02-4 - SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT E 60....

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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. 84 3 CHAPTER 2 LIMITS AND DERIVATIVES . On the leftsideofcc : 2. we need Icc— 2| < |1—70 e 2| : 3. On the right side. we need Iw— 2| < |%9 — 2| : g. For both of these conditions to be satisfied at once. we need the more restrictive of the two to hold. that is. 4 |$ e 2| < é. So we can choose 6 : . or any smaller positive number. \1 On the left side. we need |$ * 5| < I4 — 5| 2 1. On the right side. we need Ix — 5| < |5.7 — 5|: 0.7. For both conditions to be satisfied at once. we need the more restrictive condition to hold; that is. |:c — 5| < 0.7. So we can choose 6 = 0.7. or any smaller positive number. . The leftmost question mark is the solution of fl : 1.6 and the rightmost. fl : 2.4. So the values are 1.62 2 2.56 and 2.42 : 5.76. On the left side. we need |93 - 4| < |2.56 — 4| 2 1.44. On the right side. we need |:c — 4| < |5.76 i 4| : 1.76. To satisfy both conditions. we need the more restrictive condition to hold——namely. Ix — 4| < 1.44. Thus. we can choose 6 = 1.44. or any smaller positive number. 1 . The left—hand question mark is the positive solution of m2 = that is. :c : fl. and the right—hand question mark is the positive solution of $2 = that is. a: = On the left side. we need |zv e 1| < — 1| m 0.292 (rounding down to be safe). On the right side. we need |:c e 1| < e 1| m 0.224. The more restrictive of these two conditions must apply. so we choose 6 : 0.224 (or any smaller positive number). |t/4x + — 3| < 0.5 <5 25 < #450 +1 < 3.5. We plot the three parts of this inequality on the same screen and identify the x-coordinates of the points of intersection using the cursor. It appears that the inequality holds for 1.3125 g a: S 2.8125. Since |2 - 1.3125| : 0.6875 and |2 — 2.8125| : 0.8125. we choose 0 < 5 < min {0687508125} : 0.6875. |sinac — < 0.1 <=> 0.4 < sinzc < 0.6. From the graph. we see that for this inequality to hold. we need 0.42 g x g 0.64. So since |0.5 ; 0.42| : 0.08 and |0.5 i 0.64| : 0.14. we choose 0 < 6 g min {0.08.014} : 0.08. 0.7 SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT 85 9. Fore 2 1. the definition ofa limit requires that we find 6 such that [(4 + ac 2 3x3) 2 2] < 1 <=> 1< 4+:c‘3323 < 3whenever0 < |:c— 1| < 6.1fweplotthegraphsofy 2 1.y24+:c—3;t3 andy23on the same screen. we see that we need 0.86 g a: S 1.11. So since |1 — 0.86] 2 0.14 and Il — 1.11| 2 0.11. we choose 6 2 0.11 (or any smaller positive number). For E 2 0.1. we must find 6 such that |(4 + :6 ~ 3x3) — 2| < 0.1 <=> 1.9 < 4 +20 — 39:3 < 2.1 wheneverO < |;z: — 1[ < 6. From the graph. we see that we need 0.988 S ac g 1.012. So since|1~ 0.988l 2 0.012 and |1— 1.012| 2 0.012. we choose 6 2 0.012 (or any smaller positive number) for the inequality to hold. 0.981 1.02 10. Fore 2 0.5. the definition ofa limit requires that we find 6 such that e _ 1 — 1’ < 0.5 <2 :c 6“0 ~ 1 e“E — 1 0.5 < < 1.5 whenever 0 < [m 2 OI < 6. If we plot