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1. 1. CHAPTER 1 LIMITS AND CONTINUITY 1.1 RATE OF CHANGE AND LIMITS 1_ (a) Ag= f(3)-f(2)=2s-9=19 (b) Ar_f(1)-f(-1) 2-0 _í 2-: :1 Ax “'_“"3-2 1 E"_1_(_"_1) 2 2 <A1í: R(2)-R(0)= q/8+ -x/ I _1 'A9 f_ *ñ* 2 Ah_h(ãrn)_h(ã)_-l-l_ g Ah_h(§)_h(%)_0-/ §_-3/ ã 3°(a)m-_ír_- 1 -“›r (b)m~_z__zz_°- 1 --r- T-z 2 2 6 3 4_ (a)â! tã= = = _g (b)_â_2tã=8(: )_'tf_(7r”)")= =0 5. (a) Slope of PQ = &T1; QI( 10, 225) 65° : f” = 42.5 m/ sec 650 - 375 Q, (14, 375) O _14 = 45.83 m/ sec Q3(16.5,475) g5°jlêÍ5 = 50.00 m/ sec Q4(18,550) 65° ' 55° = 50.00 m/ sec 6. (a) Q¡(5,20) 30419 - 12 m/ sec Q2(7, 39) 3° * 39 - 13.7 m/ sec Q3(8.5,58) 550;” - 14.7 m/ sec é 0495,72) @Líl- 16 m/ sec (b) Approximately 16 m/ sec 7. A plot of the data shows that the slope of the secam: between t = 0.8 sec and t = 1.0 sec underestimates the instantaneous velocity (i. e., the slope of the tangent) at t: = 1.0 sec, whereas the slope of the aecant between t = 1.0 sec and t = 1.2 sec overestimates it.
2. 2. 88 Chapter 1 Limits and Continuity 10. ll. 12. 13. 14. 15. = 23.55 ft/ sec Lower bound: a = 1.0 - 0.8 Upper bound: b = = 28.85 ft/ sec v(1) z 5gb = = 26.20 ft/ sec There are many graphs that would be correct. One possible solution looks like this: distance travelad time (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x -› 1. (b) 1 (c) 0 (a) 0 (b) -1 (c) Does not exist. As t approaches O from the left, f(t) approaches -1. As t approaches 0 from the right, f(t) approaches 1. There is no single number L that f(t) gets arbitrarily close to as t -› 0. (a) True (b) True (c) False (d) False (e) False (f) True É? ) ? Fgalse Éh; 'Ffalse (c) True rue e rue lina rf¡ does not exist because à = 35-¡ = 1 if x > 0 and '-35 = _ix = -1 if x < 0. As x approaches 0 from the left, X-O à approaches -1. As x approaches 0 from the right, É¡ approaches 1. There is no single number L that all the function values get arbitrarily close to as x -› 0. 1 x - 1 from the right, the values become increasingly large and positive. There is no one number L that all the As x approaches 1 from the left, the values of become increasingly large and negative. As x approaches 1 function values get arbitrarily close to as x -› 1, so à does not exit. Nothin can be said about lim f x because the existence of a. limit as x -› x does not depend on how the func- g x-›x0 0 tion is defined at x0. In order for a limit to exist, f(x) must be arbitrarily close to a single real number L When x is close enough to x0. That is, the existence of a limit depends on the values of f(x) for x near x0, not on the definition of f(x) at x0 itself.
3. 3. Section 1.1 Rate of Change and Limits 89 16. Nothing can be aid. In order for lirrõ f(x) to exist, f(x) must close to a single value for x near 0 regardless of the value f(0) itself. x# 17. No, the definition does not require that f be defined at x = 1 in order for a limiting value to exist there. If f(1) is defined, it can be any real number, so we can conclude nothing about f( 1) from f(x) = 5. 18. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If lirrâ f(x) exists, its value may be some number other than f(1) = 5. We can conclude nothing about Imã f(x), X_0 X_¡ whether it exists or what its value is if it does exist, from knowing the value of f(1) alone. 19. (a) f(x) = (x2 - 9)/ (x + 3) x -3.1 -3.01 -3.001 -3.0001 -3.00001 -3.000001 f(x) -6.1 -6.01 -6.001 -6.0001 -6.00001 -6.000001 x -2.9 -2.99 -2.999 -2.9999 -2.99999 -2999999 f(x) -5.9 -5.99 -5.999 -5.9999 -5.99999 -5.999999 The estimate is lima f(x) = -6. X-O- (b) y fm - u' -sn/ (x + a) (c) r(x)= g= = x-3irx\$-3, and lima (x-3)= -3-3=-6. x4_ 20. (a) g(x)= (x2-2)/ (x-, /í) 1.4 2.81421 1.41 2.82421 1.414 2.82821 1.4142 2828413 1.41421 2828423 1.41421.? 2828426 The estimate is lim g(x) = Êx/ í. x-v 2 (b) : w - a' Â : m: - v! )
4. 4. 90 Chapter 1 Limits and Continuity X-Jê- x- 2) 21. (a) G(x) = (x+6)/ (x2+4x-12) (c) g<x›= 9-2 -(*+“5“*”“5)= x+~/ ãarx; e«ê, ana 1ny<x+«5›= «5+~/ ê=2«â. -5.9 -5.99 -5.999 -59999 -5.99999 -5.999999 -0.126582 -01251564 -0.1250156 401250016 -0.12500016 -0.12500002 x -6.1 -6.01 -6.001 -6.000l -6.00001 -6.000001 G(x) -0.123457 -0.1248439 -0.l249844 -01249984 -0.12499984 -0.12499998 The estimate is lima G(x) = -0.125. x-t- (b) , ou) - (x +010# + 4x - 11) x-p-G X-2 "6_ (qcwüi-_ÀÉÀ_ _§íifx; e-6,and lim 1 = 1 = _§= -0.125. x2+4x-12)_(x+6)(X'2)= x 22. (a) h(x) = (x2-2x-3)/ (x2-4x+3) x 2.9 2.99 2.999 2.9999 2.99999 2.99999!) h(x) 2052631 2.00502?) 2.000500 2000050 2000005 20000005 x 3.1 3.01 3.001 3.0001 3.00001 3000001 h(x) 1952380 1995024 1999500 1999950 1999995 1999999 The estimate is h(x) = 2.
5. 5. Section 1.1 Rate of Change and Limits 91 (b) hçg). w-zx-3)l(t'°4¡+3) _ 2-2 -3_(X-3)(X+1)_x+1. - x+1_3+1_4_ 0°) h(x>-írzzfrg-íxfm__n-r1lfx*3=wd 1% X-l-j-Ê-l 23. (a. ) g(9) = (sin 0)/0 .1 .01 .001 .0001 .00001 .000001 .998334 999983 .999999 .999999 _ .999999 .999999 0 -.1 -.01 -.001 -.0001 -.00001 -.000001 999983 .999999 .999999 .999999 999999 24. (a) G(t) = (1 -cos t)/ t2 t .1 .01 .001 .0001 .00001 .000001 G(t) .499583 .499995 .499999 .5 .5 .5 t -.1 -.01 -.001 -.0001 -.00001 -.000001 .499995 .499999 .5 G(t) = 0.5
6. 6. 92 Chapter 1 Limits and Continuity (b) «na AQUI ¡un um¡ Graph is NOT TO SCALE 25. (a) f(x)= x1/(1_x) x .9 .99 .999 .9999 .99999 .999999 f x .348678 .366032 .367695 .367861 .367878 .367879 x 1.1 1.01 1.001 1.0001 1.00001 1900001 f(x) .385543 .369711 .368063 .367898 .367881 .367880 lim f(x) z 0.36788 x-àl (b) 2.710¡ 2.11035 | I(x-| ) 04m 1.9991 2.7lIl\$ 0.9999 26. (a. ) f(x) = (3*-1)/ x .0001 .00001 .000001 1998672 1998618 1998612 1.l61231 1.104669 1999215 -.1 1940415 -.01 1992599 -.001 1998009 -.0001 -.00001 -.000001 1998551 1998606 1998611 lim f(x) as 1.0986 x-›0
7. 7. 27. 28. 29. 30. 31. 32. 33. 34. 35. (b) Step 1: Step 2: Step 1: Step 2: Step 1: Step 2: Step 1: Step 2: Step 1: Step 2: Step 1: Step 2: Step 1: Step 2: Step 1: Step 2: Step 1: Step 2: Section 1.1 Rate of Change and Limits 93 | x-5l<6=›-6<x-5<ô= ›-6+5<x<6+5 From the graph, -6+5=49 = › 620.1, or 6+5=5.1 = › 6:01; thus õ=0.1 in either case. |x-(-3)| <6=› -õ<x+3<ô= ›-6-3<x<ô-3 From the graph, -15-3: -3.1=›6=0.1,or 6-3: -2.9 à 6=0.1; thus 6=0.1. Ix-1I<6=›-õ<x-1<ó= ›-6+1<x<6+1 Fromthegraph, -6+l= -19¡= >6=TZ6, orô+1=%%= >6=-126; thus6=1z6. Ix-2I<ô= ›-6<x-2<ô= ›-6+2<x<6+2 From thegraph, -6+2=¡/ §=›6=2-¡/ §~0.2679, or5+2=(/ ã=›6=/ ã-2~0.2361; thus6=x/5-2. | (x+1)-5|<0.01 = ›Ix-4I< 0.01 = ›-0.01<x-4<0.01 = ›3.99<x<4.01 | x-4|<6=›-6<x-4<6=›-6+4<x<õ+4=›õ=0.01. | (2x-2)-(-6)| <0.02 = ›|2x+4|<0.02 = › ~o.02 <2x+4<0.02 = › -4.02 <2x< -3.98 = > -2.01<x< -199 | x-(-2)| <6=›-6<x+2<ó= -6-2<x<6-2=>6=0.01. | ¡/x+ -1|<0.1=>-0.1</ X+ -1<0.1=›0.9<¡/ x+1<1.1=›0.81<x+1<1.21 = › -o.19 <x<0.21 ' lx-0|<6=›-6<x<6=›õ=0.19. | /1'9'-“§-3|<1=-1<í-3<1=›2<§/ ñ›ÍI<4=›4<19-x<16 = ›-4>x-19>~16=›15>x>3or3<x<15 | x-10|<6=›-6<x-10<6=›-6+10<x<6+10. Then -6+10=3=›6:7,or6+10=15=›6=5;thusõ=5. §-â| <0.o5=›-0.05<§-â<o. o5=›0.2<§<0.3=›E>x>EorH<x<5. 2 3 3 | x-4|<6=›-6<x-4<6=›-6+4<x<6+4. Then -ô+4=13-0or6=§, or6+4=5or6=1;thus6=§-.
