Linford 1. Kyle Linford. Math 211. Honors Project. Theorems to Analyze: Theorem 2.4 The Limit of a Function Involving a Radical (A4)

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1 Liford 1 Kyle Liford Mth 211 Hoors Project Theorems to Alyze: Theorem 2.4 The Limit of Fuctio Ivolvig Rdicl (A4) Theorem 2.8 The Squeeze Theorem (A5) Theorem 2.9 The Limit of Si(x)/x = 1 (p. 85) Theorem 4.10 Limits t Ifiity (A11) Theorem 5.8 Preservtio of Iequlity (A12-A13) Theorem 8.4 L Hôpitl s Rule (A14) Theorem Clssifictio of Coics y Eccetricity (A17) The Itermedite Vlue Theorem

2 Liford 2 Theorem 2.4 The Limit of Fuctio Ivolvig Rdicl Let e positive iteger. The followig limit is vlid for ll c if is odd, d is vlid for c > 0 if is eve. Let c > 0 d e positive iteger ( є N). Assume tht ε > 0. We will show tht there exists δ > 0 so tht x We c write this s -ε < x Assume ε < c. c Becuse ε is greter th 0 d c lim x = c x c c < ε wheever 0 < x c < δ. < ε wheever -δ < x c < δ. is greter th ε, this implies tht 0 < c will just e the differece of umer d smller oe. Let δ e the smller of the two umers: c ( c ε) d ( c + ε) c. Usig sustitutio, we c work ckwrds to show tht -δ < x c < δ c e writte s [c ( c ε) ] < x c < ( c + ε) c. ε < c. The term c ε This simplifies to ( c ε) c < x c < ( c Addig c, the expressio simplifies to ( c + ε) c. ε) < x < ( c + ε). Usig lger to rise the expressio to the 1 power, the expressio further simplifies to c ε < x < c + ε. The iequlity c e show to still holds for this expoet y tkig <. Multiplyig ech side y gets <. Similrly, multiplyig y gets <. Thus, <. We c the tke the squre root of oth sides to get <, d the iequlity is show to still e true. Sutrctig c x c < ε. from the expressio llows us to write ε < x c < ε. This is the sme s

3 Liford 3 Theorem 2.8 The Squeeze Theorem If h(x) f(x) g(x) for ll x i ope itervl cotiig c, except possily t c itself, d if lim x c h(x) = L = lim x c g(x), the lim x c f(x) exists d is equl to L. Assume tht h(x) f(x) g(x) for ll x i ope itervl cotiig c, except possily c itself. Assume tht lim x c h(x) = L = lim x c g(x). The proof will show tht lim x c f(x) = L. We eed to verify the defiitio of the limit: for ech ε > 0 there exists δ > 0 such tht if 0 < x c < δ, the f(x) L < ε. Let ε > 0 d 0 < x c < δ so f(x) L < ε. However, sice we hve two fuctios, we use δ 1 d δ 2 so tht h(x) L < ε whe 0 < x c < δ 1 d g(x) L < ε whe 0 < x c < δ 2. By the defiitio of limit d our ssumptio tht h(x) f(x) g(x) for ll x i ope itervl cotiig c, except possily t c itself, there exists δ 3 > 0 so tht h(x) f(x) g(x) for 0 < x c < δ 3. Let δ e the smllest of the three delts. By defiitio of limit gi, ecuse ε > 0 d if 0 < x c < δ, the we c sy gi tht h(x) L < ε d g(x) L < ε. We c sy tht ε < h(x) L < ε d ε < g(x) L < ε. Addig L to the left side of the expressio, we c write L ε < h(x) < ε + L. Similrly, we c write L ε < g(x) < L + ε. Usig sustitutio with our ssumptio tht h(x) f(x) g(x), we c sy tht L ε < f(x) < L + ε. By defiitio of limit, this implies tht f(x) L < ε d lim x c f(x) = L.

