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11825-0130620262_ismappA

11825-0130620262_ismappA - Page 795 I1 2Iis less than 5...

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Unformatted text preview: Page 795 I1 + 2Iis less than 5. Appendix A APPENDIX A Introduction to the Theory of Limits 1. For any chosen value of 6 we want the As we approach 2, 4. distance between f(x) and L to be less than 6. In absolute value notation this is {f(sc) — LI<E For our function and limit we need 6. [(22 — 5) — (—3)|<6 l2zr —2I <6 2|:c — 1|< 6 So the distance between z and 1 needs to be less than 3. Therefore, let 6 = %. I31+7 —1]=3lz+2l<6sothat 6 IZ+2I < '3- Pick6=§ lim (32+ 1): 4 gé 5; the interval on the y- 1—»1 axis in the neighborhood corresponding to x = 1 can be arbitrarily restrictive. 0<|z-1|<6 must apply and the given statement is false. The actual limit is 4. lim (22 — 2)=5im_plieslz2 — 2 — 5l<e. 9. H2 Forlz— 2I<6,2 — 6<27<2+6. The doubter picks 6 = 1. The 6 inequality 10 becomes —1<12 - 7<1,6<£2<80r 2.45 < «E < 2.83. The believer has the impossible task of finding a number around a: = 2. The given statement is false. The actual limit is 2. For any chosen value of 6 we want the distance between f (z) and L to be less than 6. In absolute value notation this is 12 ”(2) — L|<6 For our function and limit we need 13 l(z2 + 2) -— (6)l<6 '22 — 4i<€ lz— 2llz+ 2l<e 11. If the distance between z and 2, written as I2 — 2| , is to be less than 6, we need In: — 2K5) < 6 = 56 . Therefore, let 6 = minimum of{1,§ . '12 — 32+2l=|(z —2)(:c —1)| Letlz — llbe less than the minimum of %or6. Then%ng%-andlx — 2I_<_%. Thus, Iz— 2IIz—- 1| < g6=€0r6=32§€ The statement is true; no matter what number the doubter selects, the believer will always be able to find '22 — 3x + 2l< 6. [f(x) — L|=[(z+3) — 5I=iz — 2|<6 Choose 6 = 6. |f(t) — L|=|(3i — 1) — 0|=|31— 1| _ 1 - 3| * 3| The statement is false. Choose 6 = 0.3 and it is not possible to find a delta, because lt—§]<§=0.1 |f(z) - L|=|(3z + 7) — 1|=|3z + 6| =3|z+2l<36;choose6 2%. [f(r) —— Ll=[(2r - 5)+3|:2iz — 1| < 26; choose6=%. [f(z) — Ll=l(22 + 2) — 6|=|z2 — 41 =lr — 2H1 + 2|; choose6=min(1, g) aslz—2l<6and|z—2I<1means —1<z<3or1<z+2<5so lz—2llz+2|<6(5)=6. mm) — LI: 1 1—”’——2" _ T ' 2m ’ choose 6 = min(1, é)‘ 1' 2'. In order for f(z‘) to be continuous at a: : 0 , yin) f(z) must equal f(0). To show that lim sin 3.13 2—40 any 6 > 0 there exists a 6 > 0 such that = 0 we need to show that for Page 796 14. 15. |f(z) - Ll < c whenlx -— 0| < 6. 16. Arbitrarily letting c = 0.5 we need to find a 6-interval about 0 such that lsin %l < 0.5. However any interval about 0 contains a point 1: = % (with n odd) where sin % -7rn = sm— = 2 :l: 1. Therefore, there does not exist a 6-interval about 0 such that lsin :15! < 0.5, and f(z) must be discontinuous at :c = 0. By hypothesis, lime f(2:) = L which means that given 61 > 0 such that |f(.7:) — LI< 61 there exists 6 for 0 <12 — cl < 6. Let 17. 