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## Calculus II

by: Chelsea Nolan MD

14

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# Calculus II MATH 1502

Chelsea Nolan MD

GPA 3.62

Alfred Andrew

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COURSE
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Alfred Andrew
TYPE
Class Notes
PAGES
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KARMA
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## Popular in Mathematics (M)

This 0 page Class Notes was uploaded by Chelsea Nolan MD on Monday November 2, 2015. The Class Notes belongs to MATH 1502 at Georgia Institute of Technology - Main Campus taught by Alfred Andrew in Fall. Since its upload, it has received 14 views. For similar materials see /class/233948/math-1502-georgia-institute-of-technology-main-campus in Mathematics (M) at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15
105 The Indeterminate Form 00 We know that 1 mm f w mm m f I I w g I 11mch 5 provided 11111 f and lim 9 exist w vc 2 quot and 39 513116 9 x 75 0 In this section7 we come up with rules to deal with the case where lim 9 0 I ve Wherever necessary we assume that f and g have derivatives that are continu ous near C Theorem 1051 L Hospitals Rule for form 00 page 611 of SHE 9th edn Suppose that f it 0 and lim g I 0 1 40 If f a it w LY then also hm amp L z m g Remarks a In other words f m f 39 0 lim 11m H g x M g39 x provided the limit on the right involving the derivatives exists as a nite real number or 00 or oo b We only apply this rule when both limch f 0 and limzc g 0 It is not in general true when these conditions are violated So we rst check Whether we need l Hospital and then differentiate the numerator f and denominator g7 and try compute the limit of f g39 c Exactly the same rule works when lim c m is replaced by lim or lim or lim or lim m ooo za oo ac vc xc d The proof uses the Cauchy Mean Value Theorem see SHE page 613 Here we do only Proof of Theorem 1051 in a special case We assume that c 0 and that f and g exist and are continuous in an open interval containing 0 and that g 0 96 0 Then f and g are continuous at 0 and Mo mm o 9 0 53512 0 By the Mean Value Theorem we can write for some a between 0 and z fwf0f am 0 and similarly for some I between 0 and 17 9y0 9 b0 If a is close enough to 0 then g b will be close to g 0 by continuity of g and so 9 b 96 0 Then as f O 90 0 this gives fx f aw f a 9w g bw 9 01 Now as c gt 07 both a b gt 0 and the assumed continuity of f g gives f m f39a 333 g 9 1 3 g b f 0 9 0 I lim f m m gt0 g I Example 1 Fmd 1 cosa 2131 7r 2 Solution Here f cosz and 7r hm 0059 cos O 55 2 l5 While g 7r 2m and 7r z11 1n7r 2av7r 2 0 We try 1quotHospital w lim sinz 1m w a 9 w 20 2 l m sinx 2 A w Emir l 2 2 quot 239 Then also 1 f w 1 1m mquota 9z 2 that is 1 cosz 1 21 13 7r 22 Example 2 Find 1 x wi sin Solution Here fz 1 andgzsin so H161 f ac 0 and limz0g 0 So we try 1 Hospital I m 11 hm 11 1H0 g w v0 cos m 2 11m z0 cos fi 2 0 cosO 039 Then also lim LE 0 WW 9 m that is 139 m 0 1m z 0 sm 16 Important Remark is violated Consider e have already said you must check that BOTH limx o f 0 and limxa0g 0 Here is an example of what can go wrong when one of these im m U 0 m mcosz sinxwl O 39 Here fwm gtoz gt0 but gz cosz sina gt 15 gt 0 If we try to apply 1quotH0spital then we see f w 1 1 g mWsinx cos t 1 0 I m 039 Thus in this case m M lim fl W0 9 ml Example 3 Sometimes apply l Hospital repeatedly Let F have two continuous derivatives at a thus F exists and F is continuous at a Prove that 4 9 70 9e hm Fam 2F aFa m F w 55 0 1 Solution We have here as a gt 0 that the numerator f satis es f Fam 2FaFa x gt Fa 2FaFa 0 Also the denominator 9 satis es gzm2 gt0aswgtO So we try 1quotHospital Hz F39ltax1 0F39az 1 g 2a F az F a w 2x 39 By 1quotHospital mm lim 11m c vO g w vO g l I Hm F 0c F a x 92 4 21 if this limit exists But mgawavaa mF F o and 133 229 z 0 So we try l Hospital again f m F a I 1 Fquot a 1 1 y x 2 II I F g F a F a as w gt0 Then l Hospital tells us that wz mw umxwwmzwwm that is F 2F F um mm w szw z ao w Example 4 Find hm e2quot 1 Team ln Solution Replace the integer variable 71 by a real variable 2 Thus we try to