the graphs ofy 2 0.5. y 2 . and y 2 1.5 on the a: at same screen. we see that we need —1.59 g m S 0.76. So since [0 * (21.59)l 2 1.59 and |0 — 0.76| 2 0.76. we 1' choose 6 2 0.76 (or any smaller positive number). For 5 2 0.1. we must find 6 such that e .7; ~1I<0.1 <2 ez~1 0.9 < < 1.1 wheneverO < — x 0] < 6. From the graph. we see that we need —0.21 g m g 0.18. So since [0 — (—0.21)[ 2 0.21 and I0 ~ 0.18l 2 0.18. we Choose 6 2 0.18 (or any smaller positive number) for the inequality to hold. :13 11. From the graph. we see that ($2 +1)($ _ NZ > 100 whenever 130 0.93 S :5 g 1.07. So since {1 ~ 0.93! 2 0.07 and [1 ~ 1.07| 2 0.07. we can take 6 2 0.07 (or any smaller positive number). 0.8 l.2 86 3 CHAPTERZ LIMITS AND DERIVATIVES 12. For M : 100. we need -0.0997 < ac < 0 or 0 < :c < 0.0997. Thus. we choose 6 : 0.0997 (or any smaller positive number) so that if0 < < 6. then cot2 a: > 100. 200 y : cotzx ll *02 :—0.0997 0 z0.0997 For .M = 1000. we need —0.0316 < m < 0 or 0 < ac < 0.0316. Thus. we choose 6 = 0.0316 (or any smaller positive number) so that if0 < < 6. then cot2 x > 1000. 13. (a) A : in? and A = 1000 cm2 => m2 = 1000 :> 7" = T T : /%‘fl [r > 0] % 17.8412 cm. (b)|A—1000|g5 ;> aging—100035 :> 1000—5gw231000+5 => £7?ng 1—29—5 => 17.7966gr317.8858.i/1—‘f—0— 2352~004466and 1 / 19795- ; t / L780 z 0.04455. So if the machinist gets the radius within 0.0445 cm of 17.8412. the area will be within 5 cm2 of 1000. (c) 1: is the radius. f(x) is the area. a is the target radius given in part (a). L tolerance in the area (5). and 6 is the tolerance in the radius given in part (b). is the target area (1000). 5 is the 202 (°C) 14. (a) T : 0.1m2 + 2.15% + 20 and T = 200 :> 017.02 + 2.15511) + 20 : 200 2 [by the quadratic formula or m from the graph] w m 330 watts (w > 0) m- 198 (watts) (b) From the graph. 199 5 T S 201 :> 32.89 < w < 33.11. (c) a: is the input power. f(1:) is the temperature. a is the target input power given in part (a). L is the target temperature (200). a is the tolerance in the temperature (1). and 6 is the tolerance in the power input in watts indicated in part (b) (0.11 watts). SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT 15. Given 8 > 0. we need 6 > 0 such that if0 < Ia: — 1I < 6. then 3“ I(2:c+3)—5I <5. ButI(2m+3)75I <5 <=> I2z—2I <5 <:> 2Ix — 1| < a <=> Ia: ~1I < 5/2. So ifwe ChOOS€6 : 5/2. thenO < Ix ~ 1| < 6 :> I(2a:+3) — 5| < 6. Thus. + 3) = 5 by the definition ofa limit. y=2x+3 16. Given 8 > 0, we need6 > 0 such that if() < Ix — (—2)I < 6‘ then I(%w+3) —2I <8. ButI(%m+3)~2I <5 <:> I§m+1|<a <=> §|z+2l<a ¢> |z—(—2)I<2e.smf we choose6 : 251thenO < Im — (—2)I < 6 => + 3) — 2I < 5. Thus, 2121712690 + 3): 2 by the definition ofa limit 17. Given 5 > 0, we need 6 > 0 such that ifO < I2 — (—3)I < 6. then I(1 —4:c) — 13I < 5. But I(1 ~ 433)713I < 5 <=> I~4x —12I< e <:> I~4I Irc +3I < e <:> Ix — (—3)I < 5/4 So ifwe choose6 : 5/4, then 0 < Ix — (~3)I < 6 => I(1 * 42:) ~ 13I < 5. Thus lim (1 — 4x) = 13 by the definition of Ifl—3 a limit. 