4 Liford 4 Theorem 2.9 The Limit of Si(x)/x = 1 We cosider circulr sector of the uit circle with degree delt. It the cptures circulr sector of this circle with degree θ i trigle. Oe leg of the trigle is log the x-xis for 1 uit. The other leg goes up the legth of t(θ), so the hypoteuse psses through the poit (cos(θ), si(θ)) t the ed of the sector. This sector is ow iscried i the trigle. However, smller trigle is formed from these poits s well. This cretes other trigle tht coects (1,0) to (cos(θ), si(θ)). This trigle is smller th the sector d the lrger trigle. We the fids the re of ech shpe idividully. Are lrger = ½ T(θ) Are sector = θ 2 Are smller = Si(θ) 2 We c the sy tht Si(θ) θ ½ T(θ) ecuse Are smller Are 2 2 sector Are lrger. We multiply the expressio y 2 Si(θ) to get 1 θ 1. Si(θ) Cos(θ) We c tke the reciprocl to sy tht Cos(θ) Si(θ) 1. θ We the tke the limits of the outer fuctios d fid they oth equl 1 s θ pproches 0. We the pply the Squeeze theorem to show tht ecuse lim θ 0 Cos(θ) = 1 d lim θ 0 1 = 1, lim θ 0 Si(θ) θ = 1 s well.

5 Liford 5 Theorem 4.10 Limits t Ifiity If r is positive rtiol umer d c is y rel umer, the lim x c x r = 0. Furthermore, if xr is c defied whe x < 0, the lim x xr = 0. The proof hs two steps. First, we will show tht lim x 1 x = 0. We eed to verify the defiitio of the limit: for ech ε > 0 there exists δ > 0 such tht if 0 < x c < δ, the f(x) L < ε. Let ε > 0. Let M = 1/ε d x > M. This implies tht x > M=1/ε so tht ε > (1/x). This suggests tht 1 0 < ε. x This fits the form for the defiitio of limit, so we c coclude tht lim x 1 x = 0. c Hvig this, we will ow show tht lim x xr = 0. Sice r is rtiol umer, it c e represeted y m/ where m є Z, d є N. Usig sustitutio, we c write lim x c m x We c divide out the c to get c [lim x ( 1 ) m ]. x We c move out the m to c [lim x 1 x. m ]. 1 We c expd the rdicl to write c ( lim x x m ). Usig wht we showed out the limit of 1 x t ifiity, we c use sustitutio to sy tht c( 0) m.

6 Liford 6 It c e see tht this is 0. Theorem 5.8 Preservtio of Iequlity 1. If f is itegrle d oegtive o the closed itervl [, ], the 0 f(x)dx. 2. If f d g re itegrle o the closed itervl [, ] d f(x) g(x) for every x i [, ], the f(x)dx g(x)dx. Proof of Prt 1: By cotrdictio, we ssume tht f(x)dx = I < 0. Let e represeted s x 0 d e represeted s x. We c the see tht x 0 < x 1 < x 2 <... < x is the itervl [, ]. Let R = i=1 f(c i ) x i e Riem sum where x i is the width of the ith suitervl [x i-1, x i ] d c i is y poit i the ith suitervl. Becuse we ssume tht 0 f(x), we c sy tht 0 R s well. We the defie to e the lrgest of the suitervls of prtitio. Whe is sufficietly smll, we c write R I < I 2. We c dd the I over to sy tht R < I I 2. By our ssumptio tht I < 0, we c sy tht R < I I 2 < 0. However, this is cotrdictio ecuse we showed tht 0 R. Proof of Prt 2: Assume tht F d g re itegrle o the closed itervl [, ] d f(x) g(x) for every x i [, ]. Assume tht f(x) g(x) for every x i [, ].

7 Liford 7 We will show tht f(x)dx g(x)dx. By the ssumptio tht f(x) g(x), we c use lger to sy tht 0 g(x) f(x). Usig the iformtio from prt 1, we c sy tht ecuse g(x) f(x) is oegtive d itegrle, 0 [g(x) f(x)]dx. We c seprte the expressio ito two itegrls to sy 0 g(x)dx We c move oe of the itegrls over to sy tht f(x)dx g(x)dx. f(x)dx.

