02-3 - SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS...

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Unformatted text preview: SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS I: 75 2.3 Calculating Limits Using the Limit Laws 2 1- (a) hm W) +. he) = gig; f(m) + 313; he) (b) gig; W)? = big; m] : (—3)? = 9 = *3 + 8 : 5 ()1' W—slimh(x)—€/§~2 (d)1im;=;=i:—l C ml—rgz x _ xaa _ _ z—m limf m) ~3 3 lim f(a:) 11m 9(w) . f(r)_wa sipj -9_($_):__w~a =1: (6)212 m * hm hm — s e s (f) 5‘32 m) 1113; f<x> —3 0 (g) The limit does not exist. since him 9 = 0 but f(27) 3A 0. 2 lim _ (mm 21%;") 7. Hz ‘ _ 2(3) __3 z—m h(ac) — 11m h(m) ~ 11_r'n 8 — (—3) 11 2. (a) limZ+g(a:)] : li1112f(z)+li1r12g(w): 2 + 0 = 2 (b) lim1 g($) does not exist since its left— and right-hand limits are not equal, so the given limit does not exist. (e) um if<x>g<x>1 : hm m2) - mm) : 0- 1.3 = 0 z—vO z—'O (d) Since lim g(:c) : O and g is in the denominator. but Iim f(:c) m—>~1 :z—>—l = —1 75 0. the given limit does not exist. (e) lim x3f(ac) = [lim m3] [$11312 mg] : 23 ‘ 2 : 16 27—»? («EH2 (f);LI111\/3+f($): 3+;Ln}f(m)=\/3+1:2 3. lim (3$4+2m2~m+1): lim 3x4+ lim 2112— lim m—l— lim 1 x—v72 x—v—Z x—>—2 z——>—2 z—>~2 [Limit Laws 1 and 2] =3 lim x4+2 lim 502— lim $+ lim 1 [3] I—i—2 16—2 m—o—Z rev—2 :3(—2)4+2(—2)2 — (—2)+(1) [9. 8. and 7] :48+8+2+1:59 ‘ 2 4 lim 2x2 +1 311312956 +1) L~ ' L 5 'z—>2m2+6$—4—lin§(m2+6$~4) [mm aw] 2 1m; x2 +1111; 1 :limx2+611mx—iim4 [z‘l‘and3] xfl2 z—>2 m—92 2(2)2 + 1 9 3 —W—fi—Z [9.7tand8] 5. 513% A 4)(;c3 + 5:17 — 1) = Iirr§(a:2 ~ 4) - lims(2:3 + 5x — 1) [Limit Law 4] _ . 2 ¥ . . . 3 . _ . - ($213“? $112334) (me +52% $1351) [1 Land 31 =(32—4)-(33+5-3a1) [7.8.and9] 25-412205 76 7. :1 CHAPTER 2 LIMITS AND DERIVATIVES 6. tango? + mm + 3)5= tgmluz + 1)3 . t§m1(t+ 3)5 [Limit Law 4] 3 5 = Liana? + 1)] ~ [thriller + 3)] [6] 3 5 : [tgmlfi +331] ~ [31191115ng 3] [1] : [(—1)2+1]‘"‘~[71+3]5 :8-322256 [9,7,and8] lim ( 1 + 39: >3 : (11m [6] Eel 1+ 4502 + 3:04 m—>1 1 + 4302 + 33:4 113110 +395) 3 £31 1 WW 3 :I12r111+41mw2+31121w4I [2’ “M31 1 +3(1) 3 4 3 1 3 1 * I1+4(1)2 +3(1)4I _ Is] _ <5) _ é [7‘8‘and 9] Uligx/u‘i +3u+6 = 7113315114 +3u+6) [111 = v (—2)4 + 3 (-2) + 6 [9, 8. and 7] : W : x/Ié : 4 113:: m 2 :31: <16 — $2) “‘1 = 11111 16— lim :32 [2I W4, Hr :t/16—(4)2:0 [7and9] 10. 11. 12. 13. 14. 15. 