YOU ARE DOWNLOADING DOCUMENT

Please tick the box to continue:

Transcript
  • INFINITE LIMITS, LIMITS AT INFINITY, ANDLIMIT RULES

    Sections 2.2, 2.4 & 2.5September 5, 2013

  • INFINITE LIMITS

    Consider the function f (x) =1x

    .

    -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

    -6-5-4-3-2-1

    123456

    As x 0+ the value off (x) = 1x grows withoutbound (i.e. 1x ).

    As x 0 the value of 1xgrows negatively withoutbound (i.e. 1x ).

    By definition a one or two-sided limit must be a real number(i.e. finite), so

    limx0+

    1x

    and limx0

    1x

    .

    do not exist.

  • INFINITE LIMITS

    However, it is convenient to describe the behavior of f (x) = 1xfrom the left or right using limit notation.

    So we will write

    limx0+

    1x

    = and limx0

    1x

    = .

    This does NOT mean that the limit is , it is just a descriptionof the behavior of our function f (x) near 0.

    We would say:

    f (x) = 1x approaches as x approaches 0 from the right.f (x) = 1x approaches as x approaches 0 from the left.

  • INFINITE LIMITS

    It may be the case that both one-sided limits are infinite andalso the same.

    For example,

    limx0+

    1x2

    = = limx0

    1x2

    .

    In this case we will again abuse the notation and write

    limx0

    1x2

    = .We would say:

    f (x) = 1x2 approaches as x approaches 0.

  • EXAMPLES

    f (x) = sec(x) =1

    cos(x)

    -5 -4 -3 -2 -1 0 1 2 3 4 5

    g(x) =1

    (x 7)2

    -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

    -4-3-2-1

    12345678

    (1) limxpi2

    f (x) =

    (2) limxpi2 +

    f (x) =

    (3) limxpi2

    f (x) = DNE

    (4) limx7

    g(x) =

    (5) limx7+

    g(x) =

    (6) limx7

    g(x) =

  • VERTICAL ASYMPTOTES

    Back to the function f (x) =1x

    .

    As noted before, theone-sided limits of f at 0are both infinite.

    Consider the line x = 0.

    The graph of f getsarbitrarily arbitrarily closeto the line x = 0, but it willnever touch it.

    We call the line x = 0 a vertical asymptote of the functionf (x) = 1x .

  • VERTICAL ASYMPTOTES

    DefinitionA vertical line x = c is a vertical asymptote of the graph of afunction f if either

    limxc+

    f (x) = or limxc

    f (x) = .

  • VERTICAL ASYMPTOTES

    How do we find vertical asymptotes?

    Check wherever a denominator is zero.

    For example,

    f (x) =1

    x 3lim

    x3f (x) =

    limx3+

    f (x) =

    Vert. Asym.:x = 3

    f (x) =x2 1x + 1

    limx1

    f (x) = 2lim

    x1+f (x) = 2

    Vert. Asym.:None

    f (x) = tan(x)

    For all odd integersn:

    limx npi2

    f (x) =

    limx npi2 +

    f (x) =

    Vert. Asym.:x = npi/2for all odd n

  • PRACTICE PROBLEMS

  • PRACTICE PROBLEMS

    Determine the following limits:

    (1) limx0

    4x2/5

    (2) lim(pi2 )

    sec() (3) limx0+

    (x2

    2 1

    x

    )

    Determine the equations of the asymptotes of the followingfunctions:

    (1) f (x) =x2 12x + 4

    (2) f (x) = sec(x) (3) f (x) =x3 + 1

    x2

  • LIMIT RULES &THE SANDWICH THEOREM

  • LIMIT RULES

    Now that were (hopefully) familiar with limits, it would benice if we had some rules for computing them more quickly.

    Let L, M c, and k be real numbers and

    limxc f (x) = L and limxc g(x) = M

    Sum & Difference Rules

    limxc (f (x) g(x)) = LM

    Product Rule

    limxc (f (x) g(x)) = L M

  • LIMIT RULES

    Let L, M c, and k be real numbers and

    limxc f (x) = L and limxc g(x) = M

    Quotient Rule

    limxc

    f (x)g(x)

    =LM

    if M 6= 0

    Power RuleIf r and s are integers with no common factor and s 6= 0

    limxc (f (x))

    r/s = Lr/s

    If s is even, we also need L > 0.