8 Liford 8 Theorem 8.4 L Hôpitl s Rule Let f d g e fuctios tht re differetile o ope itervl (, ) cotiig c, except possily t c itself. Assume tht g (x) 0 for ll x i (, ), except possily t c itself. If the limit of f(x)/g(x) s x f(x) pproches c produces the idetermite form 0/0, the lim x c = lim f (x) g(x) x c provided the limit g (x) o the right exists (or is ifiite). This result lso pplies if the limit of f(x)/g(x) s x pproches c produces y oe of the idetermite forms,,, or. We first cosider the cse whe lim x c + f(x) = 0 d lim x c + g(x) = 0. We defie the fuctios F(x) = {f(x), x c d 0, x = c} d G(x) = {g(x), x c d 0, x = c}. For y x, where c < x <, F d G re differetile o (c, x] d cotiuous o [c, x]. Usig the Exteded Me Vlue Theorem, we c sy tht there exists poit z є (c, x) such tht F (z) G (z) = F(x) F(c) G(x) G(c). Sice F(c) d G(c) re equl to 0, we c simplify this to F (z) = F(x) for just the portio f(x) where x c. G (z) G(x) This c the e writte s F (z) G (z) = f(x) g(x). By lettig x pproch c from the right, we hve z c + ecuse c < z < x. By usig sustitutio from the Exteded Me Vlue Theorem, we c sy lim x c + f(x) g(x) = lim x c + f (z) g (z). Usig sustitutio, we c sy tht lim x c + f(x) g(x) = lim z c + f (z) g (z). Agi y sustitutio, we c sy tht lim x c + f(x) g(x) = lim x c + f (x) g (x).

9 Liford 9 Theorem Clssifictio of Coics y Eccetricity Let F e fixed poit (focus) d let D e fixed lie (directrix) i the ple. Let P e other poit i the ple d let e (eccetricity) e the rtio of the distce etwee P d F to the distce etwee P d D. The collectio of ll poits P with give eccetricity is coic. 1. The coic is ellipse if 0 < e < The coic is prol if e = The coic is hyperol if e > 1. The proof opertes i cses to prove ech oe idividully. Cse 1: Assume e = 1. This implies tht the rtio from the distce from poit to the focus is equl to the distce from the poit to the directrix. By defiitio of prol, the coic must e prol. Cses 2 d 3: Assume the focus is t the origi d the directrix is t x=d to the right of the origi. Let P e poit i the ple with polr coordites (r, θ). Thus, PF = r d PQ = d rcos(θ). We the represet e = PF PQ. We c write e PQ = PF. Usig sustitutio, we c sy tht r = e(d r cos(θ)). By squrig ech side d covertig to rectgulr coordites, we c write x 2 + y 2 = e 2 (d x) 2 =e 2 (d 2 2dx + x 2 ).

10 Liford 10 Completig the squre produces (x + e2 d 1 e 2)2 + y2 = e2 d 2 1 e 2 If e < 1, this fits the form for ellipse. If e > 1, the 1 e 2 < 0 d fits the form for hyperol. (1 e 2 ) 2.

11 Liford 11 The Itermedite Vlue Theorem If f is fuctio which is cotiuous t every poit of the itervl [, ] d f() < 0, f() > 0, the f(x) = 0 t some poit x є (, ). Let X = {x є [, ] f(y) 0 for ll y є [, x]}. Here we re fidig the first poit t which the grph of f crosses the xis. It c e see tht X is o-empty sice є X. It c e see tht X is ouded ecuse X [, ]. By the Completeess Property, the set X hs lest upper oud. This oud is where f crosses the xis. Let this lest upper oud e represeted y z. The proof will show tht f(z) = 0. By cotrdictio, the proof first ssumes tht f(z) > 0. We c sy tht f(z) = ε (some vlue ove the xis). For some δ > 0, we hve f(x) > 0 for x lyig i the itervl (z δ, z + δ) However, this sys tht (z δ), which is less th z, would e upper oud for X. This is cotrdictio ecuse z is the lest upper oud y the Completeess Property. Now ssume tht f(z) < 0. We c set f(z) = -ε (some vlue ellow the xis). The for some δ > 0, we hve f(x) < 0 for x lyig i the itervl (z δ, z + δ). However, this is cotrdictio ecuse z is upper oud for X, ut there re still higher elemets i X still. Thus, f(z) = 0.

Week 13 Notes: 1) Riemann Sum. Aim: Compute Area Under a Graph. Suppose we want to find out the area of a graph, like the one on the right:

Week 13 Notes: 1) Riemann Sum. Aim: Compute Area Under a Graph. Suppose we want to find out the area of a graph, like the one on the right: Week 1 Notes: 1) Riem Sum Aim: Compute Are Uder Grph Suppose we wt to fid out the re of grph, like the oe o the right: We wt to kow the re of the red re. Here re some wys to pproximte the re: We cut the