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1 Porland Communiy College MTH 51 Lab Manual Limis and Coninuiy Aciviy 4 While working problem 3.6 you compleed Table 4.1 (ormerly Table 3.1). In he cone o ha problem he dierence quoien being evaluaed reurned he average rae o change in he volume o luid remaining in a va beween imes 4 and 4 h. As he elapsed ime closes in on 0 his average rae o change converges o 6. From ha we deduce ha he rae o change in he volume 4 minues ino he draining process mus have been 6 gal/min. The cone or Problem 3.6 Suppose ha a va was undergoing a conrolled drain and ha he amoun o luid le in he va 3/ (gal) was given by he ormula V 100 where is he number o minues ha had passed since he draining process began. Table 4.1: 4 4 V h V y h h y Please noe ha we could no deduce he rae o change 4 minues ino he process by replacing h wih 0 ; in ac, here are a leas wo hings prevening us rom doing so. From a sricly mahemaical perspecive, we canno replace h wih 0 because ha would lead o division by zero in he dierence quoien. From a more physical perspecive, replacing h wih 0 would in essence sop he clock. I ime is rozen, so is he amoun o luid in he va and he enire concep o rae o change becomes moo. I urns ou ha i is requenly more useul (no o menion ineresing) o eplore he rend in a uncion as he inpu variable approaches a number raher han he acual value o he uncion a ha number. Mahemaically we describe hese rends using is. I we call he dierence quoien in he heading or Table 4.1 rend evidenced in he able by saying he i o ha as h changes value, he value o ied number o which he value o h, hen we could describe he h as h approaches zero is 6. Please noe h changes, no he value o he i. The i value is a h converges. Symbolically we wrie h 6 Mos o he ime he value o a uncion a he number a and he i o he uncion as approaches a are in ac he same number. When his occurs we say ha he uncion is coninuous a a. However, o help you beer undersand he concep o i we need o have you conron siuaions where he uncion value and i value are no equal o one anoher. Graphs can be useul or helping disinguish uncion values rom i values, so ha is he perspecive you are going o use in he irs couple o problems in his lab. h 0 Lab Aciviies 7

2 Porland Communiy College MTH 51 Lab Manual Problem 4.1 Several uncion values and i values or he uncion in Figure 4.1 are given below. You and your group maes should ake urns reading he equaions aloud. Make sure ha you read he symbols correcly, ha s par o wha you are learning! Also, discuss why he values are wha hey are and make sure ha you ge help rom your insrucor o clear up any conusion. 6 bu 3 4 is undeined bu bu 1 does no eis Figure 4.1: 1 3 bu 1 1 The i o as approaches 1 rom he le. The i o as approaches 1 rom he righ. Problem 4. Copy each o he ollowing epressions ono your paper and eiher sae he value or sae ha he value is undeined or doesn eis. Make sure ha when discussing he values you use proper erminology. All epressions are in reerence o he uncion g shown in Figure g g g 4..5 g g 4..8 g g g g g g 4..1 g Figure 4.: g 8 L ab Aciviies

3 Porland Communiy College MTH 51 Lab Manual Problem 4.3 Values o he uncion are shown in Table 4.. Boh o he quesions below are in reerence o his uncion Wha is he value o 5? 4.3. Wha is he value o ? Table 4.: () Problem 4.4 Values o he uncion p 1 are shown in Table 4.3. Boh o he quesions below are in reerence o his uncion Wha is he value o p 1? 4.4. Wha is he value o Problem 4.5 1? 1 Table 4.3: p 1 p() Creae ables similar o ables 4. and 4.3 rom which you can deduce each o he ollowing i values. Make sure ha you include able numbers, able capions, and meaningul column headings. Make sure ha your inpu values ollow paerns similar o hose used in ables 4. and 4.3. Make sure ha you round your oupu values in such a way ha a clear and compelling paern in he oupu is clearly demonsraed by your saed values. Make sure ha you sae he i value! 10 4 sin 1 h h h Aciviy 5 When proving he value o a i we requenly rely upon laws ha are easy o prove using he echnical deiniions o i. These laws can be ound in Appendi C (pages C1 and C). The irs o hese ype laws are called replacemen laws. Replacemen laws allow us o replace i epressions wih he acual values o he is. Problem 5.1 The value o each o he ollowing is can be esablished using one o he replacemen laws. Copy each i epression ono your own paper, sae he value o he i (e.g. 5 5 ), and sae he replacemen law (by number) ha esablishes he value o he i Lab Aciviies 9