16. (a) The left—hand side of the equation is not defined for (b) Since the equation holds for all a: aé 2, it follows :3 = 2, but the right—hand side is. that both sides of the equation approach the same limit as a: —» 2. just as in Example 3. Remember that in finding lim f(:c), we never consider 3c : a. £1)—-*(l 2 _ A lim w = lim (“Mimi 2) : lim(w+3) :2+3:5 x—>2 m—2 m-v2 5572 33—?2 m $2+5m+4 hm (m+4)(m+1) hm x+1 —4+1 #3 3 1 —————— : ——’———- — ! : —— : — m-'~4 $2+3$—4 1% 4(36 I 4)(;L' 1) a: > 4:12 1 4 1 5 5 2; limgx 36:6doesnotexistsincecc—2aObutmzmm+6—>8as:c—>2. 30—» (r— 1. 332—417 hm x(mm4) #111“ an ! 4 _E m‘i‘iw2—3mi4 ‘1A4 (xm4)($+1) x~4x+1 _ 4+1 5 lim 112-9 _ hm (t+3)(t—3) _ hm t—3 # #3—3 #iszg ta732t2+7t+3 't—>73(2t+1)(t+3) 1+3 2t+1 2(—3)+1 —5 5 53—496 lim does not exist since $2 — 3:0 — 4 —> Obutcc2 — 4m —+ 5 as x ——> —1i x—Hl x2 —3:1:—4 17. 18. 19. 20. 21. 23. 24. 25. 26. ' 27. 28. SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS U 77 2 . (4+h)2—16_. (16+8h+h2)—16_- 8h+h _‘ _h<8+h>:1- h=8 0:8 m—fim—hh-m ‘I‘L‘é h 1218+) + . 553—1 ‘ (x—1)(x2+m+1) . m2+ac+1 12+1+1 3 11m =11m —11m _ _— x>lx2 1 z>1 (x 1)(l‘:1) z—vl $+1 1+1 2 . (1+h)4,1 , (1+4h+6h2+4h3+h4)—1 , 4h+6h2+4h3+h4 11m Zlmlemfi h—rO h_.0 h h—»0 h 2 3 :limmzlim(4+6h+4h2+h3):4+0+0+0:4 h—»0 h h—»0 _ (2+h)3_8 . (8+12h+6h2+h3)78 . 12h+6h2+h3 11m—:11m\:1mfi 1H0 haO h h—~0 h :}llir%(12+6h+h2):12+0+0=12 I 9—t i. (3+x/i)(3—\/i)__ _ _ Ifl31fiEfl31‘flilfl(3+m*3+fi*6 hm 1+h~1_hmx/1+h—1‘\/1+h+1~hm (1+h)—1 _lim h hao hao h x/1+h+1 h—»0h(\/1+h+1) haoh(\/1+h+1) . 1 1 1 :llth =- h~0\/1+h+1 x/T‘I-l 2 limvx+2—3_lim\/m+2—3.\/93+2+3 hm (x+2)~9 24 x—7 x—+7 w—7 ,/m+2+3 I—>7($~7)(\/x+2+3) . x—7 1 1 1 1 :hmxzhm‘: :— m—'7(a:~7)(\/;B+2+3) 297x/m+2+3 \/§+3 6 . 224—16 . ($+2)($~2)(x2+4) . . ‘ 312; 36-2 ‘33; $4 -;L'%<z+2><w2+4>=ifl<m+2>iwwz+4> =(2+2)(22+4):32 1+1 $+4 . . , 4 1 1 1 149521 4x:1L:._:_:i_ ELK—144+z 231344+x min—144x(4+:c) $13134“ 4—4) 16 1 1 ‘ (t2+t)it t2 ' 1 1 1 —— :1 ‘2. \:. —:_: 35a t2+t) 35 t(t2+t) gflt~t(t+1) ALot+1 0+1 1 . z2—81_. (:c~9)(:c+9)_. (\/_—3)(\/E+3)(x+9) factorac—Qasa 11m fillm 11m z~9fi~3 2H9 f—3 m—>9 f—3 differenceofsquares :iiirg)[(\/E+3)(ac+9)]:(\/§+3)(9+9)=6-18=108 1 1 1 _ fl _1 ‘—— _ liInM2hm3+h 3:1mmzhmi h—>0 h 11—10 h 1H0 h(3 + h)3 h—>0 h(3 + h)3 11 [ 1 J 1 1 1 :1m — :—'\:—~:i— 1H0 3(3 + h) lllmz) [3(3 + h)] 3(3 + 0) 9 78 CHAPTER 2 LIMITS AND DERIVATIVES 29.1im< 1 11>,lim1-V1H .