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Apr 07, 2018

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    LIMITS

    of

    FUNCTIONS

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    LIMITS OF FUNCTIONS

    OBJECTIVES:

    define limits;

    illustrate limits and its theorems; andevaluate limits applying the given theorems.

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    DEFINITION: Limits

    The most basic use of limits is to describe how a

    function behaves as the independent variableapproaches a given value. For example let us

    examine the behavior of the function

    for x-values closer and closer to 2. It is evident from

    the graph and the table in the next slide that the

    values of f(x) get closer and closer to 3 as the values

    of x are selected closer and closer to 2 on either the

    left or right side of 2. We describe this by sayingthat the limit of is 3 as x

    approaches 2 from either side, we write

    1xx)x(f 2 !

    1xx)x(f 2 !

    31xxlim2

    2x !p

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    2

    3

    f(x)

    f(x)

    x

    y

    1xxy 2 !

    x 1.9 1.95 1.99 1.995 1.999 2 2.001 2.005 2.01 2.05 2.1

    F(x) 2.71 2.852 2.97 2.985 2.997 3.003 3.015 3.031 3.152 3.31

    left side right side

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    1.1.1 (p. 70)Limits (An Informal View)

    This leads us to the following general idea.

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    EXAMPLE

    Use numerical evidence to make a conjecture about

    the value of

    1x

    1xlim

    1x

    p

    Although the function , this has no

    bearing on the limit.

    The table shows sample x-values approaching 1 from

    the left side and from the right side. In both cases the

    corresponding values of f(x) appear to get closer andcloser to 2, and hence we conjecture that

    and is consistent with the graph of f.

    1x

    1x)x(f

    !

    2

    1x

    1xli

    1x!

    p

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    Figure 1.1.9 (p. 71)

    x .99 .999 .9999 .99999 1 1.00001 1.0001 1.001 1.01

    F(x) 1.9949 1.9995 1.99995 1.999995 2.000005 2.00005 2.0005 2.004915

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    THEOREMS ON LIMITS

    Our strategy for finding limits algebraically has two parts:

    First we will obtain the limits of some simpler function

    Then we will develop a list of theorems that will enable us

    to use the limits of simple functions as building blocks for

    finding limits of more complicated functions.

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    We start with the following basic theorems, which are

    illustrated in Fig 1.2.1

    Theorem 1.2.1 (p. 80)

    axkk !!pp axax

    limblima

    numbers.realbekandaLetTheorem1.2.1

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    Figure 1.2.1 (p. 80)

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    33li33li33li

    exa ple,For

    a.ofvaluesallforaxaskf(x)

    whyexplainswhichvaries,xaskatfixedre ain

    f(x)ofvaluesthethenfunction,constantaiskxfIf

    x0x-25x!!!

    pp

    !

    ppp T

    Example 1.

    TT

    !!!

    pp!

    pppxxx

    If

    x-2x0xlim2lim0lim

    example,For

    .axfthattruebealsomustitathen xx,xf

    Example 2.

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    Theorem 1.2.2 (p. 81)

    The following theorem will be our basic tool for finding limits

    algebraically

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    This theorem can be stated informally as follows:

    a) The limit of a sum is the sum of the limits.

    b) The limit of a difference is the difference of the limits.

    c) The limits of a product is the product of the limits.

    d) The limits of a quotient is the quotient of the limits,provided the limit of the denominator is not zero.

    e) The limit of the nth root is the nth root of the limit.

    A constant factor can be moved through a limit symbol.

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    31

    58

    5)4(2

    5limxlim2

    5limx2lim5x2lim.1

    4x4x

    4x4x4x

    !

    !

    !

    !

    !

    pp

    ppp

    6

    12-18

    12)3(6

    12limx6lim12x6lim.23x3x3x

    !

    !!

    !ppp

    13

    131

    2)3(534

    2limxlim5xlim4lim

    2limx5limxlim4lim

    2x5limx4lim)2x5(x4lim.3

    3x3x3x3x

    3x3x3x3x

    3x3x3x

    !

    !

    !

    !

    !

    !

    pppp

    pppp

    ppp

    EXAMPLE Evaluate the following limits.

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    21

    10

    425

    52

    4lixli5

    xli2

    4lix5li

    x2li

    4x5

    x2li.4

    5x5x

    5x

    5x5x

    5x

    5x

    !

    !

    !

    !

    pp

    p

    pp

    p

    p

    3375

    15633

    6lixli3

    6lix3li

    6x3li6x3li.5

    33

    3

    3x3x

    3

    3x3x

    3

    3x

    3

    3x

    !

    !!

    !

    !

    !

    pp

    pp

    pp

    2

    3

    4

    9

    3x

    1x8li

    3x

    1x8li.6

    1x1x

    !!

    !

    pp

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    OR

    When evaluating the limit of a function at a givenvalue, simply replace the variable by the indicated

    limit then solve for the value of the function:

    22

    3

    lim 3 4 1 3 3 4 3 1

    27 12 1

    38

    x

    x xp

    !

    !

    !

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    EXAMPLE Evaluate the following limits.

    2x

    8xlim.1

    3

    2x

    p

    Solution:

    0

    0

    0

    88

    22

    82

    2x

    8xli

    33

    2x!

    !

    !

    p

    2

    2

    2

    2

    2

    3

    2

    2 2 4li

    2

    li 2 4

    2 2 2 4

    4 4 4

    12

    8li 122

    x

    x

    x

    x x x

    x

    x x

    x

    x

    p

    p

    p

    !

    !

    ! !

    !

    @ !

    Equivalent function:

    (indeterminate)

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    Note: In evaluating a limit of a quotient which

    reduces to , simplify the fraction. Just removethe common factor in the numerator and

    denominator which makes the quotient .

    To do this use factoring or rationalizing thenumerator or denominator, wherever the radical is.

    0

    0

    0

    0

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    0

    2 2 2 2 0lim

    0 0x

    x

    xp

    ! !

    0 0

    0

    0

    2 2 2 2 2 2lim lim

    2 2 2 2

    1 1 1 2

    lim 42 2 2 2 2 2

    2 2 2lim

    4

    x x

    x

    x

    x x x

    x x x x

    x

    x

    x

    p p

    p

    p

    !

    ! ! !

    @ !

    x

    22xlim.2

    0x

    p

    Solution:

    Rationalizing the numerator:

    (indeterminate)

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    9x4

    27x8lim.3

    2

    3

    0x

    p

    Solution:

    Rationalizing the numerator:

    (indeterminate)3

    2

    3

    3

    22

    38 27

    8 27 27 27 02lim

    4 9 9 9 034 9

    2

    x

    x

    xp

    ! ! !

    3 32 2

    32

    23

    2

    2

    2 3 4 6 98 27li li

    4 9 2 3 2 3

    4 6 9 9 9 9li

    2 3 3 3

    27 9 3 3 2

    6 2 22

    x x

    x

    x x x x

    x x x

    x x

    x

    p p

    p

    !

    ! !

    ! ! ! !

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    5x

    3x2xli.4

    2

    3

    2x

    p

    Solution:

    33

    222

    2 2 2 32 3lim

    5 2 5

    8 4 34 5

    15

    9

    15

    3

    x

    x x

    xp

    !

    !

    !

    !

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    EXERCISES

    5w4w

    7w7wli10.

    2x

    8xli.5

    19x9x2li9.4y

    y8y4li.4

    1y2y

    3y2y1yli8.

    1x

    4x3xli.3

    1x

    3x2x3x2li7.

    4x3x

    1x2li.2

    1x9

    1x3li6.2x5x4li.1

    2

    2

    1w

    3

    2x

    2

    134

    5x

    3

    1

    3

    2y

    2

    2

    1y3

    2

    1x

    2

    23

    1x21x

    2

    3

    1x

    2

    3x

    pp

    pp

    pp

    pp

    pp

    Evaluate the following limits.