4 Porland Communiy College MTH 51 Lab Manual Problem 5. The algebraic i laws allow us o replace i epressions wih equivalen i epressions. When applying i laws our irs goal is o come up wih an epression in which every i in he epression can be replaced wih is value based upon one o he replacemen laws. This process is shown in eample 5.1. Please noe ha all replacemen laws are saved or he second o las sep and ha each replacemen is eplicily shown. Please noe also ha each i law used is reerenced by number. Eample Limi Law A Limi Law A3 4 3 Limi Law A Limi Laws R1 and R 199 Use he i laws o esablish he value o each o he ollowing is. Make sure ha you use he sep-by-sep, verical orma shown in eample 5.1. Make sure ha you cie he i laws used in each sep. To help you ge sared, he seps necessary in problem 5..1 are oulined below. To help you ge sared, he seps necessary in problem 5..1 are oulined below. Sep 1: Apply Law A6 Sep : Apply Law A1 Sep 3: Apply Law A3 Sep 4: Apply Laws R1 and R y 7 y 3 y y cos Aciviy 6 Many is have he orm 0 0 which means he epressions in boh he numeraor and denominaor 6 i o zero (e.g. ). The orm 0 is called indeerminae because we do no know he value o he i (or even i i eiss) so long as he i has ha orm. When conroned wih is o orm 0 0 we mus irs manipulae he epression so ha common acors causing he zeros in he numeraor and denominaor are isolaed. Limi law A7 can hen be used o jusiy einaing he common acors and once hey are gone we may proceed wih he applicaion o he remaining i laws. Eamples 6.1 and 6. illusrae his siuaion. 10 L ab Aciviies

5 Porland Communiy College MTH 51 Lab Manual Eample This i does no have indeerminae orm, so we may apply he remaining i laws Limi Law A7 5 Limi Law A Limi Laws R1 and R This i does no have indeerminae orm, so we may apply he remaining i laws. Eample cos sin 1 cos 1cos 1 cos sin sin 1 cos 0 sin 1 cos 0 sin 1 cos cos cos cos cos 1 1 cos 0 1 Limi Law A7 Limi Law A5 Limi Law A1 Limi Law A6 Limi Laws R1 and R As seen in eample 6., rigonomeric ideniies can come ino play while rying o einae he orm 0. Elemenary rules o logarihms can also play a role in his process. Beore you begin 0 evaluaing is whose iniial orm is 0, you need o make sure ha you recall some o hese basic 0 rules. Tha is he purpose o problem 6.1. Lab Aciviies 11

6 Porland Communiy College MTH 51 Lab Manual Problem 6.1 Complee each o he ollowing ideniies (over he real numbers). Make sure ha you check wih your lecure insrucor so ha you know which o hese ideniies you are epeced o memorize. The ollowing ideniies are valid or all values o and y. 1 cos an 1 sin an sin y cos y sin cos There are hree versions o he ollowing ideniy; wrie hem all. cos cos cos The ollowing ideniies are valid or all posiive values o and y and all values o n. ln y ln y n ln ln e n 1 L ab Aciviies

7 Porland Communiy College MTH 51 Lab Manual Problem 6. Use he i laws o esablish he value o each o he ollowing is aer irs manipulaing he epression so ha i no longer has orm 0. Make sure ha you use he sep-by-sep, verical 0 orma shown in eamples 6.1 and 6.. Make sure ha you cie each i law used cos 1 0 cos sin 0 sin 4ln 1 ln 3 ln ln sin sin w 9 9 w w 3 Aciviy 7 We are requenly ineresed in a uncion s end behavior; ha is, wha is he behavior o he uncion as he inpu variable increases wihou bound or decreases wihou bound. Many imes a uncion will approach a horizonal asympoe as is end behavior. Assuming ha he horizonal asympoe y L represens he end behavior o he uncion boh as increases wihou bound and as decreases wihou bound, we wrie L L. The ormalisic way o read is he i o L and as approaches ininiy equals L. When read ha way, however, he words need o be aken anyhing bu lierally. In he irs place, isn approaching anyhing! The enire poin is ha is increasing wihou any bound on how large is value becomes. Secondly, here is no place on he real number line called ininiy; ininiy is no a number. Hence cerainly can be approaching somehing ha isn even here! Problem 7.1 For he uncion in Figure 7.1 (Appendi B, page B1) we could (correcly) wrie 1 igures (pages B1 and B). 1. Go ahead and wrie (and say aloud) he analogous is or he uncions in and Problem 7. Values o he uncion are shown in Table 7.1. Boh o he quesions below are in reerence o his uncion Wha is he value o 7.. Wha is he horizonal asympoe or he graph o y?? Table 7.1: () 1, , , ,000, Lab Aciviies 13