  • LIMITS OF POLYNOMIALS

    For certain functions (continuous) finding the limit is easy.Polynomials are one of those types of functions.

    Polynomial Limit RuleLet P(x) be any polynomial, then

    limxc P(x) = P(c)

    By the quotient rule, if P(x) and Q(x) are polynomial withQ(c) 6= 0, we have

    limxc

    P(x)Q(x)

    =P(c)Q(c)

  • THE ALMOST THE SAME RULE

    The following rule is very useful.

    The Almost the Same Rule

    If f (x) = g(x) for all x 6= c in some interval open containing c,then

    limxc f (x) = limxc g(x)

    Its not clear how one would use this, so lets look at anexample.

  • THE ALMOST THE SAME RULE

    Determine the following limit:

    limt1

    t2 + 3t + 2t2 t 2

    Since (1)2 (1) 2 = 0, we cannot use the quotient rule.

    However, notice that 1 is also a zero of the numerator:(1)2 + 3(1) + 2 = 0.

    This means that (t + 1) is a factor of both the numerator and thedenominator.

    limt1

    t2 + 3t + 2t2 t 2 = limt1

    (t + 1)(t + 2)(t + 1)(t 2) = limt1

    (t + 2)(t 2) =

    13

  • THE SANDWICH THEOREM

    Sometimes the limit rules arent enough to determine the limitof a function at a given point.

    The following theorem allows us to determine the limit of afunction at such a point by sandwiching it between two otherfunctions.

    The Sandwich TheoremLet f , g, and h be functions and c and L real numbers. If

    limxc g(x) = L = limxc h(x)

    andg(x) f (x) h(x)

    for all x in some open interval containing c, then

    limxc f (x) = L.

  • THE SANDWICH THEOREM

    This is a bit easier to see graphically:

    66 Chapter 2: Limits and Continuity

    aid of some geometry applied to the problem (see the proof of Theorem 7 in Section 2.4),or through methods of calculus (illustrated in Section 4.5). The next theorem is alsouseful.

    The Sandwich Theorem

    The following theorem enables us to calculate a variety of limits. It is called the SandwichTheorem because it refers to a function whose values are sandwiched between the valuesof two other functions g and h that have the same limit L at a point c. Being trapped be-tween the values of two functions that approach L, the values of must also approach L(Figure 2.12). You will find a proof in Appendix 5.

    The Sandwich Theorem is also called the Squeeze Theorem or the Pinching Theorem.

    EXAMPLE 10 Given that

    find no matter how complicated u is.

    Solution Since

    the Sandwich Theorem implies that (Figure 2.13).

    EXAMPLE 11 The Sandwich Theorem helps us establish several important limit rules:

    (a) (b)

    (c) For any function , implies .

    Solution

    (a) In Section 1.3 we established that for all (see Figure 2.14a).Since we have

    (b) From Section 1.3, for all (see Figure 2.14b), and we haveor

    (c) Since and and have limit 0 as itfollows that .limx:c (x) = 0

    x: c, sxd - sxd - sxd sxd sxd

    limu:0 cos u = 1.

    limu:0 s1 - cos ud = 0u0 1 - cos u u

    limu:0 sin u = 0.

    limu:0 u = 0,limu:0 s- u d =u- u sin u u

    limx:c (x) = 0limx:c (x) = 0

    limu:0 cos u = 1limu:0 sin u = 0

    limx:0 usxd = 1

    limx:0s1 - sx

    2>4dd = 1 and limx:0s1 + sx

    2>2dd = 1,limx:0 usxd ,

    1 - x2

    4 usxd 1 +x2

    2 for all x Z 0,

    x

    y

    0

    L

    c

    hf

    g

    FIGURE 2.12 The graph of issandwiched between the graphs of g and h.

    x

    y

    0 11

    2

    1

    y ! 1 " x2

    2

    y ! 1 # x2

    4

    y ! u(x)

    FIGURE 2.13 Any function u(x) whosegraph lies in the region between

    and haslimit 1 as (Example 10).x: 0

    y = 1 - sx2>4dy = 1 + sx2>2d

    THEOREM 4The Sandwich Theorem Suppose that forall x in some open interval containing c, except possibly at itself. Supposealso that

    Then limx:c sxd = L .

    limx:c g sxd = limx:c hsxd = L .

    x = cg sxd sxd hsxd

    y ! !

    y ! !

    y ! sin !