8. 8. 94 Chapter 1 Limits and Continuity 36. Step 1: | x°-3|<o.1=› -0.1<x2-3<0.1=>2.9<x2<3.1=› x/2.9<x<s/3.1 Step2: | x-y/ §I<õ= ›-6<x-y/ §<ô= ›-6+x/ §<x<6+x/ §. Then -6 + f . -. , /29 = › . s = Jã- ¡/2. z 0.0291, or a + Já = ,/3.1=› a = ¡/3. - 9/5 z 0.0286; thus ô = 0.0286. 37. IA - 915 0.01 = › -o. o1 5 «gy - 9 5 0.01 = › 8.99 5 13g 5 9.01 = › ; (399) 5 x2 5 ; (991) = > 2 5 x 5 29/9# or 3.384 5 x 5 3.387. To be safe, the left endpoint was rounded up and the right endpoint was rounded down. as. v= RI= §=1=| §-5|5o.1=› -o.15%-55o.1=›4.95l§'55.1 : àgzrlgõzàg: (120)(10) 51 To be safe, the left endpoint was rounded up and the right endpoint was rounded down. (120)(10) 5 R 5 _m- = › 23.53 5 R 5 24.49. 39. (a) The limit can be found by substitution. ;E f(x) = f(2) = ,M25 - = Ji = 2 (b) The graphs of y¡ = f(x), y¡ = 1.8, and ya = 2.2 are shown. [L5, 2.5] by (1.5. 2.3] The intersections of y¡ with y¡ and ya are at x z 1.7467 and x = 2.28, respectively, so we may choose any value of a. in [1.7467,2) (approximately) and any value of b in [2,2.28]. One possible answer: a = 1.75, b = 2.28. (c) The graphs of y¡ = f(x), y¡ = 1.99, and y3 = 2.01 are shown. [197, 2.03] by (1.98. 2.02] The intersections of y¡ with y¡ and y3 are at x = 1.9867 and x as 2.0134, respectively, so we may choose any value of a in [1.9867,2) and any value of b in [2,2.0134] (approximately). One possible answer: a. = 1.99, b = 2.01. "i
9. 9. Section 1.1 Rate of Change and Limits 95 _ - __ l 40. (a) _ s1n 76" _ 2 (b) The graphs of y¡ = f(x), y2 = 0.3, and y3 = 0.7 are shown. IRCYI CÍÍOI¡ 1.30 III¡ i0. l] by (0. l] The intersections of y¡ with y¡ and y3 are at. x z 0.3047 and x z 0.7754, respectively, so we may choose any value of a in [0.3047,%) and any value of b in (%,0.7754], where the inter-val endpoints are approximate. One possible answer: a. = 0.305, b = 0.775. (c) The graphs of y¡ = f(x), y¡ = 0.49, and y3 = 0.51 are shown. [o.49. 0.55] by l0.48. 0.52] The intersections of y¡ with y¡ and y3 are at x z 0.5121 and x z 0.5352, respectively, so we may choose any value of a. in [0.5121,%) and any value of b in (%,0.5352], where the interval endpoints are approximate. One possible answer: a = 0.513, b = 5.35. 41. a In three seconds, the ball falls 4.9(3)2 = 44.1 m, so its average speed is 44d = 14.7 m/ sec. T (b) The average speed over the interval from time t = 3 to time 3 + h is é; _ 4.9(3 + h)2 _ 4.9(3)2 _ 4.9(6h + h? ) At “ (3+h)-3 “Th Since lllinà (29.4 + 4.9h) = 29.4, the instantaneous speed is 29.4 m/ sec. = 29.4 + 4.9h 42. (a) y= gt*-›2o= g(42) -›g--2-0=§=1.25 (b) Average speed = í-O = 5 m/ sec. (c) If the rock had not been stopped, its average speed over the interval from time t = 4 to time t = 4 + h is Ay _ 1.25(4 + h)2 -1.25(4)' _ 1.25(8h + h2) 2T “ (4+ h) - 4 " h Since llinà (10 +1.25h) = 10, the instantaneous speed is 10 m/ sec. : 10 +1.25h
10. 10. 96 Chapter 1 Limits and Continuity 43. (a. ) x -0.001 -0.000827 -0.0001 -0.000031 -0.054402 -0.005064 (b) x 0.1 -0.054402 0.01 -0.005064 0.001 0.0001 -0.000827 -0.000031 The limit appears to be 0. 44. (a) x _0.1 _0.01 -0.001 _0.0001 0.5440 -0.5064 -0.8269 0.3056 (b) X f(x) There is no clear indication of a limit. 45. (a) x -0.1 _0.01 -0.001 -0.0001 f x 2.0567 2.2763 2.2999 2.3023 (b) x 0.1 0.01 0.001 0.0001 f(x) 2.5893 2.3293 2.3052 2.3029 The limit appears to be approximately 2.3. 0.1 -0.5440 0.01 -0.5064 0.001 0.8269 0.0001 -0.3056 -0.1 -0.0l -0.001 -0.0001 0074398 -0.009943 0.00058?) 0000021 (b) x 0.1 0.01 0.001 0.0001 -0 .074398 0009943 -0.000585 -0.00002i The limit appears to be 0. 47-50. Example CAS commands: Maple: f: =x -> (x A 4 - 81)/ (x - 3); plot (f(x), x=2.9..3.1); limit (f(x), x= -1); Mathgmatica: x0=3; f= (x A 4 - 81)/ (x - 3) Plot [f, (x, x0-0.1,x0 + 0.1]] Limit [f, x - > x0] 51-54. (values of del may vary for a specified eps): Maple: f: =¡( -> (x A 4 - 81)/ (x - 3); x0:= 'x0': eps : ='eps': L:= 'L': del: ='del':
11. 11. y1:= x -> L - eps: y2:= x -> L + eps: x0:=3: L= limit(f(x), x=x0); eps: =0.1: del: : 0.16: xmin: = x0 - 2*del: xmax : :x0 + 2*del: ymin: =L - 2-reps: ymaxz= L + 2a= epsz Section 1.2 Rules for Finding Limits 97 plot([f(x), y1(x), y2(x)], x= x0-del. .x0+del, view = [xmin. .xmax, ymin. .ymax]); Mathematica: Clear [f, x,L, eps, del] y1 : = L -eps; y2 : = L +eps; x0 = 3; f: (xA4-81)/ (x-3) Plot [f, fx, x0 - 0.2,x0 + 0.2)] L = Limit[f, x -> x0] eps = 0.1; del = 0.0015; Plot [[f, y1,y2], (x, x0 - del, x0 + del), PlotRange -> “x0 - del, x0 + del), (L - eps, L + eps)]]] 1.2 RULES FOR. FINDING LIMITS . (a) xl-Ígl_ f(x) = 3 (b) lim+ f(x) = _2 x-t3 (c) lim f(x) does not exist, because the left- x-›3 and right-hand limits are not equal. (d) f(3) = 1 . (a. ) 13x51_ f(h) = _4 (b) 1333+ f(h) = _4 (c) liga) f(h) = -4 (d) f(0) = -4 . (a) X133_ F(x) = 4 (b) lim+ F(x) = -3 x--vü (c) F(x) = does not exist because the left- and right-hand limits are not equal. (d) F(0) = 4 . (a) quotient rule (c) sum and constant multiple rules . (a) quotient rule (c) difference and constant multiple rules . (a) tgbírnl_ g(t) = 5 (b) 6113+ g(t) = 2 (c) tl_i'rn4 g(t) does not exist, because the left- and right-hand limits are not equal. (d) s(-4) = 2 . (a) x-lél-rà_ p(s) = 3 (b) lim+ p(s) = 3 x-›-2 (c) X5532 p(s) = 3 (d) p(-2) = 3 . (a) xlim_ G(x) = 1 (b) lim+ G(x) = 1 x-»Z (c) G(x) = 1 (d) G(2) = 3 (b) difference and power rules (b) power and product rules
12. 12. 98 Chapter 1 Limits and Continuity 9. 12. (a) , im f(x)s(x) = l , lLfg f(X)][ X1252 g(x)] = (5)(~2) = -10 (b) lim 2f(x)g(x) = 2[ X_VC lim f(x)][ g(x)] : - 2(5)(-2) = -20 X_5C (c) , gibi [f(x) + 3g(x)] = ,lim f(x) + a ; gb g(x) = s + s<-2) = -1 . r( ) _ ; Loc f(x) _ (d) m (a) ,133 [soou]-; ms(x›+; gx)s= -3+s= o (b) Xf(X) = x - f(x) = (4)(0) = 0 (c) ; g3 [g(x)]° = [ 33553¡ 300]” = [412 = 9 . (x) PL" g(x) _ . (a) xlirn? (22: +5): 2(-7) +5 = -14 + 5 = -9 (b) 8(t - 5)(t - 7) = 8(6 - 5)_(6 - 7) = -8 y + 2 2 + 2 4 4 1 l. íízí--É-: í-í: - (0 yg y2+5y+6 (2)°+5(2)+0 4+10+6 20 5 (d) lim 3 = 3 = 3 : ê h-›o §73b+1+1 §73(0)+1+1 §71+1 2 (a. ) lim (r3-2r2+4r+8)= (-2)3-2(-2)”+4(-2)+s = -s_s_3+s = -16 r-v-2 w : a : iã= %=% (c) ; g3 (s - yr” = [s - (4014/3 = m” = (<s›”*')4 = 2** =16 <d> (se 53-3: m e; '(3) táxi-as = tET5 = cl-l›T5 (t_2)= -5-2=_7 (b) : (1352 : xl_l. '32 É= _T2=_à (y-1)(x/ y+3+2) _ um (y-1)(y/ y+3+2)= (c) lim (y + 3) _ 4 -1 . A.