9(2) = kf(x) so that 19(2) — kLl=lkfm — kLl :lkl|f(:c) — L|<Ilc|e1 This says given 6 > 0, let 61 — L and then there isa 45 such that I“ lg(1:) —- [CL] < c if ' 0<|r — cl<6 18. This inequality is restated as lmcfiz) 2 [CL or lime km) = M. Given 6 > 0 there exists 61 and 62 such that 19. |f(a:) — L1l<gfor0<lz — c|<61 and [9(1) — L2l<§for0<lz — c|<62. Then ”(1‘)" 9(3) — (L1 — L2” = Ilf(1‘) - Ll] - [9(1) - L2H 20- SIf(I) - L1l+lg($) — Lzl <§+§=6 if0<|z — cl < min(61, 62). 21. This says that 1:21;] f(.r) —— g(z) I = L1 — L2. Appendix A If lianfflt) = L1 and lli‘rgcflz) = L2 then given any 6 > 0, If(:c) — L1|< 2%] 6 c|<6 and|g(:c) — L2I<Tb|if0<lz — Hence, laf(z) + 69(3) -— a(L1 + 6L2)| =lalf(z) “ L1] + I’M-"3) — L2H smilm) — LII+Ibugm — L2| <Ia12—fa—l+lbI—£-=e 2| bl This says that Iggclaflm bg(z>1 = a Ignore) + b 133,6 9(1)- Given 6 > 0,f(z) — 0|< ‘flfor 0<|z — c|<61 ”(1') — 0l< \/Efor 0<l1: — c|<62 So mam - 0mm — ongm — 0| < \/E \/E = c if0 <Iz - cl< min(61, 62) This says that lmcf(:c)g(z) = 0. 05lg(x) — Olglflz) —- 0l<c whenever 0<|z — c|<6. Thus,|g(z) — 0l<c for 0 <11 — cl<6 implies 1330 9(2) = 0. Let MI) = f(I) - 9(1) Since flat) 2 9(2) throughout an open interval containing c, the limit limitation theorem guarantees that Lilac be) = aggro) — gm] = lime we) — Iggc 9(1) 2 0 or Imam 2 1}an gm. 0: Hf(x)l — lLllslf(:v) — L|<e whenever 0 <|z -— c|< 6. Thus, “f(a:)| — lLl l< 6 implies llilnfllflx)! = ILI. By hypothesis, 1:3 c f (z) = L, and by Problem 20,1;gc|f(x)|=ILl, or ifs = Iz—L—l> 0 then there exists 6 so that “f(:c)' —|Ll' < lé-l for all 6 such that 0 < I z — cl < 6. Thus, —ITLI<(|f(z)[ — ILI)<'7“and AppendixA l—g—|<|f(ac)|<?’l2£-l. 22. a. lf(z)|<i2L—l, mz) )+L|<|f(z)l+|Ll<|Ll+3-l——2Ll=5ITM by Problem 21, if0 <Ia: — c| < 61. b. Given 62 > 0, there exists 6 such that [f(x) — Ll<62if0<|z — “<6 Then [[f (as)? — L2l=lf(z)+ Lllf(z) — LI <-5—l2£|€2 ifO <12: — c|<min(61, 6) = 62 c. Let 6- .. 5—l—2Llc2, thenl[f( [f(]:r:)2 — L2l< c for 0 < | 2— cl < 62 where 62: min(61, 6) 23- i—[U + g)2 - (f - 9)] =%[f2+2f9+ 92 — f2+2fg — f] = Hm + 2ng = I!) Thus, Iggy: Iggjw + 9)? — (f — gfl = £1,311.40 + g)? - (f — m by Problem 14. By Problem 15, it also equals 1 - 2 . 2 Iggy + a) — Iggy — w] by Problem 22 it becomes 3.1[(L+M)2 — (L — M)2]=LM 24. Letc=%c).r_’ lim f(z): (c) implies If ()— f(c>l< —c) for0<lz — Cl<6. —f——(—c)<f(r) — f(C)< 52—”) o < {(2—62 < f(z) < 3%; ,0<f(z) for0<l17 - cl<6. 25. 3.. Since f is continuous at L, we have lim f (w) = f(L) which means that there w—’L exists a 61 such that [f(w) — f(L) I < ‘1 Wheneverlw — LI< 61. Page 797 b. Since 19(2) — LI < 62 lime g(z) = L, by hypothesis, whenever 0 <|:r — c|< 62. Let w = g(z) in part a, and c2 = 61. Then lf[g(z)l —f(L)l < 61 whenever 0 < | 2— cl < 62. 26. Since (as it turns out) the slope at z = 3 is — 26; pick K larger than 26 works. ...
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