18. Given 5 > 0, we need 6 > 0 such that if0 < Ix — 4I < 6. then I(7 —3:12) — (—5)I < 5. But [(7 ~ 350) — (—5)I < E <=> I~3$+12I<5 <:> I—3IIw—4I<e <:> Im—4I<5/3.SO ifwe chooseé : 5/3. thenO < Izc — 4I < 6 => I(7 ~ 3m) — (—5)I < 5. Thus, lintl(7 ~ 3:0) 2 —5 by the definition ofa limit. 19. Given 5 > 0‘ we need 6 > 0 such that ifO < Im — 3I < 6. then m Ix —3I < 55. Sochooseé : 58. ThenO < Ix—3I < 6 :> x 3 t . i i . . 1L“ 3 — g < 6. By the definition ofa limit. hm — E $—>3 5 5. 87 88 3 CHAPTERZ LIMITS AND DERlVATIVES 20.Given5>0,weneed6>0suchthatif0<|m—6|<6.then|(§+3)—%|<5 <=> §igl<5 <=> film—6|<5 <=> |$i6|<45.Sochoose6=45.Then0<|$e6|<6 :> |w—6|<45 => 1 6 z 9 4 <6 :> Z—Zl<€ => |(Z+3)—§|<5.Bythedefinitionofalimit.lim(§+3)2%. maB 21. Given 5 > 0. we need 6 > 0 such that ifO < |x — (—5)| < 6. then |(4 * e 7] < 5 <=> legx—3l <5 <=> g|sc+5| < 5 <r> Ix— (75H < 25. Sochoose6 : g5. Then [93— (—5)| < 6 i [(4 — — 7| < 5. Thus. lim (4 r : 7 by the definition ofa limit. 1H-5 2 _ 22. Given 5 > 0. we need 6 > 0 such that ifO < lac — 3| < 6. then 10—332 — 7 < 5. Notice that ifO < Ia: — 3|. 3;, 2 in + —12 4 -3 thenx7é3.so———m——:fi—LD:m+4.Thus.when0<|$73|.wehave {13-3 10—3 12+m—12 ——x—_—3———7 <5 42> |($+4)*7l<5 ¢> lac—3|<5.Wetake6:5andseethat0<|ac—3|<6 $2+$e12 w2+x—12 :> ———-—77‘<5.Bythedefinitionofalimit.lim————:7. sc~3 183 310—3 23. Given 5 > O. we need 6 > 0 such that if0 < la: — al < 6. then |z — al < 5. So 6 : 5 will work. 24. Given 5 > 0. we need 6 > 0 such that ifO < la: 7 al < 6. then |c # cl < 5. But Ic * c| : 0. so this will be true no matter what 6 we pick. 25. Given 5 > 0. we need6 > 0 such that ifO < f O] < 6. then I172 — 0| < 5 <=> :02 < 5 42> < Take 6 = ye. Then 0 < [cc — 0| < 5 2 [$2 — 0| < 5. Thus. 111% :62 = 0 by the definition ofa limit. 26. Givene > 0. we need6 > Osuch that ifO < lac * 0| < 6.then —0l < 5 <=> |:c|3 < 5 <=> < Take 6 : Then 0 < [LE * 0| < 6 => [3:3 — 0| < 63 = 5. Thus. lirrifcs : O by the definition ofa limit. 27. Given 5 > 0. we need 6 > 0 such that ifO < [33 — 0] < 6. then — 0‘ < 5. But = So this is true if we pick 6 = 5. Thus. lint) : 0 by the definition of a limit. 28.Given5>0.weneed6>Osuchthatif9—6<a:<9.then‘\4/9~ —0l<5 4=> \4/9#m<5 <=> S)—ac<54 <:> 9—54<z<9.Sotake6:54.Then976<sc<9 => H79— .#Ol<5.Thus. lim {1/9 — cc : O by the definition of a limit. J:——»9’ 29. Given5 > 0.weneed6 > OsuchthatifO < [1: —2| < 6.then —4£E+5) 7 ll < 5 42> l$2e4w+4l <5 <=> l(x~2)2l <5.Sotake6:\/5.Then0<|me2l<6 <=> Im-2l<\/5 <:> |(:c r 2)2| < 5. Thus. lim2 (:62 — 4m + 5) : 1 by the definition ofa limit. 73—. 30.Given5>0,weneed6>OsuchthatifO<lxv3|<6.thenl(m2+m—4)—8l<5 <=> lm2+xel2l<5 <=> |(;c‘3)(;r.+4)|<5.Noticethatiflasfiil<1.then*1<r—3<1 :> 6<zc+4<8 => lx+4|<8.Sotake6:min{1.5/8}.Then0<lxr3|<6 <=> Km ~ + 4)‘ S — 3)| : 8- lac ‘ 3i < 86 S 5. Thus. (:02 + a: — 4) : 8 by the definition ofa limit. SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT 89 31. Given 5 > 0. we need 6 > 0 such that ifO < — (~2)| < 6. then |(:c2 — 1) — 3| < e or upon simplifying we need |m2 — 4| < 5 wheneverO < |:c + 2| < 6. Notice that if |m + 2| < 1. then —1 < m + 2 < 1 :> —5<$~2 < *3 :> |ac—2| <5. Sotake6=min{e/5.1}.Then0< |av+2| <6 => |$—2| < Sand |w + 2| < 5/5. so |(av2 ~ 1) ~ 3| : + 2)(x — 2)| = Im -I- 2| Iw — 2| < (€/5)(5) 2 5. Thus. by the definition ofalimit. lim (2:2 — 1) = 3. :c—»—2 32. Givens > 0. we need6 > Osuch that ifO < |:1: 4 2| < 6. then |23 —8| < 5. Now |m3—8| :|(:z:—2)(;22+2.2:~—4)|.