13. 13. Section 1.2 Rules for Finding Limits 99 (d) ,135 sin(à- - >= m< 144a) hm / x§+81-3= um (/ x5+8-3)(/ x§+8+3)= um (x2+8)-9 *M* x* *+1 (x+1)(Õx5+8+3) P* x+1)( x +8+3) <x+1›(x-1> _2 = à) = wing) z -0.1659 Koln- 03h- -lim ----í= lim = -- - x-l l _*"'1(x+1)(/ x2+8+3) X-°-1 / x2+8+3 3+3 3 (92+1)(o+1)(o-1)_ um (e2+1)(e+1)_(1+1)(1+1) . 94-1_ . _4 (b) àl-»Iri 93_1-«i1-›ni (9?+.9+1)(9..1) _o-. i 924.944 _ 1+1+1 _É 3-sÃ x/ Í-3 1 1' = 1' = 1' _1__= L=_ (c) x55 ã-t ; L3 (JE. .3)( t+3) n53 / §+3 / §+3 6 d Let/ í: usothat u -›0 ass-a 7r, and then rewrite and evaluate the limit as 2 ¡lliirà (7r-2u) cos(u) : lígà (w-Zm-llijfêl cos(u) = 7r-1=7r 15. (a. ) 113% (1 -§) = 1 _g = 1 and E? ) 1 = 1; by the sandwich theorem, ,130 _íz f : irão: X = 1 (b) For x ; É 0, y = (x sin x)/ (2 - 2 cos X) lies between the other two graphs in the figure, and the graphs converge as x -› 0. xsinx W575? nu›= r 16. (a. ) = lim l- lim à¡ = à- 0 : à and lim l = à; by the Sandwich theorem, x-0 2 x-o0 x-»O 2 - 1 - _ 1 111% É” X. - í (b) For all x ; É 0, the graph of f(x) = (1 - cos x)/ x2 ¡ lies between the line y = à and the pai-abala y = à- x2/ 24, and the graphs converge as x -› 0.
14. 14. 100 Chapter 1 Limits and Continuity 17 lim : Lhltlz- lim _--í1+2h+h2“1- lim M2441) . h-›O h _ h-›0 h _ . .-o 1 =1znot2+rv=2 18. lim [3(2+h)'4]'[3(2)'4]= lim @a h-›o E h-›o h 19 lim . _E3114 -2-(-2+h)_ . ..a = = ° a: _ ' _; Ê_ = _ l ' h-? O h 11.5% -ÉF 11.3% -2h(-2+h) 11.5% h(4-2h) 4 2o. lim _íV7+h'*/7= lim (ía Hh-WX” NNW): lim (7+h)'7 h-›o h h-›0 b( / ':7+h+/ i) h-. o h( / íLHHW/ ;â = lim __-h__= lim 1 = 1 h-w h(, /'-7_, _h+/ §) h-»o §77+h+§77 2 7 21. (a) False (b) True (c) False (d) True (e) True True (g) False (h) False (i) False False (k) True (l) False 22. (a) 1¡m+r(x)= §+1=2, x1ug_r(x)=3-2=1 x-o2 '* (b) No, lim f(x) does not exist because lim f(x) ; É lim_ f(x) X-t? ¡_,2+ x-›2 - _4 _ - . ..4 _ (c) xl_i5:1_f(x)-§+1-3,xh:1+f(x)_§+1_3 (d) Yes, lim f(x) = 3 because 3 = lim_ f(x) = lim f(x) x44 x_°4 x-~4+ 23. (a) No, lim+ f(x) does not exist since sin does not approach any single value as x approaches 0 x-›0 (b) xl-igi_ f(x) = xlim_ o = o -00 (c) lim f(x) does not exist because lim f(x) does not exist x-›0 x_,0+ 24. (a) Yes, lim+ g(x) = 0 by the sandwfch theorem since - x 5 g(x) 5 W when x > 0 x-oO (b) No, lixa-i_ g(x) does not exist since Ji does not exist, and therefore the function is not defined, for x < 0 x1? (c) No, ,lina g(x) does not exist since lim_ g(x) does not exist 3D c¡ ^ x 0 'JI-15. Oizcl 25. (a) domain: 0 5 x 5 2 _y- 1. mu: range: 0<y51andy=2 V 2. uz (b) f(x) exists for c belonging to (0›1)U(1›2) (c) x = 2 (d) x=0
15. 15. 26. 27. 28. 29. 30. 32. Section 1.2 Rules for Finding Limits (a) domain: -oo < x < oo range: -15 y 51 (b) f(x) exists for c belonging to (_°°1"1)U(_111)U(17°°) (c) none x, -1_<_-x<0 OI' °<*í1 - 1, x-O (d) “me y IO. x<1 or x›-1 . /x+2_ / -o.5+2_ /3_h_ ¡Jfàfs- ? +7 ' _0.5 +1 “ 1/2 “ *Á . 2x+5 _ -2 2(“2)+5 _ L_ 1313+ (X:1)(x2+x)_('2+1)((-2)§+(-2) '(2)(2)'1 um , /117+411+5-, /ã= um (s/117+411+5-/ ã) / h7+411+5+«/5 11_. o+ E 11-›o+ h ; h5+4h+5+x/ Ê : lim --_-_--(h2+4h+5)'5 = lim : ÀÊL: 0+4 = 2 h-›o+ 11(1/117+411+5+, /ã) h-›0+11(, /117+411+5+«/ ã) 15+ 5 75 um y/ §-/5h7+11h+6_ um (Jfi-/5h7+11h+6) Jã+s/5117+1111+6 h-vo' h h-°' h Jã+75h5+1111+6 = um 6-(5h2+11h+6) = lim -h(5h+11) = -(0+11)= _ 11 *“°'11(, /õ+x/5117+1111+6) *'*°'11(, /õ+x/5117+1111+6) ; M56 235 . 2 . +2 . (a) 1112+ (x+3)| ::2|= 1112+ (X-i-3)Ê: +2; (| x+2|= x+2forx>-2) X-'Ó- X-Ô- = lim (x+3)= (-2)+3=l x-›-2+ . 2 . - +2 (b) P113_ (x+3)| ::2|= x_13r_n2_ (x+3)[ (Ix+2|= -(x+2)forx<-2) = x_1§g_(x+3)(-1)= -1-2 + s) = -1 (a) = xl_i.1:1+ (lx-1I= x-1forx>1) = lim M= JE x-›1+ (b) = xÉ3íL <'xr1'= -<x-1> f°rx<1> = lim -Jí= -,/ § x-›1 _ 101
16. 16. 102 Chapter' 1 Limits and Continuity 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. f(x) exists at. those points c where x4 = x2. Thus, c4 = c¡ = › c2(1 _ g2) = 0 à c = 0, 1, or -1. Moreover, f(x) = lim x2 = 0 and liml f(x) = f(x) = 1. x-›0 x-›- Nothing can be concluded about the values of f, g, and h at x = 2. Yes, f(2) could be 0. Since the conditions of the sandwich theorem are satisñed, ;imã f(x) = -5 ; É 0. _ . f(x) _ Xl-ifx-lz f(x) _ 1312 f(x) . _ @1233-*2 í-í-%àxàêz“x>-t x-v-2 . f . f . . f . f (b) 1 = acl-IPE? :i2 = [xl-lon-lz ! E4122 = [A522 \$ à : :km2 Q = C2' (a. ) o :3-0 : [3335 “Kgõi x%(x~2)]= ,1ci3_¡z [( )(3:-2)]= ,135 [f(x)-ã] = ;E f(x)-õ = › f(x) = 5. (b) 0 -4-0 -[ li f--Q-(JÓ-ã lim (x-2) = › lim f(x) - 5 as in art (a) ~ _ x35 X - x-›2 1 x-›2 - p ' Yes. If lim+ f(x) = = X1121_ f(x), then f(x) = L. If lim+ f(x) ; é xiiãn_ f(x), then f(x) does not exist. x-m x-oa. Since f(x) = L if and only if 1im+ f(x) = L and lim_ f(x) = L, then f(x) can be found by calculating _, x-tc lim+ f(x). x ° X"'9C I= (5,5+6)= ›5<x<5+õ. Also, ¡/ x-5<c= ›x-5<c2=›x<5+ez. Chooseõ= e2 = > lim+ / X-5= O. x-›5 1=(4-6,4)=4_5<x<4. Also, ,/4-x<e= ›4-x<c“= ›x>4-c“. Choose6=c2 = > lim_ y/4-x= 0. x-v4 If f is an odd function of x, then f(-x) = -f(x). Given lim+ f(x) = 3, then lirg_ f(x) = -3. X-d x-oi) If f is an even function of x, then f(-x) = f(x). Given 1h51_ f(x) = 7 then lim f(x) = 7. However, nothing X-ó x-›-2 can be said about lim_ f(x) because we don't know lim f(x). x""2 x-t2
17. 17. Section 1.2 Rules for Finding Limits 103 The graphs suggest that lirrà g(x) = 0. X*4 (b) k(x) = sin(§) -iisxsa -g-osxs-ii The graphs suggest that lina k(x) does not exist. X_O For both g(x) and k(x), the frequency of the oscillations increases without bound as x -› 0. For g(x), the Sandwich theorem can be applied. If x > 0, -x 5 x sinGJ 5 x = > lim g(x) = 0 and if X < 0, x-›0+ x 5 x sin 5 -x = › lim g(x) = o. Therefore, ;mà g(x) = o since the left- and right-hand limits are both x-o0_ 0. For k(x), the amplitude of the oscillations remains equal to one. Therefore, k(x) cannot be kept arbitrarily close to any number by keeping x sufficiently close to 0. 44. (a. ) h(x) = x2 cos(à) The graphs suggest that linà h(x) : 0. X--ã
18. 18. 104 Chapter 1 Limits and Continuity The graphs suggest that lirrà k(x) does not exist. X--i For both h(x) and k(x), the frequency of the oscillations increases without bound as x -› 0. For h(x), the sandwich theorem can be applied: -x2 _<_ x2 coáà) < x2 à , linà g(x) = 0. For k(x), the amplitude of the oscillations remains equal to one. Therefore, k(x) cannot be kept arbitrarily close to any number by keeping x sufñciently close to 0. 1.3 LIMITS INVOLVING INFINITY Note: In these exercises we use the result lim 1 x i eo m/ n = 0 whenever 11113 > 0. This result follows immediately "' x - . - 1 _ . (1)'“/ " . 111V” ITI/ ll from Example 1 and the power rule in Theorem 7. xlniioo (Km/ n) _ xinga) Í = (xluiim í) . . 0 = 0. 1. (a) 7r (b) 1r 2. (a) à (b) à 5 5 3. (a) -ã (b) -ã 4. (a) â (b) â 5. -% 5 sinx2x 5 à = > lim = 0 by the Sandwich Theorem
19. 19. 10. 11. 12. 13. 14. 15. 17. 18. Section 1.3 Limits Involving Infinity . 2x+3_ . 2+? -¡_2 b 2 . (a. ) , (11% 5X+7-xl_1_, nà° 54-7-3 ( ) Ê (same processas part (a)) Í 1+; x 2 1' x+1_ 1° X =0 b o t (a) X320 ; +3 xa, 1 +12 ( > (sameprocessaspar (a)) . X -1--12x -1--12x . 1_123_. 2 __ . 1-12x3_ - x2 _ (a) ; (11530 4x2+Í2"xLb1o x4+_]% _ oo (b) x-lârpoo 4X2+12_x-l-1›goo 4+__2 _eo x x . 7 3 _ . 7 _ (a) 7 (same PTOCCSSaS pñft x . 2_ .3-6 . 3'-6_. 3-6_ (a) gta. 3ã, .-§“= xrn 37m (b) , nas arâ-xhwm I-_g--W 2x3 à 5 2x3+~3¡ . 2x5+3_. 2__ . 2x +3_ x_ (a) , ineo _xux-xua _1 +21: ~ °° (b) x39.. -a-_x , rx-lina _Hà -°° _2_ 2 +1 . -2x3-2x+3_ . 7 3__g (a) #LF-k amrziz-xlgo - a x (b) _É (same process as part (a)) _ 4 . ..1 (a) = ]_1_111 : -1 X °° x -7x +7x +9 * °° _¡+F+F (b) -1 (same process as part (a)) 2 1 2 í + - - +1 _ 2 x+x'1 _ (xl/ z) (x2) . 2+# . (xl/ z) _ 422., -çiraiisgo 3_7 = ° 1° xLIào 2_ x= x“~"ào “aí-Á x F75 1_ 1 3 -5 (Im-(I/ s) . ( 9/15) lim 3À/ '\$= 11!* : xñtmãilmo x =1 x-i-oo #+/ ; x-›oo1+X(/ ) 1+ 1 ¡2/15 1 -1 -4 *+7 . +x _ . _ , lago i- _x-S-xlglolo lj -°° 105
20. 20. 106 Chapter 1 Limits and Continuity 1/15 __ 1 L _ 2x5/3 _ X1/3 + 7 _ l_ 2x ¡19715 'Fxa/ s _ 19' xlLnào xs/5+3x+ x "xionolo 1+ + 1 "°° X375 xnho 22. Here is one possibility. 24. Here is one possibility. um. \$440 lxl 26.