(1—«1+t)(1+\/m) t —11m :1- _._;__ no tx/1+t t Hon/1H Ho tx/t+1(1+x/1+t) tgflth/1+t(1+\/1+t) :hm—__i__._____’1_____11 t—’0\/1+t(1+\/1+t) x/1+0(1+\/1+0)’ 2 . fiixz . fi(1—$S/2) fi(1—fi)(1+fi+x) 30. fl;— = fi— : ——T79_3——- [dlfference of cubes] :lim[fi(1+fi+x)]:im[1(1+1+1)]=3 arc—’1 Another method: We “add and subtract" 1 in the numerator. and then split up the fraction: _ 2 —1 1— 2 lim—L m :11m-—-———(‘/E )+( S")? z—pl z—rl 31. (a) 1'5 ’ —0.001 0.6661663 ‘ 41.0001 0.6666167 71 1 —0.00001 0.6666617 -- #0000001 06666662 7 .5 0 0.000001 0.6666672 lim H. z 3 0.00001 0.6666717 we 1 7 1 + 333 0.0001 06667167 0001 0.6671663 . . 2 The 11mm: appears to be 1. a: ./1+3x+1> rm $(x/1+3$+1) im x(\/1+3x+1) . _ 1 _ ______._. ((2)1113) V1+3x~1 \/1+3:c+1 w—vO (1+3$)—1 m-»0 356 : g 116%) («E—35 + 1) [Limit Law 31 :l[ I1im(1+3$)+lim1] [landll] 3 m->0 m—bO : limol+3lirrbm+1> [1.3.and7] [7 and 8] II ODIH DDIH CHI" A H + OJ C + 6—1 V SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS I: 79 32. (a) 05 (b) -- ~0.001 02886992 _1 - 1 20.0001 0.2886775 —0.00001 0.2886754 —0.000001 0.2886752 0.000001 0.2886751 0 111% “W m 0.29 0.00001 0.2886749 93* $ 0.0001 0.2886727 0001 0.2886511 The limit appears to be approximately 0.2887. (C)lim<x/3+x—\/§.\/3+x+\/§>fihm (3+:c)23 :hm 1 120 w 1/3+x+\/§ I~0m(\/3+x+\/§) I—‘Ox/m+\/§ lirr(1)1 [Limit Laws 5 and I] : lirrbx/3+zc+ lin})\/8 1 :3 [7andll] /11n%(3+ac)+x/§ 1 [17 d8] :* . .an x/3+0+\/§ 1 2m 33. Let fix) 2 2:02. g(.r) 2 .272 cos 207m: and h(a:) 2 .732. Then ~13 cos 207ch g 1 2> —502 S :62 00520701: 3 x2 2> f(a:) g g(a:) S h(;c). So since lint;J f(;c) 2 lirq) h(a:) 2 0. by the Squeeze Theorem we have lim g(.z’) 2 0. w—ro 34. Let f(ac) 2 —W. g(:c) 2 sin(7r/x). and }L(z‘) : M. Then —1 : sin(7r/;c) g 1 => —\/W 3 Wsin(7r/m) : x3 +z2 2> S g(x) S So since f(x) 2 h(x) 2 0. by the Squeeze Theorem we have lini) g($) 2 0. 35.1 g f(m) S m2+2x+2forallz. Now lim 12 land 2—121 mlingl (x2 + 2x + 2) 2 1111111 .222 + 21111131 cc + Ilirr_11 2 2 (—1)2 + 2(—1)+ 2 2 1. Therefore. by the Squeeze Theorem. lim f(;z:) 2 1. z—>—1 80 I: CHAPTERZ LIMITS AND DERIVATIVES 36. 3ng(:c)gm3+2forogxg2. Now lim3w:3and lim (x3+2) = limx3+lim2=13+2:3. 1'4?) 1H1 zv—d $.41 Therefore. by the Squeeze Theorem. lim f : 3. 2H1 37. —1 S cos(2/:i:) S 1 :> —m4 S $4 cos(2/:c) S 304. Since lim (—154) = 0 and lim $4 : 0. we have w—io z—vO lim [$4 cos(2/w)] : 0 by the Squeeze Theorem. zao 38. #1 S sin(7r/:c) S 1 => 6—1 < esmw/I) S 61 => fi/e S fieSinW/x) S fie. Since 1113+ (fi/e) : 0 and lira:L e) : 0. we have lim eSlr‘W/xl] = 0 by the Squeeze Theorem. 1%04— 39.1fm>74.then|z+4|:w+4,so lini |x+4|: lim (m+4):i4+4:0. w—>74+ mgr—4+ [fw< —4,then|w+4| : —(:c+4).so lim |w+4| : lim etc—l4) : —(74+4) :0. x—i~4_ 1345—47 Since the right and left limits are equal. lim4 [at + 4| 2 O. . . + 4] . 