微积分第二章课件.ppt

1、1Chapter 2 Limits and Derivatives2.1The tangent and velocity problems2.1.1 The tangent problemExample 1 Find an equation of the tangent line to the parabola at the point P(1,1).SOLUTION We will be able to find an equation of the tangent line t as soon as we know its slope m.The difficulty is that we

2、 know only one point,P,on t,whereas we need two points to compute the slope.But observe that we can compute an approximation to m by choosing a nearby Q(x,x2)on the parabola and computing the slope mPQ of the secant line PQ.2yx2 We choose ,then .For instance,for the point Q(1.5,2.25)we have The clos

3、er Q is to P,the closer x is to 1 and the closer mPQ is to 2.This suggests that the slope of the tangent line t should be m=2.Figure 11x 211PQxmx2.25 12.51.5 1PQmPQ32.2 Limits of Functions2.2.1 Limit of a Function f(x)as x Approaches a Definition 1 Let f be a function defined on some open interval c

4、ontaining a except possibly at a itself,and let L be a real number.We say that the limit of f(x)as x approaches a is L,and writeand say“the limit of f(x),as x approaches a,equals L”if we can make the values of f(x)arbitrarily close to L by taking x to be sufficiently close to a(on either side of a)b

5、ut not equal to a.lim()xaf xL4 In order to understand the precise meaning of a function in Definition,let us begin to consider the behavior of a function as x approaches 1.From the graph of f shown in Figure 2,we can intuitively see that as x gets closer to 1 from both sides but x1,f(x)gets closer t

6、o 3/2.In this case,we use the notation and say that the limit of f(x),as x approaches 1,is 3/2,or that f(x)Figure 2 -2-1 12-2-112/2 as x approaches 1.approaches 3,)(lim231xfx0.51,1()0,1xxf xx。.5Example 2 Guess the value of .SOLUTION The function f(x)=sinx/x is not defined at x=0.From the table and t

7、he graph in Figure 3 we guess thatThis guess is in fact correct,as will be proved in Chapter 3.0sinlimxxx0sinlim1xxxFigure 36Example 3 The Heaviside function H is defined byAs t approaches 0 from the left,H(t)approaches 0.As t approaches 0 from the right,H(t)approaches 1.There is no single number th

8、at H(t)approaches as t approaches 0.0if0()1if0tH ttFigure 472.2.2 One-Sided LimitsDefinition 2 Let f be a function defined on an open interval of the form(a,c)for some real number c,and let L be a real number.We say that the right-hand limit of f(x)as x approaches a from the right is L,and write if

9、we can make the values of f(x)arbitrarily close to L by taking x to be sufficiently close to a and x greater than a.lim()xaf xL8Similarly,we get definition of the right-hand limit of f(x)as x approaches a.Let f be a function defined on an open interval of the form(c,a)for some real number c,and let

10、L be a real number.Wesay that the left-hand limit of f(x)as x approaches a from the left is L,and write if we can make the values of f(x)arbitrarily close to L by taking x to be sufficiently close to a and x less than a.lim()xaf xL9.)(lim)(lim)(limLxfxfLxfaxaxaxifonlyandifFor example,Example 4 Use t

11、he graph of y=g(x)to find the following limits,if they exist.lim,lim1100 xxxxxxSolution This graph shows that22555lim()3,lim()1lim()2,lim()2We have lim()2(5).xxxxxg xg xg xg xg xgFigure 510For instance,since ,therefore does not exist.Example 5 Suppose that (1)Find and(2)Discuss Solution (1),.(2)Beca

12、use ,so does not exist.1100 xxxxxxlimlimxxx0lim.,)(010212xxxxxf)(limxfx0).(limxfx0).(limxfx010)(limxfx00)(limxfx10)(limxfx00)(limxfx)(limxfx011Example 6 Show that Solution Recall that We have Therefore,0lim0.xxif0if0 xxxxx0000limlim()0 and limlim0 xxxxxxxx0lim0.xx122.2.3 Infinite LimitsDefinition 3

13、Let f be a function defined on both sides of a,except possibly at a itself.Then means that the value of f(x)can be made arbitrarily large by taking x sufficiently close to a,but not equal to a.Example 6 Find if it exists.SOLUTION As x becomes close to 0,x2 also becomes close to 0,and 1/x2 becomes ve

14、ry large.(See the table on the next page.)lim()xaf x 201limxx13It appears from the graph of the function f(x)shown in Figure that the value of the f(x)can be made arbitrarily x21x10.50.20.10.010.001142510010,0001,000,000large by taking x close enough to 0.Thus201limxx Figure 614Definition 4 Let f be

15、 a function defined on both sides of a,except possibly at a itself.Then means that the value of f(x)can be made arbitrarily large negative by taking x sufficiently close to a,but not equal to a.As an example we have Similar definitions can be given for the one-sided infinite limitslim()xaf x 202lim.

16、xx lim()lim()lim()lim()xaxaxaxaf xf xf xf x 15Examples of these four cases are given in Figure 7.Figure 732lim3xxx 32lim3xxx 0lim lnxx(/2)limtanxx 16Definition 5 The line x=a is called a vertical asymptote of the curve y=f(x)if at least one of the following statements is true:For instance,the y-axis

17、 is a vertical asymptote of the curve y=1/x2 because lim()lim()lim()lim()lim()lim()xaxaxaxaxaxaf xf xf xf xf xf x 20lim1/.xx 17Example 7 FindSolution If x is close to 3 but larger than 3,then the denominator x-3 is a positive number and 2x is close to 6.So the quotient 2x/(x-3)is a large positive nu

18、mber.Thus,we see that Likewise,if x is close to 3 but smaller than 3,then x-3 is a small negative number but 2x is still a positive number(close to 6).So 2x/(x-3)is a numerically large negative number.Thus 3322limand lim.33xxxxxx32lim3xxx 32lim3xxx 18The line x=3 is a vertical asymptote.Example 8 Fi

19、nd the vertical asymptotes of f(x)=tanx.Solution Because tanx=sinx/cosx,there are potential vertical asymptotes where cosx=0.In fact,we haveThis shows that the line is a vertical asymptote.Similar reasoning shows that the lines ,where n is an integer,are all vertical asymptotes of f(x)=tanx.(/2)limt

20、anxx(/2)limtanxx/2x(21)/2xn192.3 Calculating Limits Using the Limit LawsLimit Laws If limf(x)and limg(x)both exist,then1 limf(x)+g(x)=limf(x)+limg(x),limf(x)-g(x)=limf(x)-limg(x),2 limf(x)g(x)=limf(x)limg(x),3 limcf(x)=climf(x),for any number c,4 limf(x)/g(x)=limf(x)/limg(x),provided limg(x)0.Exampl

21、e 1 Find Solution .lim)(lim)(23422 and 355212212xxxxxxx.)()()(lim)(,)(lim)(limlim)(5321341122342213932552235523552122221222xxxxxxxxxxx20Example 2 If f is a polynomial function and a is a real number,then Solution Since f is a polynomial function,we may assume that for real numbers bn,bn-1,b0 and som

22、e positive integer n.Applying limit laws for a number of times yields that).()(limafxfax,)(011bxbxbxfnnnn).(lim)lim()lim(limlimlimlim)(lim)(lim)(limafbababbxbxbbxbxbbxbxbxfnnnnaxnaxnnaxnaxnaxnnaxnaxnnaxnnaxax01101101101121A rational function f is a ratio of two polynomials:where P and Q are polynomi

23、als.The domain consists of all values of x such that .Direct Substitution Property If f is a rational function and a is in the domain of f,then()()()P xf xQ x()0Q x).()(limafxfax22Example 3 Find each limit.Solution (1)(2)Since the limit laws can not be applied to the quotient However,2 3284316(1)lim

24、(2)lim8(16/)4xxxxxxx4lim(4)440,xx.4162xx2424416(4)(4)4(4),44and lim(4)448,16so,we obtain lim lim(4)8.4xxxxxxxxxxxxxx.)/()/(lim32328168838168332328xxxx23Theorem 1 If when x is near a(except possibly at a)and the limits of f and g both exist as x approaches a,then Theorem 2(The Squeeze Theorem)Suppose

25、 f(x)h(x)g(x)for every x in an open interval containing a,except possibly at a.(Sandwich Theorem or Pinching Theorem)()()f xg xlim()lim()xaxaf xg x.)(lim,)(lim)(limLxhLxgxfaxaxax then If24Example 4 Show thatSolution First note that we cannot use because does not exist.However,since We have and By th

26、e Squeeze Theorem,we know that 201limsin0.xxx2200011limsinlimlimsinxxxxxxx01limsinxx11sin1x 2221sinxxxx2200lim0 and lim()0 xxxx201limsin0.xxxFigure 1252.4 The Precise Definition of a LimitDefinition 1 Let f be a function defined on some open interval containing a except possibly at a itself,and let

27、L be a real number.We say that the limit of f(x)as x approaches a is L,and write if for every number 0,there is a number 0 such that ,)(limLxfax.)(,Lxfthenaxif026Geometric interpretation of the limit0 x0 xLLLx0 xy)(xfy Figure 127Definition 1 can be stated as follows:1 means that the distance between

28、 f(x)and L can be made arbitrarily small by taking the distance from x to a sufficiently small(but not 0).2 means that the values of f(x)can be made as close as we please to and L by taking x close enough to a(but not equal to a).3 means that for every (no matter how small is)we can find such that i

29、f x lies in the open interval and ,then f(x)lies in the open interval lim()xaf xLlim()xaf xLlim()xaf xL00(,)aax a(,).LL28Example 1 Use Definition 1 to prove that(1)Where c is a real number.(2)Solution (1)Let f(x)=c.Since and since 0 is less than any 0,it follows from Definition 1 that f(x)has the li

30、mit c as x approaches a.Thus(2)Let f(x)=x.Since it is obvious that if chooseas，then we obtain whenever It follows that ,limccax.limaxax,)(xcccxfeveryfor0.limccax,)(axaxfaxaxf)(.ax0.limaxax29Example 2 Prove that Solution For every number 0,since to make it is sufficient to make Hence let=/2，then we o

31、btain that This implies that.)(lim3122xx,)(22312xx,22x.22xif 02,then(21)3.xx.)(lim3122xx30Definition 2 Let f be a function defined on an open interval of the form(a,c)for some real number c,and let L be a real number.We say that the right-hand limit of f(x)as x approaches a from the right is L,and w

32、rite if for every number0,there is a number 0 such that ,)(limLxfaxifthen().axaf xL31Definition 3 Let f be a function defined on an open interval of the form(c,a)for some real number c,and let L be a real number.We say that the left-hand limit of f(x)as x approaches a from the left is L,and write if

33、 for every number0,there is a number 0 such that ,)(limLxfaxifthen().axaf xL32Example 3 Prove that Solution 1.Guessing a value for.Let be a given positive number.We want to find a number such thatthat is,This suggests that we should choose2.Showing that this works.Given 0,letIf ,then This shows that

34、 0lim0.xx0whenever0 xx2whenever0 xx2.2.0 x20 xx0lim0.xx33Example 4 Prove that Solution1.Guessing a value for.Let 0 be given.We have tofind a number 0 such thatthat is,Notice that if we can find a positive constant C such that then and we can make by taking 23lim9.xx29whenever03xx33whenever03xxx3,xC3

35、33xxC x3C x3/.xC34We can find such a number C if we restrict x to lie in some interval centered at 3.In fact,since we are interested only in values of x that are close to 3,it is reasonable to assume that x is within a distance 1 from 3,that is,Then so Thus,we haveand so C=7 is a suitable choice for

36、 the constant.But now there are two restrictions on namely 31.x24,x537.x37x3,x31and37xxC35To make sure that both of these inequalities are satisfied,wetake to be the smaller of the two numbers 1 and/7.The notation for this is 2.Showing that this works.Given 0,letIf ,then We also have so This shows t

37、hat min1,/7.min1,/7.03x312437.xxx 3/7,x293377xxx23lim9.xx36Infinite LimitsDefinition 4 Let f be a function defined on some open interval that contains the number a,except possibly at a itself.Thenmeans that for every positive number M there is a positive number such that lim()xaf x()whenever0f xMxa3

38、7Example 5 Prove thatSolution 1.Guessing a value for.Given M 0,we want to find 0 such thatThat is,This suggests that we should take2.Showing that this works.If M 0 is given,letIf ,thenTherefore,by Definition 4,201lim.xx 21whenever00Mxx1whenever0 xxM1/.M1/.M00 x222211xxMx201lim.xx 382.5 Continuitylim

39、()()xaf xf aDefinition 1 A function f is continuous at a number a if39 The kind of discontinuity illustrated in part(3)of Figure 1 is called a removable discontinuity because we could remove the discontinuity by redefining at c.The discontinuity in part(2)is called a jump discontinuity because the f

40、unction“jumps”from one value to another.If f approaches+or-as x approaches c from either side,as,for example,in part(1),we say that f has an infinite discontinuity at c.Figure 140 In general,if a function f is not continuous at c,then it has a removable discontinuity at a number c if the right-hand

41、and the left-hand limits exist at c and are equal;a jump discontinuity at c if the two one-sided limits are not equal.Example 1 Where are each of the following functions discontinuous?2221if02()()()()21if02if2()()()()21if2xxxaf xbf xxxxxxxcf xdf xxxx41Solution (a)Notice that f(2)is not defined,so f

42、is discontinuous at 2.(b)We have So f is discontinuous at 0.(c)Here f(2)=1 is defined and But ,so f is not continuous at 2.(d)The greatest integer function has discontinuities at all of the integers because does not exist if n is an integer.2001lim()lim.xxf xx 22222lim()limlim(1)32xxxxxf xxx2lim()(2

43、)xf xf?()f xx limxnx42Figure 2 shows the graphs of the functions in Example 1.Figure 2(a)(c)(d)(b)。43Definition 2 A function f is continuous from the right at a number a if and f is continuous from the left at a if lim()()xaf xf alim()()xaf xf aExample 2 At each integer n,the function is continuous

44、from the right but discontinuous from the left because ()f xx lim()lim(),butlim()lim1()xnxnxnxnf xxnf nf xxnf n 44 Definition 3 A function f is continuous on an interval if it is continuous at every point in the interval.(If f is defined only on one side of an endpoint of the interval,we understand

45、continuous at the endpoint to mean continuous from the right or continuous from the left.)Example 3 Show that the function is continuous on the interval-1,1.Solution 2()11f xx 2222If 11,we getlim()lim(11)1 lim 11lim(1)11()xaxaxaxaaf xxxxaf a 45Thus,by Definition 1,f is continuous at a if-1 a 0 and y

46、=arctanx is continuous on R.Thus,y=lnx+arctanx is continuous on(0,).The denominator,y=x2-1,is a polynomial,so it is continuous everywhere.Therefore,f is continuous at all positive numbers x except where x2-1=0.So f is continuous on the intervals(0,1)and(1,).2lnarctan()1xxf xx49Theorem 7 If f is cont

47、inuous at b and then In other words,lim(),xag xblim()()xaf g xf blim()(lim().xaxaf g xfg xExample 6 Evaluate Solution Since arcsin is a continuous function,we have 11limarcsin().1xxx11111limarcsin()arcsin(lim)111arcsin(lim)11arcsin26xxxxxxxx50Theorem 8 If g is continuous at a and f is continuous at

48、g(a),then the composite function f(g(x)is continuous at a.Example 7 Where is the function continuous?Solution We know from the Theorem 6 that f(x)=lnx is continuous and g(x)=1+cosx is continuous(because both y=1 and y=cosx are continuous).Therefore,by Theorem 8,F(x)=f(g(x)is continuous wherever it i

49、s defined.Now ln(1+cosx)is defined when 1+cosx 0.So it is undefined when x=,3,.Thus,F has discontinuities when x is an odd multiple of and is continuous on the intervals between these values(see Figure 4).()ln(1 cos)F xx51Figure 4ln(1 cos)yxTheorem 9 If f(x)is one-to-one continuous function defined

50、on interval,then its inverse function f-1 is also continuous.52The Intermediate Value Theorem 10 Suppose that f is continuous on the closed interval a,b and let N be any number between f(a)and f(b),where f(a)f(b).Then there is at least one number c in(a,b)such that f(c)=N.Some properties of continuo