21. 21. Section 1.3 Limits Involving Infinity 107 2x2+x-1 27. = y x2-1 31. y=
22. 22. 108 Chapter 1 Limits and Continuity 33.y= 5 34.y= 4X x2+4 35. An' end behavior model is 2g = 2x2. (a. ) 36. An end behavior model is ; L2 = 0.5x3. (c) x 37. An end behavior model is ã: - = -2x3. (d) 38. An end behavior model is' i4¡ = -x2. (b) -x 39. (a) The function y = e” is a right end behavior model because gira* = (1 - à? ) = 1 - 0 = 1. e 1 o n x à | x (b) The function y = -2x is a left end behavior model because xiigioo e# = KB175109 (-§_x + 1) = 0 + l = 1. 2 - . . 4o. (a) The function y = x2 is e right end behavior model because xlim X +2** x = lim (1+#)= 1 - 0 = 1. _VX x Xâí X 2 - 2 (b) The function y = e** is a left end behavior model because X3909 % = ,\$900 (alí-í + 1) XEEIOCXXZCX +1): 0 + 1 = 1. 41. (a, b) The function y = x is both a. right end behavior model and a. left end behavior model because limw(x-_: -+1;"x')= lim (1+1“, 'f')=1-o= 1. x-Hk x-oioo 42. (a, b) The function y = x2 is both a. right end behavior model and a left end behavior model because 2 ~ - ' › X +sm X _ - sm x = x32; )- X3321” (1 +7 ) 1- 43. f(x) = y/ x2+x+1-x
23. 23. Section 1.3 Limits Involving Infinity 109 (a) The graph suggests that , (13130 f(x) = (b) E f x) to 6 decimal places n 1.000000 0535654 0-503731 0-500375 10000 0500037 100000 0.500004 1000000 0500000 1 The table of values also suggest that , (11% f(x) : É X_ÔOO . 1 1 alerts. , --_í+ / x = i 1+1/x+1/x2+1 Proof: xlLrà°(/ xã+x+1-x)= lim [(/ x¡+x+1-x)( )]= lim ( 1+x ) 45. At most 2 horizontal asymptotes: one for x -› oo and possibly another for x -› -oo. 46. At most the degree of the denominator, which is zero at a vertical asymptote. A polynomial of degree n has at most n real roots (or zeros).
24. 24. 110 Chapter 1 Limits and Continuity _ x _ -1 2 51. l-l. 41 by í-l. 3] The graph of y = = ¡lgel/ x is shown. f(x) = lim+ = oo lim f(x): lim r(,1-¡)= o X-V-X
25. 25. 52. 53. 54. J. . ¡-4_ 41 by l-l. 31 The graph of y = = Ale-Ux is shown. x . . 1 _ , are f(x) = xgrã; f(x) _r o - - 1 . __ xlimm f(x) = X1351_ f(x) _ oo [-3. 3] by [-2. 2] The graph of y = = x ln is shown. , (15% f(x): lim+f(¡1¡)=0 x-oO x_lii'_n°°f(x): lim r,1-¡)=0 x-oo_ l-S. 5117)' l'1-5. 1.5] ) = su; X is shown. )=1 NI** The graph of y = f ( ›<| i- X1330 f(x) = lim+ f( x-›O f KI** g/ II ›- lim f(x): lim f x"_°° x-›0- Section 1.3 Limits Involving Infinity 111
26. 26. 112 Chapter 1 Limits and Continuity 1 _ cos-_ . 9_l_ _l 55- x29» Í-olsàl- ; fia-rh (ü-x) 1+X - 1"” - z 1 56. xlLroio : zl-lfàikz =1, (Zz-í) 57. 11mm (3 +2¡)(cos¡1¡)= 31551) (3 + 20)(cos e) = (3)(1) = 3, (o = à) x-›: i: 58. im (Â2~cosà)(1+sin¡1-¡)= lim (392-co 9)(1+sin 9)= (0-1)(1+0)= -1, (0=à) *"°° x o_. .o+ 2 3 2 __x -4_ _ 3 : x -x -1= _ x-2 59.y_ x+1_1 x+x+1 60.y x2_1 x 1+x2_1 The graph of the function mimics each The graph of the function mimics each term term as it becomes dominant. as it becomes dominant. 61. The graph of the function mimics each term 62. The graph of the function mimics each term as it. becomes dominant. as it becomes dominant. yuzslnx+ -
27. 27. Section 1.4 Continuity 113 63. (a) y -› oo (see the accompanying graph) (b) y -› oo (see the accompanying graph) (c) cusps at x = :l: 1 (see the accompanying graph) 64. (a) y -› 0 and a cusp at x = 0 (see the accompanying graph) (b) y -o g (see the accompanying graph) (c) a vertical asymptote at x = 1 and contains the point (-1,2-3%Z) (see the accompanying graph) 1.4 CONTINUITY n 1. No, discontinuous at x = 2, not defined at x = 2 2. No, discontinuous at x = 3, 1 = xliiãi_ g(x) 96 g(3) = 1.5 3. Continuous on [-1,3] 4. No, discontinuous at x = 1, 1.5 = lim_ k(x) ; É lim+ k(x) = 0 x-ol x-›1 5. (a) Yes (b) Yes, lirrâ+ f(x) = 0 (c) Yes (d) Yes b. - 6. (a) Yes, f(1) = l (b) Yes, f(x) = 2 (c) No (d) No 7. (a) No (b) No
28. 28. 114 Chapter 1 Limits and Continuity 8. 9. 10. 11. 12. 13. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. l-LO) U (0› 1) U (1,2) U (2,3) f(2) = 0, since llrâl_ f(x) = -2(2) + 4 = 0 = lim+ f(x) X-4 x-›2 f(1) should be changed to 2 = f(x) The function f(x) is not continuous at x = 0 because lim f(x) = 0, f(0) = 1 and, therefore, f(x) ; É f(0). x-›0 The function f(x) is not continuous at x = l because liml f(x) does not exist since liríi_ f(x) = -1 and X* xa lim f(x) : 0. The discontinuity at x = 0 is removable because the function would be continuous there if the a1 : alue of f(0) were 0 instead of 1. The discontinuity at x = 1 is not removable because lim f(x) does not exist x-»l and the discontinuity cannot be removed by defining or redefining f(1). The function f(x) is not continuous at x = 1 because f(x) does not exist since lim f(x) = -2 and x-»l- lim+ f(x) = 0. The function f(x) is not continuous at x = 2 because lirnz f(x) = 1, f(2) = 0 and, therefore, X15 *ã )l¡im2 f(x) ; É f(2). The discontinuity at x = 1 is not removable because lim f(x) does not exist and the x-al discontinuity cannot be removed by defining or redeñning f(1). The discontinuity at x = 2 is removable because the function would be continuous there if the value of f(2) were 1 instead of 0. Discontinuous only when x - 2 = 0 à x = 2 14. Discontinuous only when (x + 2)? = 0 à x = -2 = > continuous on (-oo,2) U (2,00) = › continuous on (-oo, -2) U (-2, oo) Discontinuous only when t2 - 4t + 3 = 0 = › (t - 3)(t - 1) = 0 = > t = 3 or t = 1 = › continuous on (-oo, 1) U (1,3) U (3,00) Continuous everywhere. (| tI+ 1 ; é 0 for all t; limits exist and are equal to function values. ) Discontinuous only at 0 = 0 = › continuous on (-oo, 0) U (0,00) Discontinuous when "'70 is an odd integer multiple of g, i. e., 52-9- = (2n - 1) ã, n an integer = › 0 = 2n - 1, n an integer (i. e., 9 is an odd integer). Continuous everywhere else = › continuous on ((2n -1)1r/2,(2n+1)1r/2) for n an integer. Discontinuous when 2V + 3 < 0 or v < -â = > continuous on the interval (-g, oo). 2 Discontinuous when 3x - 1 < 0 or x < à = > continuous on the interval Baco). lim sin (x - sin x) = sin (7r - sin 7r) = sin (7r - 0) = sin 1r = 0; continuous at x = 1r X_57|' 'll-113 sin cos (tan = sin(§ cos (tan = sin co = sin: 1; continuous at t = 0 y-u] y-ol = sec 0 = 1; continuous at y = 1 lim sec(y sec2y - tanzy - 1): lim sec(y seczy - seczy) = sec((y -1)sec2y)= sec((1-1)sec21) âigà Ming cos(sin 91/3)] : tanH cos (sin(0))] = tan cos = tan: 1; continuous at 0 = 0.