8 Porland Communiy College MTH 51 Lab Manual Problem 7.3 Jorge and Vanessa were in a heaed discussion abou horizonal asympoes. Jorge said ha uncions never cross horizonal asympoes. Vanessa said Jorge was nus. Vanessa whipped ou her rusy calculaor and generaed he values in Table 7. o prove her poin Wha is he value o g? 7.3. Wha is he horizonal asympoe or he graph o y g? Jus how many imes does he curve y g is horizonal asympoe? cross Table 7.: g sin 1 g () Aciviy 8 When using i laws o esablish i values as or, i laws A1-A6 and R are sill in play (when applied in a valid manner), bu i law R1 canno be applied. (The reason i law R1 canno be applied is discussed in deail in problem 11.4) There is a new replacemen law ha can only be applied when or ; his is replacemen law R3. Replacemen law R3 essenially says ha i he value o a uncion is increasing wihou any bound on large i becomes or i he uncion is decreasing wihou any bound on how large is absolue value becomes, hen he value o a consan divided by ha uncion mus be approaching zero. An analogy can be ound in eremely poor pary planning. Le s say ha you plan o have a pizza pary and you buy ive pizzas. Suppose ha as he hour o he pary approaches more and more guess come in he door in ac he guess never sop coming! Clearly as he number o guess coninues o rise he amoun o pizza each gues will receive quickly approaches zero (assuming he pizzas are equally divided among he guess). Problem Consider he uncion. Complee Table 8.1 wihou he use o your calculaor. Wha i value and i law are being illusraed in he able? Table 8.1: 1,000 10, ,000 1,000,000 1 () 14 L ab Aciviies

9 Porland Communiy College MTH 51 Lab Manual Aciviy 9 Many is have he orm which we ake o mean ha he epressions in boh he numeraor and denominaor are increasing or decreasing wihou bound. When conroned wih a i o ype g or g ha has he orm, we can requenly resolve he i i we irs divide he dominan acor o he dominan erm o he denominaor rom boh he numeraor and he denominaor. When we do his, we need o compleely simpliy each o he resulan racions and make sure ha he resulan i eiss beore we sar o apply i laws. We hen apply he algebraic i laws unil all o he resulan is can be replaced using i laws R and R3. This process is illusraed in eample 9.1. Eample The orm o he i is now 3 0, so we can 0 5 begin o apply he i laws because he is will all eis. Limi Law A5 Limi Laws A1 and A Limi Laws R and R3 Problem 9.1 Use he i laws o esablish he value o each i aer dividing he dominan erm-acor in he denominaor rom boh he numeraor and denominaor. Remember o simpliy each resulan epression beore you begin o apply he i laws e 10e e y 5 y 5 9 y Lab Aciviies 15

10 Porland Communiy College MTH 51 Lab Manual Aciviy 10 Many i values do no eis. Someimes he non-eisence is caused by he uncion value eiher increasing wihou bound or decreasing wihou bound. In hese special cases we use he symbols and o communicae he non-eisence o he is. Figures can be used o illusrae some ways in which we communicae he non-eisence o hese ype o is. In Figure 10.1 we could (correcly) wrie k, k, and k. In Figure 10. we could (correcly) wrie 4 w, 4 w, and 4 w. In Figure 10.3 we could (correcly) wrie 3 T and 3 T shorhand way o communicaing he non-eisence o he wo sided i. There is no 3 T. Figure 10.1: k Figure 10.: w 4 3 Figure 10.3: T Problem 10.1 Draw ono Figure 10.4 a single uncion,, ha saisies each o he ollowing i saemens. Make sure ha you draw he necessary asympoes and ha you label each asympoe wih is equaion Figure 10.4: 16 L ab Aciviies

