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## Calculus I

by: August Feeney

12

0

5

# Calculus I MTH 251

August Feeney
PCC
GPA 3.98

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
5
WORDS
KARMA
25 ?

## Popular in Math

This 5 page Class Notes was uploaded by August Feeney on Monday October 19, 2015. The Class Notes belongs to MTH 251 at Portland Community College taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/224639/mth-251-portland-community-college in Math at Portland Community College.

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Date Created: 10/19/15
MTH 251 Mr Simonds class Limit practice problems 1 Draw skeTches of The funcTions y e y equot y 1nx and y 1x NoTe The coordinaTes of any and all inTercest This is someThing you should be able To do from memoryinTuiTion If you cannoT already do so spend some Time reviewing whaTever39 you need To review so ThaT you can do so Once you have The graphs drawn fill in each of The blanks below and also circle wheTher or39 noT each limiT exisTs The appropriaTe symbol for some of The blanks is eiTher 00 or39 00 Does limiT exisT lim ex Hoe yes or no lim ex Ho yes or no lim ex Heme yes or no lim lnx Hoe yes or no lim 1nx xgt0 yes or no 1 11m Him x yes or no 1 11m MN x yes or no 1 11m T H e yes or no lim 7x H e yes or no Page 1 of 5 lim ex Hugo lim e39x Hoe lim e39x Xgt0 liglnltxgt lim 1 xgtoox lim xgt0 x 39 1 11me Hoe lim i xgt7oo eI lim xgt7oo equot Does limiT exisT yes or39 no yes or39 no yes or39 no yes or39 no yes or39 no yes or39 no yes or39 no yes or39 no yes or39 no MTH 251 Mr Simonds class Limit practice problems 2 Hear me and hear me loud 00 does noT exisT Perhaps There are inTelligenT civilizaTions on oTher planeTs where They define ariThmeTic beTween Things ThaT don39T exisT or even more oddly beTween someThing ThaT exisTs and someThing ThaT does noT exisT buT here on EarTh The operaTions associaTed wiTh PEMDAS1 are only defined over The complex numbers MosT funcTions are even more resTricTive in Terms of The inpuT They will accepT Their domains are usually resTricTed To a subseT of The real numbers Since The symbol 00 does noT represenT a complex number much less a real number PEMDAS operaTions cannoT be applied To 00 nor can 00 be used as an inpuT value for a funcTion 1 When we wriTe 11m 2oo we are using unforTunaTe shorThand for The following complex xgt0 x 1 ThoughT llm 2 does naf eXIsf and The reason for ITs nonexIsTence IS ThaT as x gt 0 The xaox 1 1 value of 2 increases wiTh no limiT on how high H goes ThaT is as x gt 0 The value of 2 x x increases w muf bound Since llm 2 does noT exisT an expression such as 11m 2 11m2 IS JUST plain nuTTy The xa0x xaox xao expression is basically insTrucTing us To add 2 To someThing ThaT doesn39T exisT All of The limiT laws numbered 23 as well as limiT law 248 are sTaTed under The assumpTion ThaT all of The limiTs in The equaTion exisT This is a necessary assumpTion because if any of The limiTs do noT exisT whaTever equaTion you wriTe down isjusT plain nuTTy WiTh This in mind decide wheTher each of The following is True or false 6 lim ex 7 lim e39x e xgtroo e is 11m T OI39F 11m T OFF Him 6 hrnwe xsrw e hrnwe ex lim e 1nlt x 1im1nx T orF z T orF H11nx 11mlnx H1 ex 11m ex rel xgt1 lim 1 x1n21nx 2x113 1nx T or F hm em exam T or F N 1 lim ex lnx lim ex lim 111x T or F lim 81 8139 T or F 1 The EMDAS stand for exponents multiplication division addition and subtraction Page 2 of 5 MTH 251 Mr Simonds class Limit practice problems 1 xgtoo 3 A loT of people Think ThaT The