29. 29. 25. 26. 27. 28. 29. 30. 31. Section 1.4 Continuity 115 f(x) is continuous on [0, 1] and f(0) < 0. f(1) > 0 = › by the Intermediate Value Theorem f(x) takes on every value between f(0) and f(1) = > the equation f(x) = O has at least one solution between x = 0 and x = 1. cosx= xà(cosx)-x=0. Ifx= -§, cos(-%)-(-%)>0. Ifxzg, cos(-725)-â'-< 0. Thus cos x-x=0 I for some x between - 2 and ã according to the Intermediate Value Theorem. All live statements ask for the same information because of the intermediate value property of continuous functions. (a) A root of f(x) = x3 - 3x - 1 is a. point c where f(c) = 0. The roots are approximately x1 = -1.53, x2 = -0.347, x3 = 1.88, the points where f(x) changes sign. (b) The points where y = x3 crosses y = 3x + 1 have the same y-coordinate, or y = x3 = 3x + 1 = › y = f(x) = x3 - 3x - 1 = 0. (c) x3 - 3x = 1 à x3 - 3x - 1 = 0. The solutions to the equation are the roots of f(x) = x3 - 3x - 1. (d) The apoints where y = x3 - 3x crosses y = 1 have common y-coordinates, or y = x3 - 3x = 1 : › y = f(x) = x - 3x - 1 = 0. (e) The solutions of x3 - 3x - 1 = 0 are those points where f(x) = x3 - 3x - 1 has value 0. Answers may very. Note that f is continuous for every value of x. (a) f(0) = 10, f(1) = 13 - 8(1) + 10 = 3. Since 3 < 1r < 10, by the Intermediate Value Theorem, there exists a c so that 0 < c < 1 and f(c) = 7r. (b) f(0) = 10, f(-4) = (-4)3 - s(-4) + 10 = -22. Since -22 < -x/ ã < 10, by the Intermediate Value Theorem, there exists a c so that -4 < c < 0 and f(c) = -Jã (c) f(0) = 10, f(1000) = (1000)3 - s(1000) + 10 = 999,992,010. Since 10 < 5,000,000 < 999,992,010, by the Intermediate Value Theorem, there exists a c so that 0 < c < 1000 and f(c) = 5,000,000. ' - 2 . . . . . is discontinuous at x = 2 because it rs not defined there. However, the discontinuity can be removed because f has a limit (namely 1) as x -› 2. Answers may vary. For example, f(x) = 1 x+1 Answers may vary. For example, g(x) = has a discontinuity at x = -1 because liml g(x) does not exist. X**- ( lim g(x) = -oo and lim g(x) = +00.) xT-l xl-l Noting that r = 0 is triple zero, the polynomial can be rewritten as x3 (x2 - x - 5). Therefore, the roots of the 1 ' “ñ 1+2/ ñz2.791. quintic polynomial are r¡ = í-r- z -1.791, r¡ = t3 = t4 = 0, and t5 =
30. 30. 116 Chapter 1 Limits and Continuity 32. 33. 34. 35. 36. 37. 38. The graph shows that the polynomial has three zeros between -2 and 2, any one a candidate for r. By zooming in, the choices for r are estimated at -1.532, -0.347, or 1.879. P01) 2 . z e ' -4 (a) Suppose x0 is rational = > f(x0) = 1. Choose c = For any 6 > 0 there is an irrational number x (actually infinitely many) in the interval (x0 - 6,x0 + 6) = › f(x) = 0. Then 0 < | x -xol < 6 but [f(x) -f(x0)| = 1 > à = c, so xlim f(x) fails to exist = > f is discontinuous at x0 rational. On the other hand, x0 irrational à f(x0) = 0 and there is a rational number x in (x0 - 6,x0 + 6) à f(x) = 1. Again xl rn f(x) fails i *V 0 to exist = > f is discontinuous at x0 irrational. That is, f' is discontinuous at every point. (b) f is neither right-continuous nor left-continuous at any point x0 because in every interval (xo - 6,xo) or (xmxo + 6) there exist both rational and irrational real numbers. Thus neither limits lim_ f(x) and X"VX lim f(x) exist by the same arguments used in part (a). o X-'VXO Yes. Both f(x) = x and g(x) = x -à are continuous on [0,1]. However g3 is undefined at x = à since f x . . . = 0 = › à IS discontinuous at x = Yes, because of the Intermediate Value Theorem. If f(a) and f(h) did have different signs then f would have to equal zero at some point between a and b since f is continuous on [a, b]. Let f(x) be the new position of point x and let d(x) = f(x) - x. The displacement function d is negative if x is the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d(x) = 0 for some point in between. That is, f(x) = x for some point x, which is then in its original position. If f(0) = 0 or f(1) = 1, we are done (i. e., c = 0 or c = 1 in those cases). Then let f(0) = a > 0 and f(1) = b <1 because 0 5 f(x) 5 1. Define g(x) = f(x) - x = > g is continuous on [0,1]. Moreover, g(0) = f(0) - 0 = a > 0 and g(1) : f(1) - 1 = b - 1 < 0 = › by the Intermediate Value Theorem there is a number c in (0, 1) such that g(c) = 0 = > f(c)-c = 0 or f(c) = c. f I I > 0. Since fis continuous at x = c there is a 6 > 0 such that | x-c| < 6 à | f(x) -f(c)| < e = › f(c) - e < f(x) < f(c) + e. i If f(c) > 0, then e = ;f(c) à àüc) < f(x) < -Êf(c) = › f(x) > 0 on the interval (c ~ 6,c + 6). Letc: If f(c) < 0, then c = -àf(c) = > ãfü) < f(x) < àüc) = › f(x) < 0 on the interval (c - 6,c + 6).
31. 31. Section 1.4 Continuity 117 39. (a) Luisa's salary is \$36,500 = \$36,500(1.035)3 for the first year (0 5 t < 1), \$36,500(1.035) for the second year (1 5 t < 2), \$36,500(1.035)3 for the third year (2 5 t < 3), and so on. This corresponds to y = 3s,500(1.035)"^° °. (b) 104.931 by 55.000. 45.0001 The function is continuous at all points in the domain [0,5) except at t = 1, 2, 3, 4. 40. (a) We require: 0, x=0 1.10, 0<X51 2.20, 1<x52 3.30, 2<x53 4.40, 3<XS4 5.50, 4<x55 6.60, 5<x56 7.25, 6 <x S 24. f(x) = This may be written more compactly as _ -1.10int(-x) 0 5 x 5 6 f(x) ' ima, i 6 < x 5 24 (b) [0, 24] by [O, 91 This is continuous for all values of a: in the domain [0,24] except for : c = 0, 1, 2, 3, 4, 5, 6.
32. 32. 118 Chapter 1 Limits and Continuity 41. The function can be extended: f(0) z 2.3. 42. The function cannot be extended to be continuous at ' x = 0. If f(0) z 2.3, it will be continuous from the right. Or if f(0) z -2.3, it will be continuous from the left. 43. The function cannot be extended to be continuous 44. The function can be extended: f(0) z 7.39. at x = 0. If f(0) = 1, it will be continuous from the right. Or if f(0) = -1, it will be continuous from the left. y . Y f(x)'(1 + : um -0.01 -o. oo5 o. oos 0.01 45. x z 1.8794, 4.5321, -033473 40. x z 1.4510, -0.s54s, 0.4030 47. x z 1.7549 4a. x z 1.5596 ' 49. x z 3.5156 50. x z 4.9059, 3.5392, 0.0667 51. x z 0.7391 52. x z -1.8955, 0, 1.8955 1.5 TANGENT LINES 1. P1: m1 =1, P2: m¡ = 5 2. P1: m¡ = -2, P2: m¡ = 0 3. P1: mlzgPf mzz-à 4. P1: m1=3,P2: m2=-3
33. 33. Section 1.5 Tangent Lines 119 [4-(-1+h)”]-(4-(-1)”) _____Tí 5. m= lim h-OO 2 _ : lim *wí“(1'2h+h)“ : lim M2 “L2, h-»O h bzo h at (-1,3): y=3+2(x-(-1)) = > y=2x+5, tangent line ___ lim 2¡/1+h-2 2¡/1+h+2 h-›o h '2,/1+b+2 4(1+h)~4 = 1' = 1' 4-=1; 133% 2b(§71+b+1\$ ral-ii: ,/1+b+1 at (1,2): y = 2+ 1(x- 1) à y = x+ 1, tangent line 2,/1+b-2,/ í __. ._h_____ 6. m: lim h-DO _ 3_ _. 3 2 3 7. m: lim ( 2+h) ( 2) : lim -à-í--S-Flzh-ôh +h +8 h-›0 h-›0 h :113 (12-6h+h3)=12; at (-2,-8): y = -8 +12(x-(-2)) = > y = 12x +16, tangent line
34. 34. 120 Chapter 1 Limits and Continuity 8. 9. 14. . At a horizontal tangent the slope m = 0 = › 0 = m = !lina 1 _ 1 _ . (-2 +h)3 (-2)5 . -8-(-2+h)3 m_ 11m = 11m --_-- h-›0 h-›O . .3h(_2+h)3 -(12h-sh'+h3) 12-6h+h2 = 1% -8h(-2+h)3 = 13% 'É = _.12_ = _.3_. 8(-s) 16' at (-2,-à): y = -à--136-(x-(-2)) = › y = - 3 Ex - à, tangent line [(1 +10 - 2<1h+h›*]- <-1› z at (1,-1): y + 1 = -3(x-1), tangent line h(-3-2h)_ 3_ _h___, = lim h-oO lim (1+h-2-4h-2h3)+1 h-›0 m = lim h-› , 3 _ 2 3 _ m: um [(1+h) +3(1+h)] 4=1.1m(1+3h+3b +h +3+3h) 4: , im h(6+3h+h2)___6; h-›0 h-oü h-vO at (1,4): y - 4 = 6(t-1), tangent line _ . 3+h -2 _ . 3+h -3h+1)_ . _gh _ _ c-Hf-'i' ( ) < -m-&Lb 'E330 h(h+1) n25?, @Tr-b at (3,3): y - 3 = -2(u - 3), tangent line m: lim ______VZ3““ H1'3=1im -í_”'9+h'3.*'9+h+3= lim (3+h)'9 : lim h h-. o h h-›O h i7g+h+3 h~o0 h(;79+h+3) h-›0 h(;79+h+3) = 7613 = à; at (8,3): y - 3 = à(x - 8), tangent line 1 1 . 3+h -1 É . 2- 2+h . _ Atx=3,y= â=›m= ll¡1_rn)( )h : lili-à = )¡% 211224¡ = -à, slope h-l 11:14'” . h-1+h+1 _ ^“°= °'y= '1=”“= 155% LTÊIEÉJ( h)h+(1) Lift bb2i-1=2*°l°'°° [(x+h)2+4(x+h)-1]-(x3+4x-1) m: : = ,im (x3+2xh+h3+4x+4h-1)-(x3+4x-1)_ um (2xb+h2+4b) ! H0 h _hão h = ¡li_r_n) (2x+h+4)=2x+4; 2x + 4 = 0 à x = -2. Then f(-2) = 4 ~ 8 -1 = -5 = › (-2, -5) is the point on the graph where there is a horizontal tangent.