7 (:c + 4) . 40.1iz<—4.then +4:— 4..~ 1 '9” : l ____: _ :, Ix I (N “we- H4 .13- H4 .1131 1) 1~ . |5c ~ 2| , 9: — 2 . 41.1fx>2.then|a:-2l=m—2.so 11m : lim : 11m 1:1.If:c<2.then a5—.2+ ac — 2 1—)2+ 3c — 2 m_.2+ . — 2 . — I e 2 . . . . lac — 2| : — (so 7 2). so lim Ix l = lim L—Z : lim —1 : —1. The right and left limits are 93—;2‘ {E i 2 z—>2‘ ,T — 2 za2’ . . Ia: ; 2| . different. so 11m2 does not most. x—> LC 7 3 42.1fm > then I23: —3| : 2x — 3. so 2:52 7 3x 2x2 — 3a: m (233 — 3) 3 l‘ ———— : l' ’ : l' —— : l‘ = 1.5. If —,th $332+ |2$ 7 3| $332+ 2m # 3 $132+ 2m # 3 $133+”: “3 < 2 8“ 2$2 # 3x 2202 — 3:5 as (22: e 3) 273: —2. l' ———-: l' —————: l' ———-: l' —:—1.5. l x l 3 w SOFT?)- |2m!3| will} ~(2xv3) JE- —(2a:—3) will} it . . . . . , . 2:62 — 3x . The right and left limits are different. so 11m ———- does not ex13t. 2—4.5 |25c # 3| 1 1 2 . 43. Since : —a: for a: < 0. we have lim — : lim (— — —> : lim —. which does not most a:—>O- 1: [ml :0 —:r z-vO‘ :5 since the denominator approaches 0 and the numerator does not. 1 1 1 1 44. Since = mforx > 0. we have lim (— — ——> 2 lim (— — —> = lim 0 : 0. I_.0+ an lm] LE-vO'l' 1t :8 z—>0+ 45. (a) (b) (i) Since sgnz : 1 for (E > 0. $133+ sgnx : $138+ 1 : 1. (ii) Since sgnx : —1for:c < 0. lim $in :3 : lim -1 : —1. z—>0_ z->0‘ (iii) Since lim sgncc gé lim sgn 7:. lim sgnw does not exist. cc->O’ zao+ $—’0 (iv) Since Isgnml : 1 form 7'5 0.1irrhlsgnml = lim 1 : 1. xfi x—>0 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS 81 46. (3) lim = lim (4—302) : lim 4* lim :52 (C) x—v2‘ va’ LE—>2— :r—v?‘ :4~4:0 fix) : (:6 T1) 2 ac _ 1 :2—1:1 (b) No, ling f(nc) does not exist since lim f(x) # lirgi+ z—o2‘ 2 2 a: — 1 a: — 1 , 2 2 ac — 1 m — 1 . (“l .13?— |$ A 11 if? — (x — 1) m3?— (1’ ) (b) No. lim1 does not exist since lim F(m) 7E lim x—>1+ 1—»1’ . . V - 2 2 2 : b 48. (a) (1) $133+ h(z) i $133+ 3: 0 0 ( ) y (ii) lim h(x) : lim m = 0. so liniJ h(w) = 0. 6 :t—»O‘ xHO— 1‘" 4 (iii) 11m1 h(;c) = lim1 :02 : 12 = 1 2 (iv) Man) 2 $13; 3:2 2 22 : 4 0 2 X (V) lim h(z) lim (8—m):8~2:6 m—>2+ m—»2+ (vi) Since lim h(;r) # lim+ lirr12h(m) does not exist. I—>2_ x—v2 49. (a) (i) : —2 for —2 3 ac < ~1, so lim : lim (—2): —2 z—>72+ x—v72+ (ii) : —3 for —3 3 x < —2. so lim = lim (—3) = —3. The right and left limits are different. z—v—Q‘ I—9—2’ so lim2 does not exist. z—>fi (iii) = —3 for —3 S x < *2. so lim = lim (—3) = —3. z—>—2.4 ra—ZA (b) (i) [[27]]:n—1forn—1§$<n.so lim [M]: lim (n71):n—1. :c—>n_ z—vn.