《微积分》课程教学课件（Calculus）02. Limits and Continunity

CALCULUSTHOMAS'CALCULUSARUYTRANSCEChapter 2ES.STEWARLimits and ContinuityERMSCN2.11.AverageandInstantaneousSpeedRatesofChangeandLimitsDEFINITIONThe averagerate ofchange of y = f(x)withrespectto.rover thcinterval [xt.x2] isAyf(2)(x)_x + h)J(xt)2.AverageRatesofChangeandh*0Ar ASecant Lines

2016/11/15 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 2 Limits and Continuity Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2.1 Rates of Change and Limits Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1. Average and Instantaneous Speed Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Average Rates of Change and Secant Lines Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Investigate the behavior ofthe secants through P and nearbyThe average rate ofy-f)points Qas Q moves toward Palong the curve?change of f fromx, to x2Q(2is identical with theSecantsTangentslopeofsecantPQSecantoThnyP(xp.f(x,))Imvestigate the behavior ofathe secants through P andar=hnearby points Qas Qmoves toward Palong the0FIGURE 2.3-The tangent to the curve at Pis the line through P whose slope is the limit ofcurve?the secant slopes as O P from either sideFIGURE2.1A secant to the graphy = J(x). Its slope is Ay/Ax, theWecan picturethetangerntlinetothe curveaverage rate of change of f over theatPas a limitingpositionofsecant lines.interval [x, x2] EXAMPLE3Find the slope of the parabola y ratthepountP2.4).Wntean equaExercisestion forthetangent to the parabola at this point.Average Rates of Change Secamsopeis 2+ h2 ~4In Exercises 16,find the average rate of change of the function overy=,h+4the given interval or intervals.012 + h(2 + b))1. f(x) = + + 1angent slope = 4a. [2, 3]b. [-1, 1]2. g(x) = x2Ay= (2 + h)2-a. [-1, 1]b. [2, 0]3. h(r) =cotlAxa. [m/4, 3m/4]b. [m/6, /2]*2 +h04. g(n) =2 +costNOT TO SCALEa. [0, ]b. [,w]FIGURE2.4Finding the slope of the parabola y = x° at the point P(2, 4) as the,limit of secant slopes (Example 3).R)=3.LimitofFunctionsFIGURE2.7The graph of f isidentical with the line y = x + 1except at x =1, where f is notdefined (Example 1)

2016/11/15 2 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The average rate of change off from to is identical with the slope of secant PQ. 1 x 2 x Investigate the behavior of the secants through P and nearby points Q as Q moves toward P along the curve? Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We can picture the tangent line to the curve at P as a limiting position of secant lines. Investigate the behavior of the secants through P and nearby points Q as Q moves toward P along the curve? Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3. Limit of Functions Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Informal Definition ofLimitTABLE2.2Thecloserxgets to 1, thecloserf(x)= (x1)/(x1)seems to get to 2?Let f(x)be defined on an open interval about o(x)=Values of x below and above 1=x+1,x*1exceptpossiblyatx,itself.Iff(x)gets arbitrarily0.91.9close to Lforall xsufficienty close toXowe say1.12.1thatf(x)approachesthelimitLasxapproaches0.991.99Xo,and we write lim f(x)= L1.012.010.9991.9991.0012.0010.9999991.9999991.0000012.000001x-1lim f(x)=2, orlin+-1 x-1Thelimit valuedoesnotdependonhowthefunctionisdefinedat x-a(b) Constant func(a) identity functi(a) f() =(b) g(x) =(c) h(x) = x + 1lim f(x) = lim k = k-lim f(x)- lim X=Xx = 1FIGURE2.9The functions in Example 3FIGURE2.8The limits of f(c), g(t), and (s) llequal 2 asxapproaches 1. However,have limits at all points xoonly h(x) has the same function value as its limit at x = 1 (Example 2)lim f(x)= limg(x)= limh(x)=2Some ways that limits can fail to existExercisesr-as1. For the function g(x) graphed here, find the following limits orexplain why they do not exist.a. lim g(x)b. lim g(x)e. lim g(x)d. lim,g(x) aHtunt Lto Nese of Ccie find6(Eunple4)y=g(x)(a)There is nosinglevalueLapproachedbyU(x)asx-0(b)The values of g(x) grow arbitrarily large in absolutevalueasx--0anddon'tstayclosetoanyrealnumbel(c)The function's values oscillate between1 and -1 in everyopeninterval containing o,thevalues don'tstay closeto anyonenumberasx--0