35. 35. Section 1.5 Tangent Lines 121 16' Ozm: Em [(x+h)3-3(x+h)]-(x3-3x)= um (x3+3x2h+3xh2+h3-3x-3h)-(x3-3x) 1.. .o h h-›O E 2 2 3 = 111m3) = kliixrà(3x2+3xh+hz-3)=3xz-3; 3x2-3=0=>x= -1 orx=1. Then f(-1) = 2 and f(1) = -2 = › (-1,2) and (1,-2) are the points on the graph where a horizontal tangent exists. 1 _ 1 _ _ . (x+h)-1 X-l_ . (x-1)-(x+h-1)__ . _h _ 1 ' 17' '1"““ 11.5% h “ 33% h(x-1)(x+h-1) ' 3.5?, i+_”h-1)"í(x-1)2 = >(x-1)2=1=x2-2x=0=›x(x-2)=0=›x=0orx=2.Ifx=0,theny= -1andm= -1 : ›y= -1-(x-0)= -(x+l). Ifx=2,theny=1andm= -1ày=1-(x-2)= -(x-3). 1s. l=m= lim í”x+h'*/ í= lim Vx+h"*/ ;.V”+h+ x=1i1in -íLhkx 4 h-›0 h h-PÕ F ;7x+h+/ ; h-. o Wwíñkx/ í) =1' ___l1í=1.'rhu, l=1=› :2: =4=› =2.Tht tl' ' IL? ) h(/ m+/ í) 'a 5 4 m' [- X y e anSen meia _ -4.2h2-1-4. 7 _. 2 . 19_ 111% f(2+h) f(2)= kl1i_r? )(100 9( + )h) (00 9(2) )___ giga 4 9(4+4h+h )+49(4) = lllirxà (-19.6 - 4.9h) z: -19.6. The minus sign indicates the object is falling ggwgward at. a. speed of 19.6 m/ sec. _ 2 _ 2 2 20_ ___ = = 60 ft/ sec. _ 2 _ 2 2 _ 21_ Im f(3+hâ ; (3) = 113% = 113% 1r[9+6h+h 9]: gil? ) ? WMM = 6” _ f(2+h)_f(2) _ ? (2+h)3-4§7I(2)3 _ g[12h+6h”+h3] . 4_ 2 22' 3.11% THE ““°'“T"“*'= iíã'ê› : Hà '3'[12+6h+h]=1°” _. 2 _ 2 2 23. lim “i = lim 1'86(1+h) Lsôm = im 136 + 3-72** + 136** -1-35 = lim (3.72 +1.86h) h-›0 h h-. o E h-»O 5 Iwo = 3.72 24 um s(2 + h) - s(2) z um 11-44(2 + 11)* -11-44(2)° 45.76 + 45.76h + 11.4411* - 45.76 = lim h-›0 h-tü h-›0 = *imã (45.76 + 11.4411) = 45.76 = "|>-* __ hz sin 25. Slope at origin = = à) = h sin<%) = 0 à yes, f(x) does have a tangent at the origin with slope 0.
36. 36. 122 Chapter 1 Limits and Continuity 26_ um s(0+h)-s(0)_ “Mil 1 ! H0 h _ h = sin H. Since sin à does not exist, f(x) has no tangent at the origin. 27. lim_ = lim_ 11-5-0 = oo, and lim = lim li-Q = oo. Therefore, h-'o h hão h h-›O+ h-›O+ h llinà = oo : à yes, the graph of f has a vertical tangent at the origin. . U0+h -U0 . .. . U0+h -U0 . _. 28. hgrài_ = king_ Q-h-l = oo, and hlíràh_ = 1121314_ l-íl = 0 : à no, the graph off does not have a vertical tangent at (0,1) because the limit does not exist. f0-f-2 _-2 f3-fl 3_ 29.(a)Ê§= (_0)\$. ¡l=1_2Lzo.432 (b)Ê_í= (à_()=9_2-°z8.684 f4 -fl 1 - 3o. (a) Ã^Tfc= (fi__l(_l= n_4g= %_4zo.4sz Af _f(103) -f(100) _ln 103 -ln 100 __ i 103 _1 (b)-A-x. .-í: -103_100 _g_ 3 ln m_51n1.03 z 0.0099 31 (a) _4_f= f(37'/4)*f(7'/4) = _1-1 ' At (31r74)-(7r/4) 7r/2 (b)Ê_: = =0%/ _§= -Éz-1.654 = _à z -1.273 r-ro _ 32.(a)áÁ*-_§= .%l=1_, ,_3=_2, -o. s37 f(1r)-f-1r) . _ <b>Ê-Ê= T(«)“Lnl= ° 33. (a) 192g; : i5 3 = 0.3 The rate of change was 0.3 billion dollars per year. 3.1-2.1 _ (b) 1997-1 95'°'5 The rate of change was 0.5 billion dollars per year. (c) y = 0.0571# - 0.1514): + 1.3943 [o, 101w [0, 41
37. 37. (d) ÊÃÉ z 0.31 (e) 34. (a) Section 1.5 Tangent Lines 123 5 - 3 y(7) - y(5) , V _É ~ 0.53 According to the regression equation, the rates were 0.31 billion dollars per year and 0.53 billion dollars per year. um y(7 + h) - y(7) _ um [0.0571(7 + 11)? - 0.1514(7 + h)+1.3943] - [0.0571(7)2 - 0.1514(7) + 1.3943] h-. o h _ h-»o h . . 0.0571(14h + 117)- 0.15141¡ = 11m e h-›0 h = 115% [0.o571(14) - 0.1514 + 0.05711] z 0.65 The funding was growing at a rate of about 0.65 billion dollars per year. 17. 1815)' 10. 900] (b) Q from year 1988 1989 1990 1991 1992 1993 1994 1995 1996 Slope 440 ~ 225 N TE_ ~ 23.9 440 - 289 z 440 - 270 ~ 7771.0_ ~ 24.3 440 - 493 Q -8.8 #É z -4s. s 440 - 753 ~ _ m3 nv 80.8 z -70.3 440 - 600 z 144.0 As Q gets closer to 1997, the slopes do not seem to be approachíng a limit value. The years 1995-97 seem to be very unusual and unpredictable.
38. 38. 124 Chapter 1 Limits and Continuity 35. (a) The graph appears to have a cusp at x = 0. Y y_ ¡z/ s 10,01 l_ . f 0 h -f 0 . 2/5_ . . (b) hllrài_ = hai_ -h-hío = 1111.18_ E317 = -oo and 111-irá# 1717¡ = oo : à limit does not exist 2/5 = › the graph of y = x does not have a vertical tangent at x = 0. 36. (a) The graph appears to have a cusp at x = 0. Y y _ ¡4/5 (0,0) X . f 0 h -f 0 _ 4/5_ , , _ _ _ (b) 111351_ = hlgãl_ LH = 1113.13_ É¡ = -oo and 111-irá# ¡1-117¡ = oo = › limit does not exist = › y = x4/5 does not have a vertical tangent at x = 0. 37. (a) The graph appears to have a vertical tangent at x = 0. v com -V ' "U5 X . f 0+h -f 0 _ 1/5 _ . . (b) 1111:? ) -LH-)=1l¡1ln) h h O: 1113)) #'75 = oo à y= x1/5 hasavertical tangent at x=0. 38. (a. ) The graph appears to have a vertical tangent at x = 0. y (07 O) . Y ' 13,5 (b) água f-T“)+h)'¡(°)= lim 535-0: lim h-›O E h-›0 h? "i 3/5 = oo = › the graph of y = x has a vertical tangent 5 atx=0.
39. 39. Section 1.5 Tangent Lines 125 39. (a) The graph appears to have a. cusp at xl= O. . f(0+h)-f(0)_ . 4112/5-211_ . 4112/5-211_ . 4 _ ' - 4 (b) 4.1155*- *if-ilirã- _F-hlls- _ir--iliw- m“2-*°°““, f3;*+ has"? = eo = › limit does not exist = › the graph of y = 4x2” - 2x does not have a vertical tangent at x = 0. 40. (a. ) The graph appears to have a cusp at x = 0. v «mm-casi -_~¡_-_f(° "' h) "f(0) = lim : Ls/ a ' 5112/3 = lim 11m - L = o _ lim ; às does no: exist = › the Braph of (b) h-»O h h-»O h1/3 h-i0 y = xõ/ a - 5x2” does not have a vertical tangent. 41. (a) The graph appears to have a vertical tangent at x = 1 and a cusp at x = 0. . Y y_¡2/a_(¡_¡)1/a 2/3_' __ 1/3_ (b)x=1: lim 0+h) 0+h 1) 1: lim (1+h)2/3-h1/3- 1 h-›0 E h-»o E = -oo à y = :: H3 - (x -1)1/3 has a. vertical tangent at x = 1;
40. 40. 126 Chapter 1 Limits and Continuity 2/3_ _ 1/3__ 1/3 ___ 1/3 h (h 1)¡ (1) : um 1_(1. 1) r(0 + h) - f(0) _ ___ h-oO hl/ a h x = 0: lim . _ lim l h-»O h-›O + ll does not exist = › y = x2/3 - (x - 1)1/3 does not have a vertical tangent at x = 0. 42. (a) The graph appears to have vertical tangents at x = 0 and x = 1. yui/ :gohni/ s (b) x=0: lim (°"'h)'f(°)= hl/3 + _1)1/3 _ (_1)1/3 h-t0 h-oO E = oo= >y= x1/3+(x-1)1/3hasa vertical tangent at x = 0; x=1: lim “LÚÉQ-Í: lim (1+11)1/3+(1+11-1)1/3_1 h-vO h h-»O : eo = > y= x1/3+(x-1)1/3 hasa vertical tangent at x = 1. 43. (a) The graph appears to have a vertical tangent at x = 0. r(0+11)-1(0)_ x/ í-o . . _ . 1_ _ (b) 1111?* íEí-xl-fg* h -hggh h-oo' _,414 . f(0+h)-f(0)_ . -1/1111-0_ . _ . _ _ 111.135'- h 'W333' E '111-HT' -hl "111-Igl- ;7ih| '°° = > y has a vertical tangent at x = 0. 44. (a) The graph appears to have a cusp at x = 4.