compute ezm l t voo 11 Here fz ezm l gt eU 10asa gtoo while gzlm gtOasz aoo So apply l Hospital By l HospitaI that is and hence fl 321 HZ2 9 x 112 2e2ac I lim f w 11111 2621 2e 2 z gtoo x ooo f I hm 2 x ooo g 1 e2m 1 1320 11 W 39 e2n 1 711320 1 H 239 106 The Indeterminate Form 0000 Other Indeterminate Forms We now consider m f x z m g where lim fz 00 lim 91 1 gtC 1quotquot and other indeterminate forms Theorem 1061 L Hospitals Rule for form 00 page 616 of SHE 9th edn Suppose that lirn f 39 00 If and lim 9 00 7 lim f L we 9 I then also lim f m L ww y x Remarks a In other words fm f 1 ac zl m g 1Lch g 1 provided the limit on the right involving the derivatives exists as a nite real number or 00 or 00 b We only apply this rule when both liming f 00 and limsz 9 00 It is not in general true when these conditions are violated So we rst check whether we need l Hospital and then differentiate the numerator f and denominator g and try compute the limit of f g c Exactly the same rule works when lim Z c is replaced by lim or lim or lim or lim z voo cv b oo z rc m oc Example 1 Let a gt O Show that Solution Here as c gt oo and g z 00 So try lquotHospital f 70 190 M 1 g Dataquot1 on and hence lim 11 0 z too Ctrmo am By l Hospital mm H lim fz 0 z gtoo 1 quot x Hx g Example 2 Repeated application Let k 2 1 Show that Solution Here as x gt 00 f m wk gt 00 and g ex gt 00 So try l Hospital fr ka l 9 06 8 but if k gt 1 still both 5quot 1 gt 00 and em r 00 as c gt 00 So we keep applying 1quotHospital k k k l lim lim m z rm em z voo em lim zgt00 k k 1 mic 2 ex hm kk 1k 21 m ooo e1 1 kl 11m 0 v oo a Other Indeterminate Forms A Indeterminates of the Form 000 page 517 of SHE If gng 0 and 33960 00 then we don t know what is Lemme or if it exists It could be nite or in nite Examp e 2 1 fzz andgw then limfxOand limgzoo m 0 1H0 while 1 1 39 39 2 39 Mame mm x3 33 75 00 On the other hand if 1 fc12 andgw then linhfz03nd lirg1oo while 1 i 2 EOHEWFEW z 0 So how do we cope with this type of limit called an indeterminate of form 0 r 00 We normally write it in the form 0 fg1 form5 0139 ramifaormg and then apply the version of 1quotHospital that we ve already used We have to choose the form that keeps the derivatives simplel Example Find 133 J In a 22 Solution This has form 1133 f I 9 w where as x 0 f w x5 39 0 and gz Inst 4 00 So we rewrite lnx ll m which has the form since as x r 0 lna gt 00 and 13 gt 00 So try 1 H0spital Ina 1138 V51 138 17 d 1 lim ddx Hm 0 2 1x5 1 av 11m T z 0 x 32 12 m131614 2x 0 Remarks a We could have tried to write xi l m 1 1n c which has form but then the derivatives of 1 him get worse and worse with powers of lnz appearing So when doing thse types of limits7 we always try to ensure that the derivatives are going to simplify things b Similarly if a gt 0 lim 2 lilac 0 c 0 B Indeterminates of the Form 00 gt0 page 618 of SHE If mm ooand gimme we do not know what is gun f m 9 am This is an indeterminate of the form 00 00 We can often convert these to ratios that is quotients Example Find lim tanx sec 1 EH Solution Here as x gt 325 sin I tanx gtooand sec11 Cosac gt 00 cos x So we write sinx 1 tanx seen cos x cos ac sina l COS 6 Here as x gt g7 sinz 1 gt 0 and 60511 gt 0 so we have the form By I Hospital sinx 1 hm tanz sec x hm Eng m b cosz d E smx 1 111 A 2 dz cos 12 cos a 11m zag sum 1 C Indeterminates of the Form 00 or 10 or 000 page 619 of SHE In these types of limits7 we have f 9gm for some functions f and 9 We can take logs 1nfwg g x Inf 2 and try use our previous methods Then afterwards We take exponentials since f W exp Inf W exp g lnf 24 Since exp is continuous we then use 9w 3310 f as 33 exp 9 z Inf 22 exp iggw nf 22 if the limit inside the exp exists Example 1 Form 0 Show that lim at l z 0 Solution Here we have f 191 with f g c 0 as x gt 0 So take logs lnacz wlnr and this has form 0 00 So we proceed as for limits of the form 0 00 Write xlnx ln z 1 z giving form to which we apply lquotHospital lnx lim xlnx lim 5 0 143 c d1 lim J cAOi dimlx 110 z gtII1 lx2 z1 139111 z0 Now we go back and take exponentials SDI 11 a 133 exp In 5 exp ml 13913 1112 2 exp 0 1 Example Form 000 Find lim 1 my x OC

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