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Infinite Limits, Limits at Infinity, And Limit Rules - Sect22-24

Sep 24, 2015

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  • 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)