‘ (ii) [[mfl:nforn51:<n+1.so lim [[1]]: lim n:n‘ z—vn+ z—>n+ (0) lim exists <=> a is not an integer. I—>a 50. (a) 0 I X (b) (i) lim f(m): lim (x—flxfl): lim [x—(n—1)]:n~(n—1):1 I—rn‘ w—tn‘ 11—”).— (ii) lim f(.’l§): lim (m—flmfl): lim (x—n):n~n:0 z—vn‘l‘ m—vn.+ E—Vn (0) lim f(.z) exists <:> a is not an integer. m—ra 82 C CHAPTERZ LIMITS AND DERIVATIVES 51. The graph of f = + fl—xl] is the same as the graph of 9(33) : —1 with holes at each integer. since f (a) = 0 for any integer a. Thus. lim f(x) = —1 and limflL f = *1. so lim f(m) : *1. However. 32—»2 z—tZ— z—v2 f(2) : [[2]] + [[72]] : 2 + (—2) = 0. so 1Ln12f(m) 75 f(2). C2 UHF— 52. lim (L0 1 — v—2— ) : Lin/IT : 0. As the velocity approaches the speed of light, the length approaches 0. A left—hand limit is necessary since L is not defined for v > c. 53. Since p(ar) is a polynomial. p(a:) = a0 + an: + (122:2 + - - ~ + aux". Thus, by the Limit Laws, lim p(x) : lim ((10 + an: + (zsz + ~ - . + aux") I—Hl I——>a :ao+a1limx+a21imm2+-~+anlimxn man sic—m z_,a =ao+ala+a2a2+H-+anan=p(a) Thus. for any polynomial 1). lim p(x) = p(a). ac—>a 54. Let r(:c) = p—(fl where p(:c) and q(a:) are any polynomials. and suppose that q(a) 79 0. Thus. q(I) lim p(w) = %:—Z : [Limit Law 5] : 2% [Exercise 53] : Ma). 55. Observe that 0 g flan) S x2 for all so. and lim 0 : 0 : him :32. So. by the Squeeze Theorem, lirn f(ac) : 0. x—+O m—rO x—vo 56. Let f(:c) = and g(x) : Then lim and lim g(ac) do not exist (Example 10) but x—+3 2H3 mm) + 993)] = ligflml - M) 2 lim 0 = 0 za3 57. Let f(m) : H(cc) and g(;c) : 1 — where H is the Heaviside function defined in Exercise 1.3.59. Thus. either f or g is 0 for any value of x. Then lirrti) f(ac) and liin)g(sc) do not exist, but lim : lim 0 : 0. (IS—>0 95—»0 581. x/fi—x—Z hm(\/6—mi2 \/6'$+2 \/3—a:+1> .11’Il———'—: - . ré2x/3—zcgl 1—»? x/3—w-1 x/6—x+2 x/3#z+1 rm (s/6—x)2-22 x/3-a:+1 lim(6;:c—4 I/3—cc+1> _1 t . z~2 ./—_’3_m)2_12 x/6-a:+2 w~2 3—m—1 x/6—m+2 59. Since the denominator approaches 0 as x —> —2. the limit will exist only if the numerator also approaches 0 as :c ——> 72. In order for this to happen. we need lim2 (3m2 + am + a + 3) 2 0 <=> 2—»— 3(—2)2 + a(!2) + a + 3 = 0 <=> 12 — 211 + a + 3 = 0 <2”; (1 : 15. With a = 15. the limit becomes 2 3 3 3#2 3 hm 3m +15x+182 hm 3(m+2)($+3) _ hm £23.: 3 _ :——:—1. z 2 zglx 2 I>2(;1: 1)(:c+2) :c—»—2 m—l —2—1 —3 SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT E 60. Solution 1: First. we find the coordinates of P and Q as functions of r. Then we can find the equation of the line determined by these two points. and thus find the x-intercept (the point B). and take the limit as r —> 0. The coordinates of P are (0. 