[Hm—2|<1.thatis.1<m<3.thenx2+2$+4 < 32+2(3)+4=19 and so |x3 ~ 8| : Ix — 2| (2:2 —~2m+4) <19|m — 2|. So ifwe take6 : min{1. 139}.then 0 < [:6 — 2| < 6 :> |ac3 — 8 : Im — 2| (22 + 230 + 4) < 15—9 ~19 : 5. Thus. by the definition ofa limit. lim 9:3 : 8. x—t2 33.Givens>0.welet6:min{2.§}.If0<|x—3|<6.then|m—3|<2 => —2<a:—3<2 :> 4<$+3<8 => |x+3|<8.Also|$—3|<§.so|xz~9|=|z+3||$—3|<8-§:5.Thus. 11111 .732 : 9. 30—8 34. From the figure. our choices for 6 are 61 = 3 ~ v9 — a and 62 : \/9 + 5 — 3. The largest possible choice for 6 is the minimum value of {61. 62}: that is. 6 : min{61.62} : 62 : V9 + e — 3. 35. (a) The points of intersection in the graph are ($1. 2.6) and (232. 3.4) with :121 m 0.891 and .272 R: 1.093. Thus. we can take 6 to be the smaller of1 — x1 and 212 — 1. So 6 : $2 ~1z 0.093. (b) Solving m3 + :L‘ + 1 : 3 + 5 gives us two nonreal complex roots and one real root. which is (3(8) _ (216 +1085 +12x/336 + 3245 + 8162 )2/3 a 12 6 (216 + 108s +12x/336 + 3245 + 8152 )1/3 (c) Ifs : 0.4. then 21(5) R: 1.093 272 342 and 6 = $(E) — 1 m 0.093. which agrees with our answer in part (a). . Thus. 6 : 23(5) ~ 1. 90 CHAPTER 2 LIMITS AND DERIVATIVES 1 1 - — 5 < 6 whenever 36. I. Guessing a value for 6 Let 5 > 0 be given. We have to find a number 6 > 0 such that ac '1 27:6 Iwi2I 1 0<x-2<6.But———= = <.Wfi " —— I I x 2I I 2x |2$l 5 e ndaposmveconstantCsuch that '23:! <C :> Ice—2| . 5 . . . '2“ < C Ix — 2| and we can make C Ix — 2I < a by taking Ix — 2| < 5 2 6. We restr1ct a: to lie in the . 1 1 1 1 1 1 1 1 1ntervalm—2 <1 => 1<m<3so1>—>— :> —<— — —— —. :— I I as 3 6 2w<2 Z} IZmI <2 SOC '5 suitable. Thus. we should choose 6 = min {1, 2e}. 2. Showing that 6 works Given a > O we let 6 : min {1. 25}. HO < Ix — 2| < 6. then Ix — 2| < 1 => 1 — 2| < ___ l :c 2 I2m| 2 1 1 . 1<x<3 => % < 5(as1npartl).AlsoIcc—2I<25.so - 25 : a. This shows that lim(1/x) : z‘r2 37. ]. Guessing a valuefor 6 Given 5 > O. we must find 6 > 0 such that - «5| < 5 whenever wa that fl + \/5 > C then K—al— < lit - al < e‘ and we take Isc — aI < C5. We can find this number by fi+¢5 C restricting m to lie in some interval centered at 0 < I9: — aI < 6. But — MEI = %;a—I— < 5 (from the hint). Now if we can find a positive constant C such a.lfIac—a| < éa.then—%a<m—a< %a => éa<m< %a => fl + W > I/éa + J5. and so C : %a + «5 is a suitable choice for the constant. So Ia: * aI < (I/éa + V5): This suggests that we let 6 = min {%a, (I/%a+ fi>a}. 2. Showing that 6 works Given 5 > 0. we let 6 : min {%a. (I / %a + fi)e}. IfO < Ia: — aI < 6. then Ix—aI < go :> fi+fi> I/%a+\/E(asinpartl).Also|;v—aI< (I/%a+\/E)5.so t/ 2 + \/'d 5 Ian — aI ( a/ > . . , . . — a # < — 5. Therefore. 