11 Porland Communiy College MTH 51 Lab Manual Aciviy 11 Whenever bu 0 a side o a he value o g hese siuaions he line g 0, hen a a g does no eis because rom eiher eiher increases wihou bound or decreasing wihou bound. In a is a verical asympoe or he graph o y. g 5 For eample, he line is a verical asympoe or he uncion h. We say ha 5 5 has he orm no zero over zero. (Speciically, he orm o is 7.) Every 0 i wih orm no zero over zero does no eis. However, we requenly can communicae he 5 non-eisence o he i using an ininiy symbol. In he case o h i s prey easy o see ha 1.99 iner ha h is a posiive number whereas.01 h and h h is a negaive number. Consequenly, we can. Remember, hese equaions are communicaing ha he is do no eis as well as he reason or heir non-eisence. There is no shor-hand way o communicae he non-eisence o he wo-sided i h. Problem 11.1 Suppose ha g Wha is he verical asympoe on he graph o y g Wrie an equaliy abou g Wrie an equaliy abou g Is i possible o wrie an equaliy abou g.. 3.? I so, wrie i Which o he ollowing is eis? g, g Problem 11. Suppose ha z Wha is he verical asympoe on he graph o y z 11.. Is i possible o wrie an equaliy abou z.? I so, wrie i., and g Wha is he horizonal asympoe on he graph o y z Which o he ollowing is eis? z, z, and z Lab Aciviies 17

12 Porland Communiy College MTH 51 Lab Manual Problem 11.3 Consider he uncion 7. Complee Table 11.1 wihou he use o your calculaor. 8 Use his as an opporuniy o discuss why is o orm no zero over zero are ininie is. Wha i equaion is being illusraed in he able? Table 11.1: () Problem 11.4 Hear me, and hear me loud does no eis. This, in par, is why we canno apply Limi Law R1 o an epression like. When we wrie, say, 7, we are replacing he i epression wih is value ha s wha he replacemen laws are all abou! When we wrie, we are no replacing he i epression wih a value! We are eplicily saying ha he i has no value (i.e. does no eis) as well as saying he reason he i does no eis. The i laws (R1-R3 and A1-A6) can only be applied when all o he is in he equaion eis. Wih his in mind, discuss and decide wheher each o he ollowing equaions are rue or alse. 7 e e 0 e 0 e 0 T or F e e e e T or F e 1 ln e T or F 1 ln 1 ln ln 0 0 T or F ln 1 e sin sin ln 1 T or F e 1 sin sin T or F e e ln ln T or F e 1/ 1 e T or F 18 L ab Aciviies

13 Porland Communiy College MTH 51 Lab Manual Problem 11.5 Mindy ried o evaluae using he i laws. Things wen horribly wrong or Mindy (her work is shown below). Ideniy wha is wrong in Mindy s work and discuss wha a more reasonable approach migh have been This soluion is no correc! Do no emulae Mindy s work!! Limi Law A5 Limi Law A Limi Laws R1 and R Aciviy 1 Many saemens we make abou uncions are only rue over inervals where he uncion is coninuous. When we say a uncion is coninuous over an inerval, we basically mean ha here are no breaks in he uncion over ha inerval; ha is, here are no verical asympoes, holes, jumps, or gaps along ha inerval. To make his deiniion more precise, we begin by deining wha we mean we say ha a uncion is Deiniion 1.1 coninuous a he number a. The uncion is coninuous a he number a i and only i a. a There are hree ways ha he deining propery can ail o be saisied a a given value o a. To aciliae eploraion o hese hree manner o ailure, we can break he deining propery ino a specrum o hree properies. i. a mus be deined ii. a mus eis iii. a mus equal a Please noe ha i eiher propery i or propery ii ails o be saisied a a given value o a, hen propery iii also ails o be saisied a a. Lab Aciviies 19

14 Porland Communiy College MTH 51 Lab Manual Problem 1.1 Sae he values o a which he uncion shown in Figure 1.1 is disconinuous. For each insance o disconinuiy, sae (by number) all o he sub-properies in Deiniion 1.1 ha ail o be saisied. Figure 1.1: h 5 Aciviy 13 When a uncion has a disconinuiy a a, he uncion is someimes coninuous rom only he righ or only he le a a. (Please noe ha when we say he uncion is coninuous a a we mean ha he uncion is coninuous rom boh he righ and le a a.) Deiniion 13.1 The uncion is coninuous rom he le a a i and only i a coninuous rom he righ a a i and only i a a. a and is Some disconinuiies are classiied as removable disconinuiies. Speciically, disconinuiies ha are holes or skips (holes wih a secondary poin) are called removable. Deiniion 13. We say ha has a removable disconinuiy a a i is disconinuous a a bu a eiss. 0 L ab Aciviies