reason 11m 0 is because 11m and 1 divided by xgtoo x xgtoo x 11m x infiniTy is 0 Those people while I39m sure They have many fine qualiTies are jusT plain wrong abouT This issue InfiniTy does noT exisT and you cannoT divide 1 by ThaT which does noT exisT 1 1 The reason 11m 0 is because as x gt 00 The value of 1 rawx x Table1y approaches zero This is illusTraTed in Table 1 x We can expand This idea To oTher funcTions In facT if limfxoo or limfx oo and C is any real number Then lim 0 Here a can represenT a real number xgta C f x 00 or 00 1 1 Similarly a loT of people Think ThaT The reason 11m 2 00 Is ThaT 11m oo and xgt0 x xgt0x infiniTy plus 2 is infiniTy I39m sure These people have momenTs in Their lives where Their inTellecT and educaTion shine like The norTh sTar buT This isn39T one of Those momenTs InfiniTy does noT exisT and you cannoT add 2 To ThaT which does noT exisT 1 Table 2 y 2 x 1 The reason 11m 2 00 is because as x gt 0 The value xgt0 x 1 of 2 increases wiThouT bound ThaT is There is no limiT on x how high The value goes This is illusTraTed in Table 2 AnoTher limiT issue which geTs a liTTle complicaTed has To wiTh a siTuaTion where lim fx 0 f x W This isn39T because you are dividing by zero you can39T divide by zero This is because as x gt a buT limgx 0 In This siTuaTion someThing infiniTe is always going on wiTh lim The value of gx is geTTing closer and coser To zero and since you are dividing inTo numbers f x W illusTraTed in Table 3 wiTh lim fx 7 and lim gx 0 When you encounTer These limiTs xgt4 xgt4 noT approaching zero The absoluTe value of is increasing wiThouT bound This is and x gt 00 x gt 00 x gt 61 or x gt a you are always able To correchy wriTe eiTher ThaT The limiT 00 quot or ThaT The limiT 00quot assuming ThaT a represenTs a real number When Page 3 of 5 MTH 251 Mr Simonds class Limit practice problems The siTuaTion occurs as x approaches a real number The line xa is always a verTicaI asympToTe on The graph of y f x gx Table 3 y f x gx Where39s The quesTion Here you go DeTermine The appropriaTe symbol To wriTe afTer an equal Sign following each of The following limiTs In each case The appropriaTe symbol is eiTher a real number 00 or 00 Also sTaTe wheTher or noT each limiT exisTs Several Thing you should keep in mind are IisTed on page 5 CI lim 5 1 xgt4 x 4 x2 4 limz rafx 4x4 3x 7239 11ms1n ram 28 46 x2 12x35 11m xgt5 5x l 2 lim X gt no elX lim lnx lnx6 m 71W lim hgt0 5h2 3 11m he02 3h2 Page 4 of 5 43h2 53h 21 h C x2 4 11m 2 r x 4 3x3 2x 1m 3 raiw3x 2x hgt0 h MTH 251 Mr Simonds class Limit practice problems If a is a real number and fa C Then you can preTTy much puT iT in The bank ThaT lim C AbouT The only Times This is noT True is when you are working wiTh a piecewise defined funcTion or a sTepfuncTion If 1imfx 00 or limfx oo and C is any real number Then lim 0 This is True wheTher a is a real number 00 or 00 C Ham If you are finding lim fgx and boTh lim fx 0 and lim gx 0 Then The xgta g x xgta Ha f x so ThaT The imiT no x firsT Thing you need To do is algebraically manipulaTe 0 longer has form This is True wheTher a is a real number 00 or 00 Ifyou are finding lim or lim and boTh fx gtioo and wgltxgt Mm gx gt i 00 Then divide from boTh The numeraTor and denominaTor The Term facTor ThaT dominaTes The denominaTor This is The mosT rapidly changing facTor of The Term in The denominaTor whose absoluTe value increases mosT rapidly For example 3ezx4 3ezx4 xix 11m 1m ew3ex 8e H 35 86 y e 3 1 2E3 38 lt7 8 6 30 0 8 Page 5 of 5

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