −∞ (b) +∞ (a) (b) (a) (b) −1 (a) does not exist (b) (a) + · (−5) = −12 (e) √ (b) − · (−5) + = 21 (f ) 3/(−5) = −3/5 5+3=2 (c) · (−5) = −15 (d) (−5)2 = 25 (g) (h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t √ (a) · − (−6) = 20 (b) · + · (−6) = (c) +∞ (d) −∞ (e) −42 (f ) −6/7 (h) −7/12 (g) x 10 f (x) 0.953463 100 0.995037 1000 0.999500 10000 100000 0.999950 0.999995 1000000 0.9999995 The limit appears to be x −10 f (x) −1.05409255 −100 −1000 −1.00503781 −1.00050037 −10000 −1.00005000 The limit appears to be −1 The limit is −∞, by the highest degree term 10 The limit is +∞, by the highest degree term 11 The limit is +∞ 12 The limit is +∞ 13 The limit is 3/2, by the highest degree terms 14 The limit is 5/2, by the highest degree terms 15 The limit is 0, by the highest degree terms 16 The limit is 0, by the highest degree terms 17 The limit is 0, by the highest degree terms 18 The limit is 5/3, by the highest degree terms 19 The limit is −∞, by the highest degree terms 20 The limit is +∞, by the highest degree terms 21 The limit is −1/7, by the highest degree terms 22 The limit is 4/7, by the highest degree terms √ 23 The limit is −5/8 = − /2, by the highest degree terms Full file at −100000 −1.0000050 −1000000 −1.00000050 Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at Exercise Set 1.3 24 The limit is 25 26 √ √ 27 29 30 3/2 , by the highest degree terms − x22 √ 5x2 − = when x < The limit is − x+3 −1 − x 5− x2 x 5x2 − = x+3 1+ 2−y − y2 + 7+ 6y 2−y 28 + 6y = = y2 y √ √ when y < The limit is 1/ +6 −1 y2 when x > The limit is +6 √ when y > The limit is −1/ √ + x13 √ 3x4 + x = when x < The limit is x2 − − x2 √ + x13 √ 3x4 + x = when x > The limit is x2 − − x2 √ x2 + + x 31 lim ( + − x) √ = lim √ = 0, by the highest degree terms x→+∞ x2 + + x x→+∞ x2 + + x √ x2 − 3x + x −3x 32 lim ( x2 − 3x − x) √ = lim √ = −3/2, by the highest degree terms x→+∞ x2 − 3x + x x→+∞ x2 − 3x + x x2 33 False; if x/2 > 1000 then 1000x < x2 /2, x2 − 1000x > x2 /2, so the limit is +∞ 34 False; y = is a horizontal asymptote for the curve y = ex yet lim ex does not exist x→+∞ 35 True: for example f (x) = sin x/x crosses the x-axis infinitely many times at x = nπ, n = 1, 2, 36 False: if the asymptote is y = 0, then lim p(x)/q(x) = 0, and clearly the degree of p(x) is strictly less than the x→±∞ degree of q(x) If the asymptote is y = L = 0, then lim p(x)/q(x) = L and the degrees must be equal x→±∞ 37 It appears that lim n(t) = +∞, and lim e(t) = c t→+∞ t→+∞ 38 (a) It is the initial temperature of the potato (400◦ F) (b) It is the ambient temperature, i.e the temperature of the room 39 (a) +∞ 40 (a) 41 (b) −5 (b) −6 lim p(x) = +∞ When n is even, lim p(x) = +∞; when n is odd, lim p(x) = −∞ x→−∞ x→+∞ 42 (a) p(x) = q(x) = x 43 (a) No (b) p(x) = x, q(x) = x2 x→+∞ (c) p(x) = x2 , q(x) = x (d) p(x) = x + 3, q(x) = x (b) Yes, tan x and sec x at x = nπ + π/2 and cot x and csc x at x = nπ, n = 0, ±1, ±2, 44 If m > n the limit is zero If m = n the limit is cm /dm If n > m the limit is +∞ if cn dm > and −∞ if cn dm < Full file at Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at 10 Chapter 45 (a) If f (t) → +∞ (resp f (t) → −∞) then f (t) can be made arbitrarily large (resp small) by taking t large enough But by considering the values g(x) where g(x) > t, we see that f (g(x)) has the limit +∞ too (resp limit −∞) If f (t) has the limit L as t → +∞ the values f (t) can be made arbitrarily close to L by taking t large enough But if x is large enough then g(x) > t and hence f (g(x)) is also arbitrarily close to L (b) For lim the same argument holds with the substitutiion ”x decreases without bound” instead of ”x increases x→−∞ without bound” For lim− substitute ”x close enough to c, x < c”, etc x→c 46 (a) If f (t) → +∞ (resp f (t) → −∞) then f (t) can be made arbitrarily large (resp small) by taking t small enough But by considering the values g(x) where g(x) < t, we see that f (g(x)) has the limit +∞ too (resp limit −∞) If f (t) has the limit L as t → −∞ the values f (t) can be made arbitrarily close to L by taking t small enough But if x is large enough then g(x) < t and hence f (g(x)) is also arbitrarily close to L (b) For lim the same argument holds with the substitutiion ”x decreases without bound” instead of ”x increases x→−∞ without bound” For lim− substitute ”x close enough to c, x < c”, etc x→c 47 t = 1/x, lim f (t) = +∞ t→+∞ 48 t = 1/x, lim f (t) = t→−∞ 49 t = csc x, lim f (t) = +∞ t→+∞ 50 t = csc x, lim f (t) = t→−∞ , so x−2 The only vertical asymptote is at x = 51 After a long division, f (x) = x + + 15 lim (f (x) − (x + 2)) = and f (x) is asymptotic to y = x + x→±∞ y y=x+2 –12 –6 –3 x 15 x=2 –9 –15 , so x The only vertical asymptote is at x = 52 After a simplification, f (x) = x2 − + y y = x2 – 1 –4 –2 lim (f (x) − (x2 − 1)) = and f (x) is asymptotic to y = x2 − x→±∞ x –2 , so lim (f (x)−(−x2 +1)) = and f (x) is asymptotic to y = −x2 +1 x→±∞ x−3 The only vertical asymptote is at x = 53 After a long division, f (x) = −x2 +1+ Full file at Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ 28 Chapter y c/2 y c/2 x x –0.5 0.5 – c/2 13 (a) – c/2 (b) 14 42 = 22 + 32 − 2(2)(3) cos θ, cos θ = −1/4, θ = cos−1 (−1/4) ≈ 104◦ 15 (a) x = π − sin−1 (0.37) ≈ 2.7626 rad (b) θ = 180◦ + sin−1 (0.61) ≈ 217.6◦ 16 (a) x = π + cos−1 (0.85) ≈ 3.6964 rad (b) θ = − cos−1 (0.23) ≈ −76.7◦ 17 (a) sin−1 (sin−1 0.25) ≈ sin−1 0.25268 ≈ 0.25545; sin−1 0.9 > 1, so it is not in the domain of sin−1 x (b) −1 ≤ sin−1 x ≤ is necessary, or −0.841471 ≤ x ≤ 0.841471 18 sin 2θ = gR/v = (9.8)(18)/(14)2 = 0.9, 2θ = sin−1 (0.9) or 2θ = 180◦ − sin−1 (0.9) so θ = 21 sin−1 (0.9) ≈ 32◦ or θ = 90◦ − 21 sin−1 (0.9) ≈ 58◦ The ball will have a lower parabolic trajectory for θ = 32◦ and hence will result in the shorter time of flight 19 20 lim sin−1 x→+∞ x − 2x = sin−1 x x→+∞ − 2x lim = sin−1 − π =− lim cos(2 tan−1 x) = cos( lim tan−1 x) = cos(2(π/2)) = −1 x→+∞ x→+∞ 21 False; the range of sin−1 is [−π/2, π/2], so the equation is only true for x in this range 22 False; it is the interval −π/2 < x < π/2 23 True; the line y = π/2 is a horizontal asymptote as x → ∞ and as x → −∞ 24 Let g(x) = f −1 (x) and h(x) = f (x)/x when x = and h(0) = L Then lim h(x) = L = h(0), so h is continuous x→0 at x = Apply Theorem 1.5.5 to h ◦ g to obtain that on the one hand h(g(0)) = L, and on the other h(g(x)) = f (g(x)) x , x = 0, and lim h(g(x)) = h(g(0)) Since f (g(x)) = x and g = f −1 this shows that lim −1 = L x→0 x→0 f g(x) (x) 25 lim x→0 x sin x = = lim sin−1 x x→0 x tan−1 x x x = lim = ( lim cos x) lim = x→0 x→0 tan x x→0 x→0 sin x x 26 tan(tan−1 x) = x, so lim sin−1 5x 5x = lim = x→0 x→0 sin 5x 5x 27 lim sin−1 (x − 1) x−1 lim = lim = x→1 x + x→1 x−1 x→1 sin(x − 1) 28 lim Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ Exercise Set 1.7 c 29 y c/2 y c/2 x x –10 29 (a) 10 (b) The domain of cot−1 x is (−∞, +∞), the range is (0, π); the domain of csc−1 x is (−∞, −1] ∪ [1, +∞), the range is [−π/2, 0) ∪ (0, π/2] 30 (a) y = cot−1 x; if x > then < y < π/2 and x = cot y, tan y = 1/x, y = tan−1 (1/x); if x < then π/2 < y < π and x = cot y = cot(y − π), tan(y − π) = 1/x, y = π + tan−1 x (b) y = sec−1 x, x = sec y, cos y = 1/x, y = cos−1 (1/x) (c) y = csc−1 x, x = csc y, sin y = 1/x, y = sin−1 (1/x) 31 (a) 55.0◦ (b) 33.6◦ 32 θ = α − β, cot α = (c) 25.8◦ x x x x and cot β = so θ = cot−1 − cot−1 a+b b a+b b a ␪ b ␤ ␣ x 33 (a) If γ = 90◦ , then sin γ = 1, 0.93023374 so h ≈ 21.1 hours − sin2 φ sin2 γ = − sin2 φ = cos φ, D = tan φ tan λ = (tan 23.45◦ )(tan 65◦ ) ≈ (b) If γ = 270◦ , then sin γ = −1, D = − tan φ tan λ ≈ −0.93023374 so h ≈ 2.9 hours 34 (b) θ = sin−1 R 6378 = sin−1 ≈ 23◦ R+h 16, 378 35 y = when √ x2 = 6000v /g, x = 10v −1 θ = tan (3/ 30) ≈ 29◦ √ √ 60/g = 1000 30 for v = 400 and g = 32; tan θ = 3000/x = 3/ 30, 36 (a) Let θ = sin−1 (−x) then sin θ = −x, −π/2 ≤ θ ≤ π/2 But sin(−θ) = − sin θ and −π/2 ≤ −θ ≤ π/2 so sin(−θ) = −(−x) = x, −θ = sin−1 x, θ = − sin−1 x (b) Proof is similar to that in part (a) 37 (a) Let θ = cos−1 (−x) then cos θ = −x, ≤ θ ≤ π But cos(π − θ) = − cos θ and ≤ π − θ ≤ π so cos(π − θ) = x, π − θ = cos−1 x, θ = π − cos−1 x (b) Let θ = sec−1 (−x) for x ≥ 1; then sec θ = −x and π/2 < θ ≤ π So ≤ π − θ < π/2 and π − θ = sec−1 sec(π − θ) = sec−1 (− sec θ) = sec−1 x, or sec−1 (−x) = π − sec−1 x 38 (a) sin−1 x = tan−1 √ x (see figure) − x2 Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ 30 Chapter x sin –1 x !1 – x (b) sin−1 x + cos−1 x = π/2; cos−1 x = π/2 − sin−1 x = π/2 − tan−1 √ 39 tan(α + β) = tan α + tan β , − tan α tan β tan(tan−1 x + tan−1 y) = so x+y tan(tan−1 x) + tan(tan−1 y) = − xy − tan(tan−1 x) tan(tan−1 y) tan−1 x + tan−1 y = tan−1 40 (a) tan−1 41 sin(sec x+y − xy 1 1/2 + 1/3 + tan−1 = tan−1 = tan−1 = π/4 − (1/2) (1/3) (b) tan−1 tan−1 x − x2 1 1/3 + 1/3 = tan−1 + tan−1 = tan−1 = tan−1 , 3 − (1/3) (1/3) 1 3/4 + 1/7 + tan−1 = tan−1 + tan−1 = tan−1 = tan−1 = π/4 7 − (3/4) (1/7) −1 −1 x) = sin(cos (1/x)) = x 1− = √ x2 − |x| Exercise Set 1.8 (a) −4 (b) (a) 1/16 (c) 1/4 (b) (c) 1/3 (a) 2.9691 (b) 0.0341 (a) 1.8882 (b) 0.9381 (a) log2 16 = log2 (24 ) = (b) log2 (a) log10 (0.001) = log10 (10−3 ) = −3 (a) −0.5229 10 (a) (b) = log2 (2−5 ) = −5 log10 (104 ) = (c) log4 = (c) ln(e3 ) = (d) log9 = log9 (91/2 ) = 1/2 √ (d) ln( e) = ln(e1/2 ) = 1/2 (b) −0.3011 (a) 1.3655 (a) ln a + 32 (b) 1.1447 1 ln b + ln c = 2r + s/2 + t/2 2 ln c − ln a − ln b = t/3 − r − s Full file at https://TestbankDirect.eu/ (b) ln b − ln a − ln c = s − 3r − t (b) (ln a + ln b − ln c) = r/2 + 3s/2 − t Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ Exercise Set 1.8 11 (a) + log x + log(x − 3) ln(x2 + 1) log |x + 2| − log | cos 5x| when x < −2 and cos 5x < or when x > −2 and cos 5x > 12 (a) 1 ln(x2 + 1) − ln(x3 + 5) 2 (b) 13 log 24 (16) = log(256/3) 14 log √ 15 ln (b) ln |x| + ln(sin x) − 31 √ x − log(sin3 2x) + log 100 = log √ 100 x sin3 2x x(x + 1)2 cos x 16 + x = 103 = 1000, x = 999 17 √ x = 10−1 = 0.1, x = 0.01 18 x2 = e4 , x = ±e2 19 1/x = e−2 , x = e2 20 x = 21 2x = 8, x = 22 ln 4x − ln x6 = ln 2, ln √ 4 = ln 2, = 2, x5 = 2, x = x5 x 23 ln 2x2 = ln 3, 2x2 = 3, x2 = 3/2, x = 24 ln 3x = ln 2, x ln = ln 2, x = 3/2 (we discard − 3/2 because it does not satisfy the original equation) ln ln 25 ln 5−2x = ln 3, −2x ln = ln 3, x = − ln ln 26 e−2x = 5/3, −2x = ln(5/3), x = − ln(5/3) 27 e3x = 7/2, 3x = ln(7/2), x = ln(7/2) 28 ex (1 − 2x) = so ex = (impossible) or − 2x = 0, x = 1/2 29 e−x (x + 2) = so e−x = (impossible) or x + = 0, x = −2 30 With u = e−x , the equation becomes u2 − 3u = −2, so (u − 1)(u − 2) = u2 − 3u + = 0, and u = or Hence x = − ln(u) gives x = or x = − ln Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ 32 Chapter y x 31 (a) Domain: all x; range: y > −1 –2 y x –4 –4 (b) Domain: x = 0; range: all y y x –2 32 (a) Domain: x > 2; range: all y y x (b) Domain: all x; range: y > –2 y x -4 -4 33 (a) Domain: x = 0; range: all y y x (b) Domain: all x; range: < y ≤ Full file at https://TestbankDirect.eu/ -2 Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ Exercise Set 1.8 33 y x –1 –10 34 (a) Domain: all x; range: y < y x –1 (b) Domain: x > 1; range: all y 35 False The graph of an exponential function passes through (0, 1), but the graph of y = x3 does not 36 True For any b > 0, b0 = 37 True, by definition 38 False The domain is the interval x > 39 log2 7.35 = (log 7.35)/(log 2) = (ln 7.35)/(ln 2) ≈ 2.8777; log5 0.6 = (log 0.6)/(log 5) = (ln 0.6)/(ln 5) ≈ −0.3174 10 40 –5 41 –3 42 (a) Let X = logb x and Y = loga x Then bX = x and aY = x so aY = bX , or aY /X = b, which means loga b = Y /X loga x loga x Substituting for Y and X yields = loga b, logb x = logb x loga b (b) Let x = a to get logb a = (loga a)/(loga b) = 1/(loga b) so (loga b)(logb a) = Now (log2 81)(log3 32) = (log2 [34 ])(log3 [25 ]) = (4 log2 3)(5 log3 2) = 20(log2 3)(log3 2) = 20 43 x ≈ 1.47099 and x ≈ 7.85707 44 x ≈ ±0.836382 Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ 34 Chapter √ (b) y = ( 2)x 45 (a) No, the curve passes through the origin (c) y = 2−x = (1/2)x √ (d) y = ( 5)x –1 46 (a) As x → +∞ the function grows very slowly, but it is always increasing and tends to +∞ As x → 1+ the function tends to −∞ y x (b) –5 47 log(1/2) < so log(1/2) < log(1/2) 48 Let x = logb a and y = logb c, so a = bx and c = by First, ac = bx by = bx+y or equivalently, logb (ac) = x + y = logb a + logb c Second, a/c = bx /by = bx−y or equivalently, logb (a/c) = x − y = logb a − logb c Next, ar = (bx )r = brx or equivalently, logb ar = rx = r logb a Finally, 1/c = 1/by = b−y or equivalently, logb (1/c) = −y = − logb c 49 1−0 − ex = = x→−∞ + ex 1+0 lim 50 Divide the numerator and denominator by ex : − ex e−x − 0−1 = lim = = −1 x→+∞ + ex x→+∞ e−x + 0+1 51 Divide the numerator and denominator by ex : 1+0 + e−2x = = x→+∞ − e−2x 1−0 lim 52 Divide the numerator and denominator by e−x : lim e2x + 0+1 = = −1 x→−∞ e2x − 0−1 lim 53 The limit is −∞ 54 The limit is +∞ 55 56 x+1 (x + 1)x = + , so lim = e from Figure 1.3.4 x→+∞ x x xx 1+ x −x = 1 + x1 x, so the limit is e−1 57 t = 1/x, lim f (t) = +∞ t→+∞ Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ Exercise Set 1.8 35 58 t = 1/x, lim f (t) = t→−∞ 59 t = csc x, lim f (t) = +∞ t→+∞ 60 t = csc x, lim f (t) = t→−∞ 61 Let t = ln x Then t also tends to +∞, and t + ln ln 2x = , so the limit is ln 3x t + ln 62 With t = x − 1, [ln(x2 − 1) − ln(x + 1)] = ln(x + 1) + ln(x − 1) − ln(x + 1) = ln t, so the limit is +∞ 63 Set t = −x, then get lim 1+ t→−∞ 64 With t = x/2, lim x→+∞ 1+ t t = e by Figure 1.3.4 x x = lim [1 + 1/t] t→+∞ 65 From the hint, lim bx = lim e(ln b)x x→+∞ x→+∞ t = e2  if b < 1,    if b = 1, =    +∞ if b > 66 It suffices by Theorem 1.1.3 to show that the left and right limits at zero are equal to e (a) lim (1 + x)1/x = lim (1 + 1/t)t = e x→+∞ t→0+ (b) lim (1 + x)1/x = lim (1 + 1/t)t = e x→−∞ t→0− v 200 160 120 80 40 t 67 (a) 12 16 20 (b) lim v = 190 − lim e−0.168t = 190, so the asymptote is v = c = 190 ft/sec t→∞ t→∞ (c) Due to air resistance (and other factors) this is the maximum speed that a sky diver can attain 68 (a) p(1990) = 525/(1 + 1.1) = 250 (million) 500 P 250 t (b) 1920 2000 2080 Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ 36 (c) lim p(t) = t→∞ Chapter 525 = 525 (million) + 1.1 limt→∞ e−0.02225(t−1990) (d) The population becomes stable at this number 69 (a) n −n + 10 1.01 + 10n 101 −n 1+10n (1 + 10 ) 2.7319 The limit appears to be e 1.001 1001 2.7196 1.0001 10001 2.7184 1.00001 1.000001 1.0000001 100001 1000001 10000001 2.7183 2.71828 2.718282 (b) This is evident from the lower left term in the chart in part (a) (c) The exponents are being multiplied by a, so the result is ea 70 (a) f (−x) = 1− (b) lim x x→−∞ 1+ −x x = x−1 x −x x = lim f (−x) = x→+∞ = x x−1 x , f (x − 1) = x x→+∞ x − lim x x−1 x−1 = x−1 x f (−x) lim f (x − 1) = lim f (x − 1) = e x→+∞ x→+∞ 71 75e−t/125 = 15, t = −125 ln(1/5) = 125 ln ≈ 201 days 72 (a) If t = 0, then Q = 12 grams (b) Q = 12e−0.055(4) = 12e−0.22 ≈ 9.63 grams (c) 12e−0.055t = 6, e−0.055t = 0.5, t = −(ln 0.5)/(0.055) ≈ 12.6 hours 73 (a) 7.4; basic (b) 4.2; acidic (c) 6.4; acidic (d) 5.9; acidic 74 (a) log[H + ] = −2.44, [H + ] = 10−2.44 ≈ 3.6 × 10−3 mol/L (b) log[H + ] = −8.06, [H + ] = 10−8.06 ≈ 8.7 × 10−9 mol/L 75 (a) 140 dB; damage (b) 120 dB; damage (c) 80 dB; no damage (d) 75 dB; no damage 76 Suppose that I1 = 3I2 and β1 = 10 log10 I1 /I0 , β2 = 10 log10 I2 /I0 Then I1 /I0 = 3I2 /I0 , log10 I1 /I0 = log10 3I2 /I0 = log10 + log10 I2 /I0 , β1 = 10 log10 + β2 , β1 − β2 = 10 log10 ≈ 4.8 decibels 77 Let IA and IB be the intensities of the automobile and blender, respectively Then log10 IA /I0 = and log10 IB /I0 = 9.3, IA = 107 I0 and IB = 109.3 I0 , so IB /IA = 102.3 ≈ 200 78 First we solve 120 = 10 log(I/I0 ) to find the intensity of the original sound: I = 10120/10 I0 = 1012 · 10−12 = (2/3)n 2 W/m Hence the intensity of the n’th echo is (2/3)n W/m and its decibel level is 10 log = 10−12 11 10(n log(2/3) + 12) Setting this equal to 10 gives n = − ≈ 62.5 So the first 62 echoes can be heard log(2/3) 79 (a) log E = 4.4 + 1.5(8.2) = 16.7, E = 1016.7 ≈ × 1016 J (b) Let M1 and M2 be the magnitudes of earthquakes with energies of E and 10E, respectively Then 1.5(M2 − M1 ) = log(10E) − log E = log 10 = 1, M2 − M1 = 1/1.5 = 2/3 ≈ 0.67 80 Let E1 and E2 be the energies of earthquakes with magnitudes M and M + 1, respectively Then log E2 − log E1 = log(E2 /E1 ) = 1.5, E2 /E1 = 101.5 ≈ 31.6 Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ Chapter Review Exercises 37 Chapter Review Exercises (a) (b) Does not exist (h) 2 (a) Use (a) (d) (e) 0.001 4.0000213 0.01 4.0021347 (f ) (g) (i) 1/2 x f (x) 2.00001 0.250 For x = 2, f (x) = (b) (c) Does not exist x f (x) 2.0001 0.250 2.001 0.250 2.01 0.249 2.1 0.244 2.5 0.222 , so the limit is 1/4 x+2 -0.01 4.0021347 -0.001 4.0000213 -0.0001 4.0000002 0.0001 4.0000002 sin 4x sin 4x tan 4x = = · ; the limit is x x cos 4x cos 4x 4x x f (x) -0.01 0.402 -0.001 0.405 -0.0001 0.405 0.0001 0.406 0.001 0.406 3.01 5.564 3.1 5.742 0.01 0.409 y 0.5 x (b) -1 x f (x) 2.9 5.357 The limit is For x = 1, 2.99 5.526 2.999 5.543 3.001 5.547 (−1)3 − (−1)2 = −1 − x3 − x2 x3 − x2 = x2 , so lim = x→1 x − x−1 If x = −3 then 3 3x + = with limit − x2 + 4x + x+1 The limit is −∞ By the highest degree terms, the limit is 10 √ 25 32 = 3 √ √ x2 + − x2 + + x2 x2 + − 1 √ √ √ · = = , so lim = lim √ = 2 2 2 x→0 x→0 x x x +4+2 x ( x + + 2) x +4+2 x +4+2 11 (a) y = 12 (a) √ 5, no limit, (b) None √ 10, √ (c) y = 10, no limit, +∞, no limit (b) −1, +1, −1, −1, no limit, −1, +1 Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ 38 Chapter 13 If x = 0, then sin 3x = cos 3x, and the limit is tan 3x 14 If x = 0, then x sin x + cos x x · = (1 + cos x), so the limit is − cos x + cos x sin x 15 If x = 0, then sin(kx) 3x − sin(kx) =3−k , so the limit is − k x kx − cos θ θ 16 lim tan θ→0 = tan − cos θ θ→0 θ lim = tan − cos2 θ θ→0 θ(1 + cos θ) lim = tan lim θ→0 sin θ sin θ · θ (1 + cos θ) = 17 As t → π/2+ , tan t → −∞, so the limit in question is 18 ln(2 sin θ cos θ) − ln tan θ = ln + ln cos θ, so the limit is ln 19 1+ x 20 1+ a x −x 1+ x = bx = 1+ a x x/3 (−3) x/a (ab) , so the limit is e−3 , so the limit is eab 21 $2,001.60, $2,009.66, $2,013.62, $2013.75 23 (a) f (x) = 2x/(x − 1) 10 y x 10 (b) 24 Given any window of height centered at the point x = a, y = L there exists a width 2δ such that the window of width 2δ and height contains all points of the graph of the function for x in that interval 25 (a) lim f (x) = x→2 (b) δ = (3/4) · (0.048/8) = 0.0045 26 δ ≈ 0.07747 (use a graphing utility) 27 (a) |4x − − 1| < 0.01 means 4|x − 2| < 0.01, or |x − 2| < 0.0025, so δ = 0.0025 (b) 4x2 − − < 0.05 means |2x + − 6| < 0.05, or |x − 1.5| < 0.025, so δ = 0.025 2x − (c) |x2 − 16| < 0.001; if δ < then |x + 4| < if |x − 4| < 1; then |x2 − 16| = |x − 4||x + 4| ≤ 9|x − 4| < 0.001 provided |x − 4| < 0.001/9 = 1/9000, take δ = 1/9000, then |x2 − 16| < 9|x − 4| < 9(1/9000) = 1/1000 = 0.001 28 (a) Given (b) Given > then |4x − − 1| < > the inequality provided |x − 2| < /4, take δ = /4 4x2 − −6 < 2x − Full file at https://TestbankDirect.eu/ holds if |2x + − 6| < , or |x − 1.5| < /2, take δ = /2 Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ Chapter Review Exercises 39 29 Let = f (x0 )/2 > 0; then there corresponds a δ > such that if |x − x0 | < δ then |f (x) − f (x0 )| < , − < f (x) − f (x0 ) < , f (x) > f (x0 ) − = f (x0 )/2 > 0, for x0 − δ < x < x0 + δ 30 (a) x f (x) 1.1 0.49 1.01 0.54 1.001 0.540 1.0001 0.5403 1.00001 0.54030 1.000001 0.54030 (b) cos 31 (a) f is not defined at x = ±1, continuous elsewhere (b) None; continuous everywhere (c) f is not defined at x = and x = −3, continuous elsewhere 32 (a) Continuous everywhere except x = ±3 (b) Defined and continuous for x ≤ −1, x ≥ (c) Defined and continuous for x > 33 For x < f is a polynomial and is continuous; for x > f is a polynomial and is continuous At x = 2, f (2) = −13 = 13 = lim f (x), so f is not continuous there x→2+ 35 f (x) = −1 for a ≤ x < a+b a+b and f (x) = for ≤ x ≤ b; f does not take the value 2 36 If, on the contrary, f (x0 ) < for some x0 in [0, 1], then by the Intermediate Value Theorem we would have a solution of f (x) = in [0, x0 ], contrary to the hypothesis 37 f (−6) = 185, f (0) = −1, f (2) = 65; apply Theorem 1.5.8 twice, once on [−6, 0] and once on [0, 2] √ 38 (a) x = f (y) = (ey )2 + 1; f −1 (x) = y = ln x − = ln(x − 1) (b) x = f (y) = sin − 2y y ; f −1 (x) = y = + sin−1 x 1−x −2 The range of f consists of all x < or > , so this is also 3x 3π − 3π + 1−x −2 the domain of f −1 Hence f −1 (x) = tan ,x< or x > 3x 3π − 3π + (c) x = ; y = tan + tan−1 y 39 Draw right triangles of sides 5, 12, 13, and 3, 4, Then sin[cos−1 (4/5)] = 3/5, sin[cos−1 (5/13)] = 12/13, cos[sin−1 (4/5)] = 3/5, and cos[sin−1 (5/13)] = 12/13 12 (a) cos[cos−1 (4/5) + sin−1 (5/13)] = cos(cos−1 (4/5)) cos(sin−1 (5/13) − sin(cos−1 (4/5)) sin(sin−1 (5/13)) = − 13 33 = 13 65 (b) sin[sin−1 (4/5) + cos−1 (5/13)] = sin(sin−1 (4/5)) cos(cos−1 (5/13)) + cos(sin−1 (4/5)) sin(cos−1 (5/13)) = + 13 12 56 = 13 65 Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ 40 y Chapter c/2 c/2 y y x x x 1 y c/2 x –c/2 40 (a) (b) (c) (d) 41 y = ft = 60 in, so 60 = log x, x = 1060 in ≈ 1.58 × 1055 mi 42 y = 100 mi = 12 × 5280 × 100 in, so x = log y = log 12 + log 5280 + log 100 ≈ 6.8018 in 43 ln e2x (ex )3 + exp(ln 1) = ln e2x + ln(ex )3 + · = 3(2x) + (3 · 3)x + = 15x + 44 Y = ln(Cekt ) = ln C + ln ekt = ln C + kt, a line with slope k and Y -intercept ln C y x –2 45 (a) (b) The curve y = e−x/2 sin 2x has x−intercepts at x = −π/2, 0, π/2, π, 3π/2 It intersects the curve y = e−x/2 at x = π/4, 5π/4 and it intersects the curve y = −e−x/2 at x = −π/4, 3π/4 v 20 t 46 (a) (b) As t gets larger, the velocity v grows towards 24.61 ft/s (c) For large t the velocity approaches c = 24.61 (d) No; but it comes very close (arbitrarily close) (e) 3.009 s N 200 100 t 47 (a) 10 30 50 (b) N = 80 when t = 9.35 yrs Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ Chapter Making Connections 41 (c) 220 sheep 48 (a) The potato is done in the interval 27.65 < t < 32.71 (b) The oven temperature is always 400◦ F, so the difference between the oven temperature and the potato temperature is D = 400 − T Initially D = 325, so solve D = 75 + 325/2 = 237.5 for t, so t ≈ 22.76 49 (a) The function ln x − x0.2 is negative