15 Porland Communiy College MTH 51 Lab Manual Problem 13.1 Reerring o he uncion h shown in Figure 13.1, sae he values o where he uncion is coninuous rom he righ bu no he le. Then sae he values o where he uncion is coninuous rom he le bu no he righ. Problem 13. Reerring again o he uncion h shown in Figure 13.1, sae he values o where he uncion has removable disconinuiies. Aciviy 14 Figure 13.1: h Now ha we have a deiniion or coninuiy a a number, we can go ahead and deine wha we mean when we say a uncion is coninuous over an inerval. 5 Deiniion 14.1 The uncion is coninuous over an open inerval i and only i i is coninuous a each and every number on ha inerval. The uncion is coninuous over he closed inerval ab, i and only i i is coninuous on ab,, coninuous rom he righ a a, and coninuous rom he le a b. Similar deiniions apply o hal-open inervals. Problem 14.1 Wrie a deiniion or coninuiy over he hal-open inerval ab,. Problem 14. Reerring o he uncion in Figure 14.1, decide wheher each o he ollowing saemens are rue or alse h is coninuous on 4, h is coninuous on 4, h is coninuous on 4, h is coninuous on 1, h is coninuous on 1, h is coninuous on, h is coninuous on, 4 Figure 14.1: h 5 Lab Aciviies 1

16 Porland Communiy College MTH 51 Lab Manual Problem 14.3 Several uncions are described below. Your ask is o draw each uncion on is provided ais sysem. Do no inroduce any unnecessary disconinuiies or inerceps ha are no direcly implied by he saed properies. Make sure ha you draw all implied asympoes and label hem wih heir equaions Draw ono Figure 14. a uncion ha saisies all o he ollowing properies and Figure 14.: Draw ono Figure 14.3 a uncion ha saisies all o he ollowing properies. g g g 0 4, g 3, and g 6 0 g is coninuous and has consan slope on 0, Draw ono Figure 14.4 a uncion ha saisies all o he ollowing properies. Figure 14.3: g The only disconinuiies on m occur a 4 and 3 m has no -inerceps m m 3 m m m has a consan slope o over, 4 m is coninuous over 4,3 Figure 14.4: m L ab Aciviies

17 Porland Communiy College MTH 51 Lab Manual Aciviy 15 Disconinuiies are a lile more challenging o ideniy when working wih ormulas han when working wih graphs. One reason or he added diiculy is ha when working wih a uncion ormula you have o dig ino your memory bank and rerieve undamenal properies abou cerain ypes o uncions. Problem Wha would cause a disconinuiy on a raional uncion (a polynomial divided by anoher polynomial)? Wha is always rue abou he argumen o he uncion, u, over inervals where he uncion y ln u is coninuous? Name hree values o where he uncion y an Wha is he domain o he uncion k 4? Wha is he domain o he uncion g 3 4? is disconinuous. Aciviy 16 Piece-wise deined uncions are uncions where he ormula used depends upon he value o he inpu. When looking or disconinuiies on piece-wise deined uncions, you need o invesigae he behavior a values where he ormula changes as well as values where he issues discussed in Aciviy 15 migh pop up. Problem 16.1 This quesion is all abou he uncion shown o he righ. Answer quesion a each o he values 1, 3, 4, 5, 7, and 8. A he values where he answer o quesion is yes, go ahead and answer quesions ; skip quesions a he values where he answer o quesion is no Is disconinuous a he given value? Is coninuous only rom he le a he given value? Is coninuous only rom he righ a he given value? Is he disconinuiy removable? 4 i i i i 7 8 Lab Aciviies 3

18 Porland Communiy College MTH 51 Lab Manual Problem 16. Consider he uncion g shown o he righ. The leer C represens he same real number in all hree o he piece-wise ormulas Find he value or C ha makes he uncion coninuous on,10. Make sure ha your reasoning is clear Is i possible o ind a value or C ha makes he uncion coninuous over,? Eplain. C i g C 3 i 10 C 4 i 10 Problem 16.3 Consider he uncion shown o he righ. Sae he values o where each o he ollowing occur. I a saed propery doesn occur, make sure ha you sae ha (as opposed o simply no responding o he quesion). No eplanaion necessary A wha values o is disconinuous? A wha values o is coninuous rom he le bu no he righ? 5 i i i A wha values o is coninuous rom he righ bu no he le? A wha values o does have removable disconinuiies? 4 L ab Aciviies