    !

    1

    1

    " "

    y

    (a)

    y ! !

    y ! 1 # cos !

    !

    y

    (b)

    2

    2

    1

    112 0

    FIGURE 2.14 The Sandwich Theoremconfirms the limits in Example 11.

  • EXAMPLES

    We can use the sandwich theorem to prove two importantlimits:

    limx0

    sin(x) = 0 and limx0

    cos(x) = 1

    In order to apply the sandwich theorem we need to find tofunctions to serve as the bread.

    In Chapter 1 the following two inequalities are proven(you wouldnt be expected to know these):

    |x| sin(x) |x| and 1 |x| cos(x) 1.

    So, by the sandwich theorem we have

    limx0|x| = 0 = lim

    x0|x| = lim

    x0sin(x) = 0

    limx0

    1 |x| = 1 = limx0

    1 = limx0

    cos(x) = 1.

  • EXAMPLES

    Use the sandwich theorem to determine the following limit:

    limx0

    x2 sin(x)

    (Note that we could find this limit using the product rule.)

    In order to apply the sandwich theorem we need to find tofunctions to serve as the bread.

    Since 1 sin(x) 1 for all x 6= 0, we know thatx2 x2 sin(x) x2

    So, by the sandwich theorem we have

    limx0x2 = 0 = lim

    x0x2 = lim

    x0x2 sin(x) = 0

  • EXAMPLES

    This is a bit easier to see graphically:

  • PRACTICE PROBLEMS

  • PRACTICE PROBLEMS

    Determine the following limits:

    (1) limx2

    x3 2x2 + 4x + 8

    (2) limt6

    8(t 5)(t 7)

    (3) limy2

    y + 2y2 + 5y + 6

    (4) It can be shown that 1 x2

    6 x sin(x)

    2 2 cos(x) 1.Use this fact to determine

    limx0

    x sin(x)2 2 cos(x)

  • LIMITS AT INFINITY

  • LIMITS AT INFINITY

    Up to this point weve only considered the limit of a function ata real number c, that is, as x c.We can also consider the limits of functions as x growspositively or negatively without bound, i.e.

    limx f (x) and limx f (x)

    Limits at Infinity (intuitive)We say that a real number L is the limit of a function f as xapproaches (resp., ), if the values of f (x) can be madearbitrary close to L by taking x to be sufficiently large (resp.,sufficiently negative).

  • EXAMPLES

    Conisder the function f (x) = 1x again.

    61. A function discontinuous at every point

    a. Use the fact that every nonempty interval of real numberscontains both rational and irrational numbers to show that thefunction

    is discontinuous at every point.

    b. Is right-continuous or left-continuous at any point?

    62. If functions (x) and g(x) are continuous for couldpossibly be discontinuous at a point of [0, 1]? Give rea-

    sons for your answer.

    63. If the product function is continuous at must (x) and g(x) be continuous at Give reasons for youranswer.

    64. Discontinuous composite of continuous functions Give an ex-ample of functions and g, both continuous at for whichthe composite is discontinuous at Does this contra-dict Theorem 9? Give reasons for your answer.

    65. Never-zero continuous functions Is it true that a continuousfunction that is never zero on an interval never changes sign onthat interval? Give reasons for your answer.

    66. Stretching a rubber band Is it true that if you stretch a rubberband by moving one end to the right and the other to the left,some point of the band will end up in its original position? Givereasons for your answer.

    67. A fixed point theorem Suppose that a function is continuouson the closed interval [0, 1] and that for every x in[0, 1]. Show that there must exist a number c in [0, 1] such that

    (c is called a fixed point of ).scd = c

    0 sxd 1

    x = 0. ! gx = 0,

    x = 0?x = 0,hsxd = sxd # g sxd

    (x)>g (x) 0 x 1,sxd = e1, if x is rational

    0, if x is irrational

    2.6 Limits Involving Infinity; Asymptotes of Graphs 97

    68. The sign-preserving property of continuous functions Let be defined on an interval (a, b) and suppose that at somec where is continuous. Show that there is an interval

    about c where has the same sign as (c).

