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# Class Note for MATH 1432 at UH

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This 6 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 13 views.

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Date Created: 02/06/15

Example 2 1 3n472n2170 1 174n7 7 4 ngIolo n5 7 3n3 7 i n7 12n 7 7 i 4 7 2 lim w does not exist naoo n3 7n Pinching Theorem Pinching Theorem Suppose that for all n greater than some integer N7 an S bn S Cn If lim an lim an L7 then lim bn L1 Suppose that bn S an Vn gt N for some N1 If an A 07 then In A 01 Example 3 casn A 07 since COS S l and l A 01 TL TL TL TL 2 Some Important Limits 21 Some Limits Some Important Limits 1 I hm i 07 a gt 0 naoo n04 Proof Va gt 07 3106 N siti 1plt 11 Then 1 1 0 1 1 0 lt i lt7gt lt lt7gt n n n Since 1 A 0 and ulp is continuous at 07 we have n 1 117 1 117 iim 7 iim lt7 iim ulp 0117 01 new n naoo n ugt0 By the pinching theorem7 1 lim i 07 a gt 01 naoo n0t Some Important Limits 2 A hm Iquot 17 zgt01 ngtoo Proof Note that VI 1 in 15271n140 asnaooi Since e is continuo s at 76 we have L A 1r u 0 hmznhmen hme e 11 ngtoo naoo ugt0 Some Important Limits 3 lim 1 0 if111lt1i neoo The limit does not exist if gt 1 or z 71 Proof Note that for any 1 st 0 lt lt1 we have ln z lt 0 and 1n z nn1n z eeoo asneooi Since lim 6 0 we have ueew 11m lim lim enlnm lim 6 01 new new new ueew Thereforev lim z 0 if m lt 1 neoo The sequence 1 gt 1 is unbounded thus divergenti Some Important Limits 3 lim 1 0 if111lt1i neoo The limit does not exist if gt 1 or z 71 Proof Note that for any 1 st 0 lt lt1 we have ln z lt 0 and 1n z nn1n z eeoo asneooi Since lim eu 0 we have neew 11m 1 lim lim enlnm lim 6 01 new new new neew Since 711711 S I S 11 by the pinching theorem we have lim In 0 if111lt1i neoo n The sequence 1 gt 1 is unbounded thus divergenti Alembert s Rule Alembert s Rule Let un il an 0 Vn such that lim neoo an1 an un il converges to 0 and if p gt 1 it divergesi pi If p lt 1 then the sequence Proof p lt 1 1P 1P an1 n Since 0 lt 7 lt 1 lim 0 Since lim p 3N1 st 2 new 2 new an Vn gt N1 an1 S p tip thus 1an11 S lgp an i Then for n gt N1 n 1 1 2 1 N1 w 2p1an11ltTpgt 1an721 lt7 mm and therefore 1imnn00 an 01 Since fian S an S an by the pinching theorem we have limnn00 an 01 Proof p gt 1 6 36 gt 0st p gt16i Since 1im an1 p 3N2 siti Vn gt N2 an1 2 p771 ngtoo an n 2 6 6 6 As p7 E gt1 1an112 1 E an i Then for n gt N1 6 6 2 6 niNz m1 2 1 mm 2 lt1 gt mm 22 1 1am and therefore an is unbounded thus divergenti Some Important Limits 4 n 1im i 0 naoo 7 Proof Note that an1 zin1 n1 7 A 0 as n A 00 a n11 z n1 By Alembert7s Ruie 1 11m i 01 naoo 7 Some Important Limits 5 17 hm T 0 gt 11 ngtoo 1 Proof Note that an1 n 1 1 1 1 7 if i 1 7 i an I n1 np n gt asnaoo Since i lt 1 by Alembert7s Rule 17 iim 7 0 zgt1i HOO In Some Important Limits 6 l lim 2 0 new 7 Proof l 1 dt 1 dt 2 02 ege 77 71e0asneoi n n 1 t n 1 By the pinching theorem7 lim 2 0 Alternative proof is done by L7Hopital7s new 7 Rule 10157 1 1 1 lim 2 lim 2 hm iOi new 7 ugtoo u ugt0ltgt 1 1 hm n 11 new Proof lim n1 lim EMT limeu 60 11 new new ueo Some Important Limits 7 lnn hm 7 07 a gt 0 neoo 7 Proof By L7Hopital7s Rule 10157 l l 1 lim hm hm lime0 aw no ueoo ua uew aua ugtoo auu Some Important Limits 8 lim nlnlt1 E I new 7 Proof l 1 g fl 1 lim nlnlt1 E 1 lim new 7 new E I 1imln1hiln1 neo h I hm lnuh 7 lnu neo h uil 1 zlnu1 Iglu1 1 Alternative proof is done by L7Hopital7s Rule 1057 naoo ugtoo ugtoo z ln1zu 11zu7Iu2 hm nlnlt1ggt hm U u hm W 1 I n hm lt1 7 emi ngtoo 7 Proof I n hm lt17gt hm enln1 hm eu 675 71400 n naoo uax Outline Contents 1 Limit of Sequence 11 Properties of Limits Some Limits 21 Some Limits i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 2 Lecture 18Section 103 Limit of Sequence Section 104 Some Important Limits Jiwen He 1 Limit of Sequence 11 Properties of Limits Properties of Limits Properties of Limits 1 Let lim an L and lim bn Mi Then naoo naoo 0 lim can CL naoo 0 lim anbnLM 0 lim anbn LM OJLIEOZiZforM7 Oi 0 lim an for f being continuous at L Example 1 If lim an L7 then lim bn EL Where In eak Limit of Sequence De ned by Rational Function Properties of Limits Let an apnpap11np71 a0 With 11 y 07 and In qnq q11nqil o With q 0 Then an oprltqthen1im 70 naoo n a a o prq7 then lim in it new bn q a a o Hp gt 417 then lim in does not exist7 and the sequence in diverges naoo n bn

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