11m f = \/(—l by the definition of a 11m1t. I\/_ fi+\/E ( fa/_2+\/a> m—va 38. Suppose that lim H(t) : L. Given 5 : there exists 6 > 0 such that 0 < ItI < 6 :> IH(t) — LI < % <=> tao L—g<H(t)<L+§.For0<t<6.H(t):Lso1<L+§ 2 L>§.For—6<t<0.H(t):0.so L e % < 0 :> L < This contradicts L > Therefore. H(t) does not exist. 39. Suppose that lirgflm) 2 L. Given 5 : there exists 6 > 0 such that 0 < ImI < 6 => If(av) — LI < Take any rational number 7“ with 0 < IT'I < 6. Then f(r) : 0. so I0 — LI < so L 5 ILI < Now take any irrational numbers with 0 < IsI < 6. Then f(s) : 1. so I1 7 LI < Hence. 1 e L < so L > This contradicts L < so f(:c) does not exist. SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT 91 40. First suppose that lim f(;r) : L. Then. given a > 0 there exists 6 > 0 so that 0 < [m — a! < 6 => 41. 42. zaa |f(;c)iL|<e.Thena—6<:c<a :> 0<|m—a|<6solf($)iL|<e.Thus. lim f(a:):L.Also z—>a.‘ a<m<a+6 => O<|w—a|<6so|f(:c)—L[<5.Hence. lim f(3:)=L. 1—>a+ Now suppose lim z—va’ = L = lim f(:c). Let a > 0 be given. Since lim flan) : L. there exists 61 > 0 so + m—ra zflai thata — 61 < x < a :> |f(:c) — L| < 5. Since lim+ f(x) : L. there exists 62 > 0 so thata < :1: < a + 62 2‘”). => |f(x)—L[<5.Let6bethesmallerof61and62.ThenO<lx—a]<6 => a—61<ac<aor a < a: < a + 62 so 7 LI < 5. Hence. lim f(:c) : L. So we have proved that lim f(:c) : L <:> z—wz IB—ba lim f(x):L: lim I—Oai Z—>CL+ fi > 10000 c> (w+3)4 < 1 e |m+3l < ; <:> [2244,» < i (:1: + 3)4 ‘ 10.000 310.000 10 1 GivenM > O.weneed6> OsuchthatO < |$+3| <6 => 1/(.93+3)4 > M. Now m > M <=> (073—3)4 < i <=> lx+3| < 1 .Sotake6: 1 .ThenO < |$+3| < 6: 41 => M {7M {‘/M M 1 1 h A . l’ = . (z n 3)4 > 1 so wing?) (at + 3)4 DO . Given M < 0 we need 6 > 0 so that lnm < M whenever 0 < :e < 6; that is. 1: 2 e1” < 6M wheneverO < m < 6. This suggests that we take 6 : eM. IfO < ac < 6M. then Inac < 1n eM = M. By the definition ofa limit. lim Ina: : —oo. z—vO’l' (a) Let M be given. Since lim f(:c) = 00. there exists 61 > 0 such that 0 < In: — al < 61 => f(:1:) > M +1— 6. Since lim g(x) : 0. there exists 62 > 0 such that 0 < [:c — al < 62 => — cl <1 => g(x) > c — 1. Let6be the smaller of61 and 62. Then 0 < Ia: ~ al < 6 :> f(ac)+g(x) > (M+1—c)+(c— 1): M. Thus. lim [f(2:)—I—g(m)] : 00. (b) Let M > 0 be given. Since lim g($) 92—”). = c > 0. there exists 61 > 0 such that 0 < lac — al < 61 => [9(33) ~ cl < 6/2 :> 9(00) > c/2. Since lim f(ac) (II—>11, 2 00. there exists 62 > 0 such that 0 < Ix — al < 62 :> f(a:)>2M/c.Let6:min{61.62}.Then0<[ac—a[<6 => f(m)g(a:)>2—§ c M.so gig; fix) you) = 00. (c) Let N < 0 be given. Since Iim g(m) Zi—Hl : c < 0. there exists 61 > 0 such that 0 < [x ~ (1] < 61 3 Ig(:c) — cl < —c/2 :> g(x) < c/2. Since lim f(.’1:) : 00. there exists 62 > 0 such that 0 < In: — al < 62 => f(:z:) > 2N/c. (Note that c < 0 and N < 0 :> 2N/c > 0.) Let 6 : 2N c min {(51, Then 0 < [33 — a] < 6 => f(x) > 2N/c => f(.7:)g(ac) < : N. so Iim f($)g($): —oo. 2 I—>a ...
View Full Document

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.

Page1 / 9

02-4 - SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT E 60....

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online