Revisit: Limits at Infinity

Transcription

1 Revisit: Limits t Infinity Limits t Infinity: Wewrite to men the following: f () =L, or f ()! L s! + Conceptul Mening: Thevlueoff () willbesclosetol s we like when is su Forml Definition: Forny"> 0(nomtterhowsmll)thereeistsnM > 0suchtht f () L <" whenever > M. ciently lrge. Figure: Grph eventully trpped inside ny rbitrrily thin bnd round the H.A. University Clculus II (University of Clgry) Winter 208 /

2 Emple! 0 s! +. The following tble suggests tht the vlue of is rbitrrily smll whenever is su ciently lrge The following grphs lso seem to confirm this observtion: Figure: 0 < < 0. whenever > 0, 0 < < 0.05 whenever > 20 Generlly, for ny ">0, no mtter how smll, 0 = <"whenever > ". University Clculus II (University of Clgry) Winter /

3 Infinite Limits t Infinity: Wewrite to men the following: f () =+, or f ()! + s! + Conceptul Mening: Thevlueoff () willbeslrgeswelikewhen is su Forml Definition: FornyB > 0(nomtterhowlrge)thereeistsM > 0suchtht f () > B whenever > M. ciently lrge. Figure: Grph eventully rises bove ny horizontl br t n rbitrrily high ltitude. University Clculus II (University of Clgry) Winter /

4 Emple p! + s! +,. We consider list of incresing vlues of p p will be rbitrrily lrge when is su ciently lrge. p is lrger thn 00 whenever is lrger thn 0000: p > 00 () > p is lrger thn 000 whenever is lrger thn : p > 000 () > In generl, for ny B > 0, no mtter how lrge, p is lrger thn B whenever is lrger thn B 2 : p > B () > B 2. University Clculus II (University of Clgry) Winter /

5 Clculting Limits t Infinity: Someusefulfcts:As! +, n! if n > 0 nd! 0 if n > 0. n! + if > nd! 0 if 0 < <. ln! +. tn! 2. sin nd cos do not hve it. University Clculus II (University of Clgry) Winter /

6 Clculting Limits t Infinity: Someusefuldominncefcts:As! +, e domintes ny positive power of : Any positive power of domintes ln : n e! 0. ln n! 0. University Clculus II (University of Clgry) Winter /

7 Clculting Limits t Infinity: Limit Lws: Iff ()! A nd g()! B eist s! +, then(providedtheseredefined), f () +g()! A + B, f () g()! A B, f ()g()! AB, Squeeze Theorem: Iff () pple g() pple h() forll su! +, theng()! L s! +. l Hospitl s Rule: Suppose f () g()! A, (provided B 6= 0). B ciently lrge, then if f ()! L nd h()! L s f () g() is n indeterminte form, 0 0 or,then f () g() = provided the right hnd side eists or is + or is. f 0 () g 0 (), Emple We cn prove the dominnce of the eponentil function over the power functions by repeted pplictions of l Hospitl s rule: n e = n n n(n ) n 2 n(n )(n 2)... 2 = =... = =0. e e e University Clculus II (University of Clgry) Winter /

8 Revisit: Improper Integrls Improper Integrls over unbounded domins [, +): Z + f () d mens Z b f () d. b!+ Figure: Improper integrl defined s the it of (right br dvnces continuously, not step-by-step)... Note: If f is continuous on [, +) nd < c, then Z + f () d converges if nd only if Z + c f () d does. Furthermore, Z + f () d = Z c f () d + Z + f () d c University Clculus II (University of Clgry) Winter /

9 Emple The p-integrls: Afcttoremember Z + d =+ if p pple nd p Z + d = p p if p >. Solution. For the cse p =, Z + d = Z b b!+ d = b ln b = ln b =+. b!+ For the cses p > ndp <, the ntiderivtive is given by the power rule: Z + p d = b!+ Z b p d = b!+ p+ p + b = b!+ After this point, the it di ers depending on whether p < orp >. b p p + p If p <, then p > 0. So, b p! + s b! +. So,theintegrldivergesto+. If p >, then p < 0. So, b p! 0sb! +. So,theintegrlconvergesto p. University Clculus II (University of Clgry) Winter /!.

10 Improper Integrls - Comprison: If0pple f () pple g() forll in [, +), then Z + Z + g() d < + =) f () d =+ =) Z + Z + f () d < +. g() d =+. Improper Integrls - Limit Comprison: Iff () > 0ndg() > 0forll in [, +), nd if f () g() = L, where L is positive rel number, then the two improper integrls Z + f () d nd Z + Reson: The it condition implies tht if is su m < f () g() g() d either both converge or both diverge to +. ciently lrge, then f () g() is so close to L tht < M where (m, M) issmllopenintervlcontiningl nd m > 0. It follows tht f () < Mg() ndg() < m f (). Thus, the convergence of one of the two integrls implies the convergence of the other. University Clculus II (University of Clgry) Winter /

11 Emple Discuss the convergence or divergence of the improper integrl Z + d. Solution. For 2, we hve 2 > 0. So, tking reciprocls, we get 0 < pple. 2 Since Emple Z + 2 It follows tht d converges, so does 2 Z + Z + d converges too. 2 d. Discusee the convergence or divergence of the improper integrl Z d. Solution. Letf () denotethegivenintegrndndg() =.Then, 2 f () > 0ndg() > 0forll su ciently lrge. As! +, f () g() = !. By it comprison, the integrls of f () ndg() eitherbothconvergeorbothdiverge. Since the integrl of g() converges,sodoesthtoff (). University Clculus II (University of Clgry) Winter 208 /

12 Emples nd Eercises Find the following its , e +cos sin, +sin, Evlute the following improper integrls. Z + 2ln +3 p, tn e, etn, (2 +3) 4 d, Z + 0 e 4 d, Z + e Z + (ln ) 2 + d, e + e +, 3/2 d, Z + 0 Z + e + c ln d, 2 + d Use comprison rgument to determine the convergence or divergence of the following improper integrls. Z d, Z + 2+sin 2 d, Z + e 2+ln d, Z + 0 e cos d, Z e d University Clculus II (University of Clgry) Winter /