7‘). The point Q is the point of intersection of the two circles :02 + y2 = r2 and (an e 1)2 + y2 : 1. Eliminating y from these equations. we get 72 — 3:2 = 1 — (a: — 1)2 <=> r2 : 1+ 220 *1 <=> m : $13. Substituting back into the equation of the shrinking circle to find the y—coordinate. we get (ér2)2 + 312 : r2 42> y2 : r2(1 — irz) ¢> y : r, /1 — it"? (the positive y—value). So the coordinates of Q are (579, r 1 — fiT‘Z The equation of the line joining P and Q is thus 1“ 1~— it” — 7" y i 7" — 1 2 0 (cc ~ 0). We set y = 0 in order to find the w-intercept. and get 5T * 1 2 —lt~2(,/1—lr2+1) h T 2’“ 2 4 2(./1—§r2+1). r((/1—%r271) l‘iflil Nowwetakethelimitasr—>0+: lim $2 lim 2((/1—%r2+1>: lim 2(x/1+1) :4. 14—10+ T—vO‘l” r—bO‘l' So the limiting position of R is the point (4. 0). Solution 2: We add a few lines to the diagram. as shown Note that ZPQS : 90O (subtended by diameter PS). So ZSQR = 900 : ZOQT (subtended by diameter 0T). It follows that AOQS : ZTQR. Also ZPSQ 2 900 — ASPQ = ZORP. Since AQOS is isosceles. so is AQTR. implying that QT 2 TR. As the circle 02 shrinks. the point Q plainly approaches the origin. so the point R must approach a point twice as far from the origin as T. that is. the point (4.0). as above. 2.4 The Precise Definition of a Limit 1. (a) To have 52: + 3 within a distance of 0.1 of 13. we must have 12.9 3 5:0 + 3 S 13.1 :> 9.9 3 590 S 10.1 : 1.98 g 90 S 2.02. Thus. :0 must be within 0.02 units of2 so that 530 + 3 is within 0.1 of 13. (b) Use 0.01 in place of 0.1 in part (a) to obtain 0.002. 29.99 g 63: g 30.01 :> 4.9983 g a: 5 5.0016. Thus. w must be within 0.0016 units of5 so that 6$ ~ 1 is within 0.01 of 29. (b) As in part (a) with 0.001 in place of 0.01. we obtain 0.00010. (c) As in part (a) with 0.0001 in place of 0.01, we obtain 0.000010. ...
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This note was uploaded on 12/08/2009 for the course MATH 101 taught by Professor Dr.tahir during the Fall '08 term at King Fahd University of Petroleum & Minerals.

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02-3 - SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS...

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