    69. Prove that is continuous at c if and only if

    70. Use Exercise 69 together with the identities

    to prove that both and are continuousat every point

    Solving Equations GraphicallyUse the Intermediate Value Theorem in Exercises 7178 to prove thateach equation has a solution. Then use a graphing calculator or com-puter grapher to solve the equations.

    71.

    72.

    73.

    74.

    75.

    76.

    77. Make sure you are using radian mode.

    78. Make sure you are using radianmode.2 sin x = x sthree rootsd .cos x = x sone rootd .x3 - 15x + 1 = 0 sthree rootsd2x + 21 + x = 4xx = 2xsx - 1d2 = 1 sone rootd2x3 - 2x2 - 2x + 1 = 0x3 - 3x - 1 = 0

    x = c .g sxd = cos xsxd = sin x

    sin h sin ccos h cos c -cos sh + cd =cos h sin c ,sin h cos c +sin sh + cd =

    limh:0 sc + hd = scd .

    sc - d, c + dd

    scd Z 0

    2.6 Limits Involving Infinity; Asymptotes of GraphsIn this section we investigate the behavior of a function when the magnitude of the inde-pendent variable x becomes increasingly large, or . We further extend the conceptof limit to infinite limits, which are not limits as before, but rather a new use of the termlimit. Infinite limits provide useful symbols and language for describing the behavior offunctions whose values become arbitrarily large in magnitude. We use these limit ideas toanalyze the graphs of functions having horizontal or vertical asymptotes.

    Finite Limits as

    The symbol for infinity does not represent a real number. We use to describe thebehavior of a function when the values in its domain or range outgrow all finite bounds.For example, the function is defined for all (Figure 2.49). When x ispositive and becomes increasingly large, becomes increasingly small. When x isnegative and its magnitude becomes increasingly large, again becomes small. Wesummarize these observations by saying that has limit 0 as or

    or that 0 is a limit of at infinity and negative infinity. Here are precise definitions.

    sxd = 1>xx: -q , x: qsxd = 1>x1>x1>x x Z 0sxd = 1>x

    qsq d

    x:

    x: ;q

    T

    y

    0

    1

    111 2 3 4

    2

    3

    4

    x

    1xy !

    FIGURE 2.49 The graph of approaches 0 as or .x: -qx: q

    y = 1>x

    Notice that as x, thevalue of f (x) 0. So,

    limx

    1x

    = 0

    Similarly, as x , thevalue of f (x) 0. So,

    limx

    1x

    = 0

  • EXAMPLES

    For any integer n > 0, the graph of f (x) =1xn

    is given by:

    n evenn odd

    So we see that

    limx

    1xn

    = 0

    for all integers n 1.

  • EXAMPLES

    The limit of a function at need not exist:

    f (x) = cos(x)

    limx cos(x) = DNE

    f (x) = x3 x

    limx x

    3 x =

    limx x

    3 x =

    f (x) = ex

    limx e

    x =lim

    x ex = 0

  • LIMIT RULES

    The limit rules also hold for limits at :

    Limit RulesLet L and M be real numbers and suppose

    limx f (x) = L and limx g(x) = M

    Then:lim

    x(f (x) g(x)) = LM

    limx(f (x) g(x)) = L M

    limx(f (x)/g(x)) = L/M if M 6= 0

    limx(f (x))

    r/s = Lr/s

    (r, s are integers with no common factors, and L > 0 if s is even)

  • THE SANDWICH THEOREM

    The sandwich theorem also hold for limits at :

    The Sandwich Theorem (for limits at )Let f , g, and h be functions and L real numbers, with

    limx g(x) = L = limx h(x)

    andg(x) f (x) h(x)

    for all sufficiently large x, then

    limx f (x) = L.

    (The analogous statement holds for limits at .)

  • EXAMPLE

    Use the sandwich theorem to determine the following limit

    limx

    sin(x)x

    We need to find functions that we can sandwich sin(x)/xbetween for sufficiently large and sufficiently negative x.

    Since 1 sin(x) 1, for all x 6= 0 we have

    1x sin(x)

    x 1

    xSo, by the sandwich theorem

    limx

    sin(x)x

    = limx

    1x

    = 0.

  • EXAMPLE

    Graphically, we have

  • LIMITS OF RATIONAL FUNCTIONS

    Even with our limit rules, determining limits at can bedifficult.

    However, for rational functions finding these limits is relativelyeasy.

    For example, suppose f (x) =x 1

    4x2 + 2and we want to

    determine limx f (x).

    First, we rewrite f (x) by dividing the numerator and thedenominator by the term involving the largest power of x. In thiscase the term 4x2:

    limx

    x 14x2 + 2

    = limx

    x4x2 14x21 + 12x2

    .

  • LIMITS OF RATIONAL FUNCTIONS

    Next we can apply the limit rules, specifically the quotient rule:

    limx

    x 14x2 + 2

    = limx

    x4x2 14x21 + 12x2

    =lim

    xx

    4x2 lim

    x1

    4x2

    limx 1 + limx

    12x2

    =0 01 + 0

    = 0

    Notice: The degree of the denominator was greater than that ofthe numerator and the limit was 0.

  • LIMITS OF RATIONAL FUNCTIONS

    Suppose f (x) =x2 1

    4x2 + 2.

    We rewrite f (x) by dividing the numerator and thedenominator by the term involving the largest power of x andemploy the quotient rule:

    limx

    x2 14x2 + 2

    =lim

    x14 lim

    x1

    4x2

    limx 1 + limx

    12x2

    =14 01 + 0

    =14

    Notice: The degree of the denominator was equal to that of thenumerator and the limit was a non-zero constant.

  • LIMITS OF RATIONAL FUNCTIONS

    Suppose f (x) =x3 1

    4x2 + 2.

    We rewrite f (x) by dividing the numerator and thedenominator by the term involving the largest power of x anduse the quotient rule:

    limx

    x3 14x2 + 2

    =lim

    x 1 limx1x3

    limx

    4x+ lim

    x2x3

    =1

    limx

    4x+ lim

    x2x3

    =

    Notice: The degree of the denominator was less than that of thenumerator and the limit was infinite.

  • LIMITS OF RATIONAL FUNCTIONS

    More generally we have:

    TheoremIf f is a rational function, with

    f (x) =anxn + + a1x + a0bmxm + + b1x + b0 .

    Thenlim

    x f (x) = limxanxn

    bmxm.

    Note that if m n

    limx f (x) = limx

    anxn

    bmxm=

    {0 m > n

    anbm m = n

  • EXAMPLES

    Determine the following limits:

    limx

    7x3

    x3 3x2 + 6x = limx7x3

    x3= lim

    x 7 = 7

    limx

    3x + 7x2 2 = limx

    3xx2

    = limx

    3x

    = 0

    limx

    2x + x2

    x 4 = limxx2

    x= lim

    x x =

    limx

    x4x2 + 7x4 7x2 + 9 = limx

    x47x4

    = limx

    17

    = 17

  • HORIZONTAL ASYMPTOTES

    DefinitionWe say that a function f has a horizontal asymptote of y = L ifeither

    limx f (x) = L or limx f (x) = L.

    For example, as was just shown limx

    7x3

    x3 3x2 + 6x = 7.

    So f (x) = 7x3

    x33x2+6x has ahorizontal asymptote ofy = 7.

    -1

    01234567891011121314

  • PRACTICE PROBLEMS

  • PRACTICE PROBLEMS

    For each of the functions below, determine the limit as x and the equation of any horizontal asymptotes:

    (1) f (x) =3 (2/x)4 + (1/x2)

    (2) g() =cos()

    (3) h(y) =y + 2

    y + 5 + 6y2

    (4) q(x) =7 + 3x2

    x + 6

    Infinite LimitsInfinite LimitsInfinite LimitsExamplesVertical AsymptotesVertical AsymptotesVertical AsymptotesPractice ProblemsLimit RulesLimit RulesLimits of PolynomialsThe ``Almost the Same'' RuleThe ``Almost the Same'' RuleThe Sandwich TheoremThe Sandwich TheoremExamplesExamplesExamplesPractice ProblemsLimits at InfinityExamplesExamplesExamplesLimit RulesThe Sandwich TheoremExampleExampleLimits of Rational FunctionsLimits of Rational FunctionsLimits of Rational FunctionsLimits of Rational FunctionsLimits of Rational FunctionsExamplesHorizontal AsymptotesPractice Problems


Related Documents