41. 41. Chapter 1 Practice Exercises 127 _ f(4+h)-f(4)_ . / |4*(4+h)| '°_ . y/ Íhl_ . 1 _ _ (b) hm Ê--íh _ 11m í--í _ 11m -h _ lim _ oo, h-›O+ 1140+ h-›0+ h-›0+ h . f(4+h)-f(4)_ . 4-(4+h)| _ . y/ ihl_ 111153' E _ by? " h _ 111-l? " _; Fba “ "°° = > y = v4 - x does not have a vertical tangent at x = 4. 45-48. Example CAS commands: Maple: f: =x -> cos(x) + 4*sin(2*x); x0:= Pi: dq: =h -> (f(x0 + h) - f(x0))/ h; slope: =limit(dq(h), h=0); L: =x -> f(x0) + slope*(x - x0); y1:= f(x0) + dq(3)*(x - x0)_: y2:= f(x0) + dq(2)›r(x - x0): y3:= f(x0) + dq(1)*(x - x0): plot ([f(x), y1,y2,y3,L(x)], x = x0 -1.. x0 + 3); Mathemgtica: Clear [f, m,x, y] x0 = Pi; f[x_] : = Cos[x] + 4 Sin[2x] Plot[ f[x], [x, x0 - 1,x0 +13) ] dq[h_] : : (f[x0 +h] - f[x0])/ h m = Limit[ dq[h], h -> 0] y : = f[x0] + m (x -x0) y1 z: f[x0] +dq[1] (x -x0) y2 : = f[x0] + dq[2] (x - x0) y3 : = f[x0] +dq[3] (x - x0) Ploti ifixl›y›ylay2›y3la ix›x0 " lixo + l CHAPTER 1 PRACTICE EXERCISES 1. Atx= -1: lim_ f(x): lim fx =1=> lim f(x *"'1 x-›-1+ ( ) x"'1 ) 1, xs-l = l = f(-1) = > f is continuous at x = -1. 'x' 'd <x <° ! CÚ- l. x-O Atx = 0: lirg_ f(x) = 7"' o<x<l x=0 lim+ f(x) = 0 = › f(x) = 0. ¡_ x a ¡_ x-›0 But f(0) = 1 ; E f(x) = › f is discontinuous at. x = 0. At x = l: lim_ f(x) = -1 and lim f(x) = 1 = › f(x) x-›1 x_,1+ does not exist = › f is discontinuous at x = 1. 2. At x = -lz lim_ f(x) = 0 and lim f(x) = -1=> lim f(x) x""'1 x-v-l xá-l does not exist à f is discontinuous at x = -1.
42. 42. 128 Chapter 1 Limits and Continuity At x = 0: lim_ f(x) = -oo and lim+ f(x) = oo = › f(x) , W à Í x-«O X_, D does not exist = > f is discontinuous at x = 0. At x = 1: xlim_ f(x) = lim+ f(x) = 1 = > f(x) = 1. But x-ol f(1) = 0 ; É f(x) = › f is discontinuous at x = 1. 3. (a) th? , (arm) = 3 339o f(t) = 3(-7) = -21 (b) ,590 (w = (3390 f(0)” = <-7)° = 49 (c) 339o (f(t) -s(t)) = ,ligo f(t) -ggggo g(t) = (4)(0) = o &= = '-7 : ñ-'Í : l “t0 g(t) _ 7 : liilo (g(t) _ 7) clí-rilo gm _ ella¡ 7 _ (e) "lima cos (g(t)) = cosçlimo g(t)) = cos 0 = 1 (f) tl_'1_. l§1°| f(t)| =Itl_i4I§1o f(t)| =I-7I = 7 (s) , Ego (f(t) + g(t)) = ,E130 f(t) + ¡ljgrgo g(t) = -7 + 0 = -7 (h) , Eç0(1/f(t))= = :17 = -à "' O 4. (a) ; g5 -s(x) = - ; ano g(x) = MI (b) lim (g(x) -f(x)) - lim g(x) - lim f(x) = _é x--oü - x-›0 x-oO 2 2 (c) 315g, (f(x) + g(x)) = ,935 f(x) + ; gnu g(x) = à+ «ã - 1 _ 1 _1- (d) x90 (e) ; gn°<›<+r(x»= ;gr; ,x+; grn, ,r<x›= o+à= § f(x) _cos x lim f(x) - cos x . _p_ . ..o _ ___l (013%, x-1 - -7›_*1- 2 5. Since x = 0 we must have that (4 - g(x)) = 0. Otherwise, if (4 - g(x)) is a. fmite positive number, we would have llràl_ [4 _fool = -oo and lim+ [4 -fool = oo so the limit could not equal 1 as x” x-›0
43. 43. 10. Chapter 1 Practice Exercises 129 x -› 0. Similar reasoning holds if (4 - g(x)) is a finite negative number. We conclude that g(x) = 4. 2 = [x , P43 s<›= >]= [935 s<›= >]= -4 : e sw] = -4 332a so» (since g(x) is a constant) = › g(x) = à = -à (a) f(x) = x1/ 3 = c1/3 = f(c) for every real number c = › f is continuous on (-oo, oo) (b) lim g(x) = X314 3/4 X-UC = c = g(c) for every nonnegative real number c = › g is continuous on [0, oo) (c) h(x) E x4” = É = h(c) for every nonzero real number c = › h is continuous on (-oo,0) and (0›°°) (d) k(x) = : Fl/ s = C117 = k(c) for every positive real number c = › h is continuous on (0,00) . (a) U «n _Em (n +à)7r), where I = the set of all integers. nel (b) U (mr, (n + 1)1r), where I = the set of all integer-s. n e I (c) (-°°. °°) (d) (-°°›0) U (0- 0°) x'-4x+4 . _ (x-2)(a: -2) _ . x_2 _ . . . (a) x +5X2_I4X- *gta xo¡ +7)(x_2)- lulu, rx +7), x;ê2, the limit does not exist because - X-2 = l- X-2 = _ , llfâ- x(x+7) °°°“d , ,132 ›T+_7) °° . x*~4x+4 __ (x-? Xx-Z) __ - x-2 _ o _ (b) Em e- , .% x(x+7)(x-2)_›lc1-›Ir¡2 x(x+7)'x#2_ã'§í_g . x2+x _ . x(x+1) _ . x+1 __ . 1 _ (a) 1% ma? ? 1% _Ff3(x2+2x+1)_'1'LP° x<x+1)<x+1)'›1=% “"x2<x+1›'**°““** 1' ' 1 = d ° A: l' x2+x = . Now : :El-CEL x (x+1) com X2214' x2(x+1) co# *EP0 x3+2x4+x3 oo 2 . x +x _ . X(X+1) _ . 1 _ . . (b) X1313¡ -, ¡1_1_@1 xoa(xz+zx+l)-xlj_ff_ll mnwéüandx# 1. The limit does not 'b 1'-¡__1=- dl' T-1=. exist ecausexJínl_ x(x+1) ooan *Mu-raw x (x+1) oo l-W l-W _ 1 1 ' = l' = l = - : lrl-rortll l-X xl-Igl (1- XXI-i' x) : :l-org 1+3): 2 . x2_a2=1. (x2-a2) = l. 1 : L : lag-H x4_a3 : em (x2+a2)(x2 a2) x53 X5415 2a?
44. 44. 130 Chapter 1 Limits and Continuity 13 14. 15. 16. 17. 19. 20. 21. 23. 24. 25. 26. 27. 29 2_ 2 2 2 _ 2 _11m : km) (__: ___. __X HhXJh) X = ¡1_%(2x+h)=2x . x+h2-x2 . 2+2h +h7 - 2 . m, í~+= m híxh-*Lí= l%<2x+h>= h 1 *l 2 (2+ ) . 2+x 2_. - X _. _1__l 15% X ',135 2x(2+x) “Íãííà 4+2x'” 4 . (2+x)3_s . (x3+6 ? +12 +s)_a . Em, X _\$136 = \$%(›<2+6x+12)=12 um 2x+3__lim 2+Ê=2_+g: g 18 um 2x2+3_ 1¡ 2+%_2+0_g x-ooo 5X+7 x-voo +0 5 . x-›-oo 5x2+7-x-›-oo -5 x . x2-4x+s_ . 1 4 8 _ _ X290., 3x3 -xk9oo(5-3_x2'+g; )-°'°+°-° L 2 lim 1 : lim X = 0 =0 -r _ -' 7 1 l-O 0 x °° x 7x+1 x °° 1_¡+F + x2-7x= 1- x-7 = _ 22 l- É: x+1 = x-›-oo x+1 x-1›E1oo(1+§) 0o x-l-crãolo 12X3+128 X1520 12+lf3§ oo lim | .sinx|5lim . l :0 sinceintx-vooasx-›w X-'oo mtx x-»oo mtx lim 'í°°s9”“< lim : E0 ô-wo 0 _O-ooo 9 . 1+SÍ§§X+~Ã um x+smx+2/ x_ lim x/ Í _1+0+0_1 x-wo x+sinx _x-*oo 1+sincx _ 1+0 _ 2/3 _1 -5/3 . x +x _ . 1+x _1+0_ '(11% x2/3+cos2x_"ll%° 1+cos2x _1*õ~1 xz/ a lim 5x2: lim L2=0 28. Lettingu= §gives lim el/ x: lim_ e": 1. X-¡OO x-»oo ex x-›-oo “ao . (a) f(-l) = -1 and f(2) = 5 = > f has a root between -1 and 2 by the Intermediate Value Theorem. (b), (c) root is 1132471795724
45. 45. Chapter 1 Additional Exercises-Theory, Examples, Applications 131 30. (a) f(-2) = -2 and f(0) = 2 = > f has a root between -2 and 0 by the Intermediate Value Theorem. (b), (c) root is -1.76929235424 CHAPTER 1 ADDITIONAL EXERCISES-THEORY, EXAMPLES, APPLICATIONS 1. (a) x 0.1 0.01 0.001 0.0001 0.00001 xx 0.7943 0.9550 0.9931 0.9991 0.9999 Apparently, lim xx = 1 x-›0 (b) y 1 0.6 y - x' 0.2 1 l/ (lnx) (Y) 0.3678 0.3678 0.3678 1 1 / (ln x) Apparently, xliârgo = 0.3678 : à (b) , 0.4 0.2 2 4 6 B - - V2 VEÍLV C2 3.v1i_r3_L= v1_1.T_L0 1-? =Lo 1- c¡ : Lo 1-c-, =0 The left-hand limit was needed because the function L is undefined if v > c (the rocket cannot move faster than the speed of light).