at x = and positive at x = 4, so by the intermediate value theorem it is zero somewhere in between (b) x = 3.654 and 332105.108 y –1 –1 x –3 50 (a) –5 If xk = ex then k ln x = x, or ln x = The steps are reversible x k (b) By zooming it is seen that the maximum value of y is approximately 0.368 (actually, 1/e), so there are two distinct solutions of xk = ex whenever k > e (c) x ≈ 1.155, 26.093 51 (a) The functions x2 and tan x are positive and increasing on the indicated interval, so their product x2 tan x is also increasing there So is ln x; hence the sum f (x) = x2 tan x + ln x is increasing, and it has an inverse y π/2 -1 y=f (x) y=x x π/2 y=f(x) (b) The asymptotes for f (x) are x = 0, x = π/2 The asymptotes for f −1 (x) are y = 0, y = π/2 Chapter Making Connections Let P (x, x2 ) be an arbitrary point on the curve, let Q(−x, x2 ) be its reflection through the y-axis, let O(0, 0) be the origin The perpendicular bisector of the line which connects P with O meets the y-axis at a point C(0, λ(x)), whose ordinate is as yet unknown A segment of the bisector is also the altitude of the triangle ∆OP C which is isosceles, so that CP = CO Using the symmetrically opposing point Q in the second quadrant, we see that OP = OQ too, and thus C is equidistant from the three points O, P, Q and is thus the center of the unique circle that passes through the three points Full file at https://TestbankDirect.eu/ Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at https://TestbankDirect.eu/ 42 Chapter Let R be the midpoint of the line segment connecting P and O, so that R(x/2, x2 /2) We start with the 2 Pythagorean Theorem OC = OR + CR , or λ2 = (x/2)2 + (x2 /2)2 + (x/2)2 + (λ − x2 /2)2 Solving for λ we obtain λx2 = (x2 + x4 )/2, λ = 1/2 + x2 /2 Replace the parabola with the general curve y = f (x) which passes through P (x, f (x)) and S(0, f (0)) Let the perpendicular bisector of the line through S and P meet the y-axis at C(0, λ), and let R(x/2, (f (x) − λ)/2) 2 be the midpoint of P and S By the Pythagorean Theorem, CS = RS + CR , or (λ − f (0))2 = x2 /4 + 2 f (x) + f (0) f (x) + f (0) − f (0) + x2 /4 + −λ , 2 x2 which yields λ = f (0) + f (x) + f (x) − f (0) (a) f (0) = 0, C(x) = + 2x2 , x2 + (y − 18 )2 = (b) f (0) = 0, C(x) = 12 (sec x + x2 ), x2 + (y − 12 )2 = 2 (c) f (0) = 0, C(x) = x2 + |x|2 , x + y = (not a circle) |x| (d) f (0) = 0, C(x) = x(1 + sin2 x) ,x + y − sin x (e) f (0) = 1, C(x) = x2 − sin2 x , x + y = cos x − (f ) f (0) = 0, C(x) = x2 g(x) + ,x + y − 2g(x) 2g(0) (g) f (0) = 0, C(x) = 2 = = 2g(0) + x6 , limit does not exist, osculating circle does not exist x2 Full file at https://TestbankDirect.eu/ ... be Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at Exercise Set 1.2 3.5 (c) – 0.000001 0.000001 2.5 The limit does not exist Exercise Set 1.2 (a) By Theorem... does not exist (c) (b) 33 True, by Theorem 1.2.2 x2 = x→0 x 34 False; e.g lim Full file at (c) Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at Exercise Set... limit is −5/8 = − /2, by the highest degree terms Full file at −100000 −1.0000050 −1000000 −1.00000050 Solution Manual for Calculus Early Transcendentals 11th Edition by Anton Full file at Exercise