2016/11/15 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2 1 1 1 lim ( ) 2, lim 2 x x 1 x f x or   x     Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Informal Definition of Limit Let be defined on an open interval about , except possibly at itself. If gets arbitrarily close to L for all sufficiently close to , we say that approaches the limit L as approaches , and we write f x() 0 x 0 x f x() x 0 x 0 x f x() x 0 lim ( ) x x f x L   Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1 1 1 lim ( ) lim ( ) lim ( ) 2 x x x f x g x h x       The limit value does not depend on how the function is defined at x=a. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 0 0 0 lim ( ) lim x x x x f x x x     0 0 lim ( ) lim x x x x f x k k     Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley (a)There is no single value L approached by U(x) as x→0 (b)The values of g(x) grow arbitrarily large in absolute value as x→0 and don't stay close to any real number (c)The function's values oscillate between 1 and -1 in every open interval containing 0, the values don't stay close to any one number as x→0 Some ways that limits can fail to exist Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises

ExercisesExercises4, Which of the following statements about the function y = f(x)2. For the function f(0) graphed here, find the following limits or exgraphed here are true, and which are false?plain why they do not exist. lim, f(r) does not existh jim, 0)a.lim,f(o)eJim)d.tim,r()1)=2lim J(x) does Bot existC.d.lim fux) exists ateery point xg in (1, 1)f(x) exists at every point xo in (1, 3)=f(r)ZT2.21.PropertiesofLimitsFinding Limits and One-SidedLimitsIM1m=1Iim eia) = M, thenadPower Ride:limf(x" L",na positive integer1. Sae Awlim(o)+aom)The lima orthe sctions is tfhe sun.oftheir limillim Vfa)VL- L,napositiveintegerRoot Rule.2. D9Rlin() -Thelimitofthedins is the diffaofheirlimits(Ifn is even, we assume that lim f(x) = L> 0.)lim()-g(a))I+1Thelimitofapsaeproductofthrirlimit4elim(k-f(s)) k+LThs the limit of thofu05. Cholosine gla)The linnlimitprovidedofaquotient oftao fntoftCtioni is theqooti6,PrPalcIfrandareinlegerxwifhso commonfictorands O,theJmofor-tprovided hat ov in a real minher. (fs is even, we assune that . > 0.)n isthrpowerof thelimitof theferowidedthe lat

2016/11/15 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2.2 Finding Limits and One-Sided Limits Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1. Properties of Limits Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Exercises51., Suppose limo J(x) - 1 and limog(s) = 5. Name therules in Theorem I that are used to accomplish steps (a), (b), and(c) of the following calculation.img (2/() 8()2f(x) - g(x)(a)mc)+ 7mim (f(t) + 7)/2.MethodsofEvaluatingLimitslim 2/(x) lim g(r)(b)(1im () + 7)2 lim f(x) lim g(x)(c)(m a)+ Jimg7)n(2)(1) (5)(1 + 7)2/9Limitsofsomefunctions canbefoundby(b)RationalFunctionssubstitutionIf thedenominatorisnot zeroatthe limit point.(a)PolynomialsFunctionslimitsofrational functions canbefound bysubstitution.THEOREM 2-Limits of PolynomialsIf P() = ar + a--1 + .+ o, thenTHEOREM3-Limits of Rational FunctionsIf P(x) and O(x) are polynomials and Q(c) * 0, thenlim P(x)=P(c) =ac*+ar-1caP()Pc)0Identifying Common FactorsItcanbeshownthatif O(x)isapolynomial and Q(c) = 0, then(x c) is a factor of Q(x).Thus, ifthenumeratorand denominator ofarational function of xareboth zero atx=c,theyhave (x-c)asa commonfactor.5.±2?If the limit ofdenominatoris zero,limits ofrational(1.3)FIGURE2.11Thegraph offunctions can be found byeliminating zerof() (+x-2)/(x)indenominators algebraically and then substitutingpart (a) is the same as the graph ofx*+x-2g(x) = (r + 2)/x in part (b) except atExample3:Evaluatelim,whereisundfindThefunctionx-xhave the same limit as x → 1 (Example 7),h

2016/11/15 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Methods of Evaluating Limits Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley • Limits of some functions can be found by substitution. (a) Polynomials Functions Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley If the denominator is not zero at the limit point, limits of rational functions can be found by substitution. (b) Rational Functions Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley • If the limit of denominator is zero, limits of rational functions can be found by eliminating zero denominators algebraically and then substituting. Example 3: Evaluate 2 2 1 2 lim x x x  x x    Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Exercises-If the limit ofdenominatoris zero,limits ofFind the limits in Exercises 11-22.functions can befound by creating and cancelling a12. lim(y2 + 5x 2)11. lim (2x + 5)commonfactor.13. Jim 8(t 5)( 7)14. im,( 2r* + 4r + 8)V2+h-J2Example4:Evaluatelim15.hh-t16, 3(2 - 1)x+ 6y+ 217. lim, 3(2x 1)218.jp+5y+619. lim,(5-y/20. lim (22 8)/V5h + 4-222.lim21. JmV3h +1 +1AhExercisesExercisesLimits of quotieats Find the limits in Exercises 23-42.V-34x35.lim36. limx + 349x-924jim7 + 4x +3N2-VxV2+8-325m102 7x + 1037. lim38. lim26. Jmx+5x+1X - 2+1Vx+3-272 +1 2r2 + 3t + 2V+12- 427. 1m+.228. linm39.lim40.jmV+5-37-117-1-2x 225y3 + 8y230.jm.3y-16y2-V2-54-41.lim42.limx+3-45-Vx2+9-1古31.32, limT-1.833#号34. lim2416Limits of some functions can be found byExercisesSandwichTheorem.~an.indirectwayLimits with trigonometric functions Find the limits in Exercises4350.44. lim sin2x43. lim (2 sinx 1)45. lim sec.x46. lim tan x+x+sinx47. lim48. lim (2 1)(2 cos x)3cosx49. lim Vx + 4 cos (x + m) 50. lim V7 + sec x0FIGURE2.12Thegraph of f issandwiched between the graphs of g and h

2016/11/15 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley • If the limit of denominator is zero, limits of functions can be found by creating and cancelling a common factor. Example 4: Evaluate 0 2 2 lim h h  h   Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley • Limits of some functions can be found by Sandwich Theorem. ~ an indirect way

THEOREM4The Sandwich TheoremTHEOREMSIf f(x) g(x) for all x in some open interval containing c, exceptSuppose that g(x) s f(x) s h(x) for all x in some open interal containing c,possibly at x = c. itself, and the limits of f and g both exist as x approaches cexcept possibly at x c itself, Suppose also thatthenin g(c) - lin Mx) - L.lim f(x) s lim g(x)Then lim, f(x) L.EXAMPLE10Given that≤)≤1+号1-forallx *o.find lim, u(r), no matter how complicated ir is.=-10l(b)(a)FIGURE 2.14The Sandwich Theorem showsthat (a) limeo sin e 0 and(b) lime-o (1 -- cos 0) - 0 (Example 11).FIGURE 2:13Anyfunction (x) whosegraph lies in the region betweny = 1 + (x*/2) and y = 1 - (x /4) haslimit 1as x→0 (Example 10).Thefunctionhaslimity=:x1asxapproaches0from the right, andlimit -1from theleft.3.One-Sided Limitslim f(x)=1 andlim f(x)=-1x0X-→0FIGURE2.24Different right-hand andleff-hand limits at the origin

2016/11/15 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3. One-Sided Limits Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The function has limit 1 as x approaches 0 from the right, and limit -1 from the left. 0 0 lim ( ) 1 lim ( ) 1 x x f x and f x       

DefinitionRight-Hand and Leff-Hand LimitsLetf(x) be defined on an interval (c,a), where d<a.MIf f(x) approaches arbitrarily close to M/as xapproachesa from within the interval,thenwe saythatfhas left-hand limttMata,(b)m-Mandwewriterirnf(x)=M(b)im-MFIGURE2.25(a)Righ-handlimitasapproacbesc (b)Lf-handlimitas(a)im-1approachFIGURE2.25(a) Right-hand limit as x approacbes e. (b) Leff-hand limit as xLet f(x) be defined on an interval (a,b), where a<b.approaches c.Iff(x)approaches arbitrarilycloseto Las xapproaches a from within that interval, thenwe say that f has right-hand limit L ataandwewritelim(x)=LV4-rTHEOREM6Afunction f(oy has alimit asx approaches cif and only ifithas lef-hand andright-hand limits there and these one-sided limits are equalm Fu) = Llim, f(x) = Llim (x) = L.eandlimV4-x2-0andFIGURE2.26lim,V4-2=0(Example1)ExercisesFinding Limits Graphically1. Which of the following statemenits about the function y f(x)graphed here are truc, and which are false?-fUxs0im, /Gx) 1alim_ f(x) = 0eFIGURE2.27Graph of the functionlimn f(x)existsfjimf(x)=0in Example 2.lim)=1lim(x)=1gi.lim yx)-o()-2f(x) dolim, (x) = 0

2016/11/15 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Definition Right-Hand and Left-Hand Limits Let f(x) be defined on an interval (a,b), where a<b. If f(x) approaches arbitrarily close to L as x approaches a from within that interval, then we say that f has right-hand limit L at a, and we write lim ( ) x a f x L    Let f(x) be defined on an interval (c,a), where c<a. If f(x) approaches arbitrarily close to M as x approaches a from within the interval, then we say that f has left-hand limit M at a, and we write lim ( ) x a  f x M   Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises

2. Which of the following statements about the fiunction y = f(x)ExercisesX2y=f()+lim,)-1h. lim flux) does not exist.)-2apFla)-2lim. x) -1. fim fia) does not esist.a. Find lim,- J(x) and lim- f(x).m-gfh. Does lim, f(r) exist? If so, what is it? If not, why not?f(a)istsryeoal (,)c, Find lim,-- f(x) and lim,-+ f(x),L ()eistsaeryn e imel(,)d. Does lim,αt f(x) exist? If so, what is it? If not, why nor?,F)-0k. lim, F(x) does not existx 2.Findthelimits inExercises11-18.+2x-112.21.msV13 鸟()()14 ()()(=)a. Find lim,~2 f(x), lim,2- f(x), and f(2)Vh+4h+5- V515. ,m.b, Does lim,~2 f(x) exist? If so, what is it? If not, why not?he. Find lime- f(x) and lin,-- f(t),d. Does lim,- J(x) exist? If se, what is it? If not, wty not?ExercisesV6-V5h +11h + 616.lim/17. ±,m,(+ + 3)2Ix + 2b. lim.(x + 3)X+2x + 24.TwoSpecial CasesV2x (r 1)V2x(x = 1)b. lim18. a ,lim[x 1]x11

2016/11/15 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4. Two Special Cases

.Thefunction Oscillating TooMuctsin e (radians)0NOT TO SCALEFIGURE 2.32 The graph of J(6) = (sin 0)/0 shows that the right-and left-hand limits as f approaches O are both 1.singlim.Theorem=1(0inradians)6FIGURE 2.31The function y= sin (1/x) has neither a6-+0right-hand nor a lef-hand limit as x approaches zero(Example 4).ExercisesUsting Jin sinoFind the limits in Exercises 21-42.21.JimiVesinkr22.(k constant)0V28sin 3yA23.m4y24.sim3h26m盒25. lm en2t27. lim ese228. lim 6r'(cot x)(csc 2t)A(1, 0)0cos5xx + sin x29. lim+x005x30. lim2xaosinxcosxFIGURE 2.33 Thefigurefor the proofofTheorem 7. TA/O4 - tane, but O4 1.31.mc05832.lim-.00sx0sin26so TA -tang.0sin 3xExercisessin8sin5x35.36. limlim0sin 20+0 sin 4x37.lim0cos638. lim sin cot 202.3sin 3y cot5ytan3x39.lim40.lim-0sin&xycot4y0Limits Involving Infinity

2016/11/15 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley • The function Oscillating Too Much Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley • Theorem 0 sin lim 1 ( ) in radians       Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2.3 Limits Involving Infinity