46. 46. 132 Chapter 1 Limits and Continuity 4. llz-í- l<0.2=›-0.2<ç-1<0.2=›0.8<-)§E-<1.2=1.6<Á<2.4=›2.56<x<5.76. x/ í W -Í--11<0.1à-0.1<T-1<0.1=›0.9<ç<1.1=>1.8<W<2.2=›3.24<x<4.84. 5. |10 + (t - 70) x 10"* - 10| < 0.0005 = › I(t - 70)x10'4|< 0.0005 = › -00005 < (t - 70) x 10"* < 0.0005 à -5<t-70<5 : › 65°<t<75°= ›Within 5°F. 6. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero point, 0, on the equator = > 0 + 1rR represents the midnight point (at the same exact time). Suppose x1 is a point on the equator “just after” noon = › x¡ + WR is simultaneously “just after” midnight. It seems reasonable that the temperature T at a. point just after noon is hotter than it would be at the diametrically opposite point just after midnight: That is, T(x1) -T(x¡ + 1rR) > 0. At exactly the same moment in time pick x2 to be a point just before midnight = > x¡ + 1rR is just before noon. Then T(x2) -T(x2 + vrR) < 0. Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c between 0 (noon) and 1rR (simultaneously midnight) such that T(c) - T(c + 7rR) = 0; i. e., there is always a pair of antipodal points on the earth's equator where the temperatures are the same. . .1 , / _. ,/ _. _ 7_ (a) Atx=0¡ l%¡+(a)= l% __. l.'. .ã. .1_ÍÉ= lim 1 , /1+a) n-»o -1_§71+a : Em 1-(1+a) z -1 :1 nâo a(-1-; ;1+a) -1-; ;1+0 É 1-(1+a) _ At = -1: r =1' : r v -a = 1 =1 x , JÍUJQ . ..Tu a(-1-§71+a) n35 a(-1-§71+a) -1- o __ . _ . -1-, /1+a_ . (-1-¡/1+a) -1+, /1+a (b) ^°*-°- .1i%'-'-(“)'. '.'. 'à*- a 2.13:# _r- : ma: 1-(1+a) = ' z: a' : _l : b h .123- . .H r 1+. ) . rss- _Twnn + 1+. ) .38- *T + 1+. °°< “ ° denominator is always negative); lim r_(a) = lim "l = -oo (because the denominator + -1 + V1 + a. a-»O a-›0+ is always positive). Therefore, r_(a) does not exist. At x = -1: lim+ r_(a) = lim+ --¡í (c) .1 -0.5 0.5 'l Grlph not to scaie
47. 47. 10. 11. Chapter 1 Additional Exercises-Theory, Examples, Applications 133 (d) (a) Since x -› 0+, 0 <x3 <x< 1 = › (x3-x) -o 0' = › lim fixa-x): llIàl_ f(y) = B where y = x3-x. x-»O y* (b) Since x -› 0', -1< x < x3 < 0 = > (xa-x) -v 0'* = › llfàl_ f(x3-x)= lim+ f(y) = A where y : xa-x. x-› y_¡o (c) Since x -v 0+, 0 < x4 < x2 <1=>(x2 -x4) -› 0+ = > lim+ f(x2 -x4) = lim+ f(y) = A where y = x2 -x4. x-0 y-›O (d) Sincex-›0', -1<x<0=›0<x4<x2<l= ›(x2-x4)-›0+à lim+f(xz-x4)= Aasinpart (c). x-›0 (a) True, because if (f(x) + exists then (f(x) +g(x)) - f(x) = [(f(x) + g(x)) -f(x)] = g(x) exists, contrary to assumption. (b) False; for example take f(x) = à and g(x) = -¡1¡. Then neither f(x) nor g(x) exists, but 335% (f(x) + g(x)) = ;gua (à - à) = lim o = o exists. x-›0 (c) True, because g(x) = 1x| is continuous à g(f(x)) = [f(x) | is continuous (it is the composite of continuous functions). 1, x 5 0 = › f(x) is discontinuous at x = 0. However (f(x) | = 1 is 1, x > 0 (d) False; for example let f(x) = (- continuous at x = 0. f(x) = x+2 cos x = › f(0) =0+2 cos 0 =2 >0 and f(-1r) = -1r+2 cos(-7r) = -7r-2 < 0. Since f(x) is continuous on [-1r,0], by the Intermediate Value Theorem, f(x) must take on every value between [-7r - 2,2]. Thus there is some number c in [-7r,0] such that f(c) = 0; i. e., c is a solution to x + 2 cos x = 0. Show ,133 f(x) = ,im (x2 -7)= -6 = r(1). Step 1: | (X2-7)+6l<t'= -6<X2-1<6=>1-E<X2<1+6à/ l-E<X</1+c. Step2: | x-1I<6=›-6<x-1<6=›-6+1<x<ô+1. Then -6+1=¡/1-cor ó+1=, /i+c. Choose6=min(1-y/1-e, ¡/1+e-1), then
48. 48. 134 Chapter 1 Limits and Continuity 12. 13. 14. 15. 0 < Ix - 1 I < 6 = > | (x2 - 7)- 6' < e and f(x) = -6. By the continuity test, f(x) is continuous at x = 1. Show linâ g(xã= lim] à=2=g(â) “z x": Stepl: -1--2 <s= ›-e<-1--2<c= :›2-c<i-<2+e= > 1 >x> 1 2x 2x 2x 4-26 4+2e Step2: lx-âi<ô= >-6<x-â<6=>-6+â<x<6+à. _ l: 1 = l_ 1 : à 1:; = 1 _l_:6 __ Th” 5+4 4+2c: ›ô 4 4+2: 4(2+5)'°r5+4 4-2535 4-2e 4'4(2_c)' Chooseô=4-(#), the smaller ofthe two values. Then0<| x-H<6à| à-2l<cand lim à=2. X-Vz By the continuity test, g(x) is continuous at x = Show h(x) = y/ ÊX- = 1 = h(2). (1-6)2+3<x<(1+c)2+3. Stepl: |y/2x-3-1|<6=-c< 2x-3-1<c= >1-e<y/2x-3<1+e= > 2 2 Step2: Ix-2I<ô= >-6<x-2<õor -6+2<x<6+2. (1-22+3_1-(1-c)2_f_§23 °r6+2_(1+e)*+3 í__-T--_ , _m_ 1 2 -1 2 Lil: = e + Choose 6 = e - 5;, the smaller of the two values. Then, __ 2 Then-6+2= à6= - _(1+c)2+3 _ = ›ó_: -í-: -2_ 2 0 < Ix-2 I < 6 = > l y/2x-3 -1|< c, so y/ Zx- 3 = 1. By the Continuity test, h(x) is continuous at x = 2. lim y/9-X=2=F(5). x--›5 Show lim F(x) : x-oâ Step 1: Im-2|<E= -6<m-2<C= >9-(2-C)2>X>9-(2+6)2. Step2: 0<Ix-5I<5=>-ô<x-5<6=>-6+5<x<6+5. Then -ó+5=9-(2+e)2=>6=(2+e)2-4=e2+2e, or6+5=9-(2-c)2=›6=4-(2-e)2=c2-2e. Choose6=e2-2c, the smaller ofthe two values. Then,0<Ix-5|<6=>| /9-: -2I<c, so lim m : 2. By the continuity test, F(x) is continuous at x = 5. x-›5 (a. ) Let e > 0 be given. If x is rational, then f(x) = x = › | f(x) -01= IX-Ol < e à Ix-0I< E; i. e., choose 6 = c. Then 1x- OI < 6 = > [f(x) -0 I < e for x rational. Ifx is irrational, then f(x) = 0 = › [f(x) -0|< e c› 0 < e which is true no matter how close irrational x is to 0, so again we can choose 6 = e. In either case, given e > 0 there is a. 6 = e > 0 such that 0 < Ix - 0 l < ô = > | f(x) -- 0| < c. Therefore, f is continuous at x = 0. (b) Choose x = c > 0. Then within any interval (c - 6,c + ô) there are both rational and irrational numbers. If c is rational, pick e = No matter how small we choose 6 > O there is an irrational number x in (c - 6,c + 6) = > | f(x) - f(c) | = I0 -cI = c > É = e. That is, fis not continuous at any rational c > 0. On
49. 49. 16. (a) (b) (c) Chapter 1 Additional Exercises-Theory, Examples, Applications 135 the other hand, suppose c is irrational = > f(c) = 0. Again pick e = No matter how small we choose 6 > 0 3c Then | f(x) -f(c)| =Ix-0I there is a rational number x in (c-6,c+6) with | X-c| <%: e <= › §< x <-í-. = |x| > É = e = › f is not continuous at any irrational c > 0. If x = c < 0, repeat the argument picking e = IÉ-l = Therefore f fails to be continuous at any nonzero value x = c. Let c z: % be a rational number in [O, 1] reduced to lowest terms = > f(c) z Pick e = No matter how small 6 > 0 is taken, there is an irrational number x in the interval (c - 6,c + 6) = › [f(x) -f(c) | = | 0 -ll V n = à > à; = e. Therefore f is discontinuous at x = c, a rational number. Now suppose c is an irrational number = › f(c) = O. Let e > 0 be given. Notice that à is the only rational number reduced to lowest terms with denominator 2 and belonging to [0,1]; à and ã the only rational with denominator 3 belongíng to [0,1]; à and ã- ã, :à [0,1]; etc. ln general, choose N so that 1%¡- < e = › there exist only ñnitely many rationals in [O, 1] having with denominator 4 in [0, 1]; à, and Ê with denominator 5 in denominator 5 N, say r1,r2,. .., rp. Let 6 = mÍnHc -r¡| : i = 1,. .., p). Then the interval (c -6,c+6) contains no rational numbers with denominator 5 N. Thus, 0 < | x - c | < 6 à 'f(x) -f(c) | = [f(x) - OI = |f(x) | 5 à < e = > f is continuous at x = c irrational. The graph looks like the markíngs on a typical ruler when the points (x, f(x)) on the graph of f(x) are connected to the x-axis with vertical lines. y. 1 0.8 0.6 0.4 0.2 x 0 0.2 0.4 0.6 0.a 1 l/ n ¡lx-m/ u h ¡numa! numberinlowertunn¡ m) " [o u: u irrational
50. 50. 136 Chapter l Limits and Continuity NOTES: