15. L'Hôpital's Rule

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    Guillaume De l'Hpital

    1661 - 1704

    LHpitals Rule

    Actually, LHpitals Rule wasdeveloped by his teacher Johann Bernoulli. De lHpital

    paid Bernoulli for privatelessons, and then published thefirst Calculus book based on

    those lessons.

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    Johann Bernoulli

    1667 - 1748

    LHpitals Rule

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    Zero divided by zero can not be evaluated, and is anexample of an indeterminate form .

    2

    2

    4lim

    2 x

    x

    xConsider:

    If we try to evaluate this by direct substitution, we get:00

    In this case, we can evaluate this limit by factoring andcanceling:

    2

    2

    4lim

    2 x x x

    2

    2 2lim

    2 x x x

    x

    2

    lim 2 x

    x 4

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    If we zoom in far enough,the curves will appear as

    straight lines.

    2

    2

    4lim

    2 x

    x

    xThe limit is the ratio of the numerator

    to the denominator

    as

    x approaches 2.

    -5

    -4

    -3

    -2

    -10

    1

    2

    3

    4

    -3 -2 -1 1 2 3x

    2 4 x

    2 x -0.05

    0

    0.05

    1.95 2 2.05x

    lim x a

    f x

    g x

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    2

    2

    4lim

    2 x

    x

    x

    lim x a

    f x

    g x

    -0.05

    0

    0.05

    1.95 2 2.05x

    f x

    g x

    f x

    g x

    As 2 x

    becomes:

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    2

    2

    4lim

    2 x

    x

    x

    lim x a

    f x

    g x

    -0.05

    0

    0.05

    1.95 2 2.05x

    As 2 x

    f x

    g x

    becomes:

    df

    dg

    df dg

    dx

    dx

    df

    xdgd

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    2

    2 4lim 2 x x x lim x a f x

    g x

    2

    2

    4lim

    2 x

    d x

    dxd x

    dx

    22lim1 x x 4

    LHpitals Rule:

    If is indeterminate, then:

    lim x a

    f x

    g x

    lim lim x a x a

    f x f x

    g x g x

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    We can confirm LHpitals rule by working backwards, and

    using the definition of derivative:

    f a

    g a

    lim

    lim

    x a

    x a

    f x f a

    x ag x g a

    x a

    lim x a

    f x f a

    x ag x g a

    x a

    lim x a

    f x f a

    g x g a

    0lim

    0 x a f x

    g x

    lim x a

    f x

    g x

    Let f and g be defined on an interval where f(x) 0 and

    g(x) 0 as x a, but tends to a finite limit L.

    xg

    x f

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    Example:

    20

    1 coslim x

    x x x 0

    sinlim

    1 2 x x x

    0

    If its no longer indeterminate, thenSTOP !

    If we try to continue with LHpitals rule:

    0

    sinlim

    1 2 x

    x

    x0

    coslim

    2 x

    x 1

    2

    which is wrong,

    wrong, wrong!

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    On the other hand, you can apply LHpitals rule asmany times as necessary as long as the fraction is still

    indeterminate:

    20

    1 12lim

    x

    x x

    x

    12

    0

    1 11

    2 2lim 2 x

    x

    x

    0

    0

    0

    0

    0

    0not

    12

    20

    11 1

    2lim x

    x x

    x

    32

    0

    11

    4lim 2 x

    x 14

    2

    1

    8

    (Rewritten inexponentialform.)

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    When applying LHpitals

    Rule, one has to differentiate the

    numerator and the denominator separately .

    A common mistake is to differentiate the whole expression f(x

    ) / g(x)

    This leads to tedious, unnecessary, and, most importantly,

    wrong computations.

    Correct calculation:

    2

    1

    2cos2lim

    2sinlim

    2sinlim

    000

    x

    xdxd

    xdxd

    x

    x x x x

    Incorrect calculation:

    ????2sin2cos2lim2sinlim 200 x x x x

    x x

    x x

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    LHpitals rule can be used to evaluate other indeterminate

    00

    forms besides .

    The following are also considered indeterminate:

    0 10

    00

    The first one, , can be evaluated just like .0

    0The others must be changed to fractions first.

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    What makes an expression indeterminate?

    lim 1000 x x

    Consider:

    We can hold one part of the expression constant:

    1000lim 0 x x

    There are conflicting trends here. The actual limitwill depend on the rates at which the numerator anddenominator approach infinity, so we say that anexpression in this form is indeterminate .

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    Lets look at another one:

    0

    0lim1000 1 x

    x

    Consider:

    We can hold one part of the expression constant:

    0.1

    lim x x

    Once again, we have conflicting trends, so this formis indeterminate .

    0.1

    lim 0 x x

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    Here is an expression that looks like it might be

    indeterminate :

    0

    lim .1 0 x

    x

    Consider:

    We can hold one part of the expression constant:

    lim .1 0 x

    x

    The limit is zero any way you look at it, so theexpression is not

    indeterminate .

    1000

    0lim 0 x x

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    Finally, here is an expression that looks like it

    should NOT be indeterminate :

    We can hold one part of the expression constant:

    Consider: 1

    461000411000

    100010001000

    -101.759.0;102.471.1

    :

    01

    lim;11;1

    lim

    ex

    x x

    x x

    Once again, we have conflicting trends, so this form is

    indeterminate .

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    Here is the standard list of indeterminate forms:

    0

    1 00 0

    00

    There are other indeterminate forms using complexnumbers, but those are beyond the scope of this class.

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    Example

    x 1

    1 1

    x 1 1

    Let's return to our former example: lim1

    1. Check to see if L'Hopital's rule applies:lim( ) 0 and lim( 1) 0

    L'Hopital's rule does apply.

    ( )lim lim lim

    1 ( 1)

    x

    x

    x x

    x x

    x

    e e x

    Stepe e x

    d e ee e dx

    d x xdx

    1

    1

    x

    x

    ee

    a

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    Example:

    1lim

    1lim

    1lim 2

    2

    2

    2

    2 x x

    x

    x

    x

    x x x x

    1lim

    2 x

    x x

    ;1

    lim1

    lim 22

    2

    2

    x x

    x x

    x x

    Determine the limit:

    Rewrite:

    Use the Power Law: The limit of a functions positive integer power is the power of the functions limit:

    OR Ratio of the leading powers:

    1

    1lim

    1lim

    2

    2

    2

    2

    xdxd

    xdxd

    x x

    x x

    111

    1lim

    1lim 2

    2

    2

    2

    x x

    x x

    x x

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    1lim sin x

    x x

    This approaches00

    1sin

    lim

    1 x

    x

    x

    This approaches 0

    We already know that0

    sinlim 1 x

    x

    x

    but if we want to use LHpitals rule:

    2

    2

    1 1cos

    lim1 x

    x x

    x

    1sinlim

    1 x x

    x

    1lim cos x x

    cos 0 1

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    1

    1 1lim

    ln 1 x x x

    If we find a common denominator and subtract, we get:

    11 lnlim1 ln x

    x x x x

    Now it is in the form00

    This is indeterminate form

    1

    11lim

    1ln

    x

    x x

    x

    x

    LHpitals rule applied once.

    0

    0

    Fractions cleared. Still1

    1lim

    1 ln x x

    x x x

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    1

    1 1lim

    ln 1 x x x

    11 lnlim1 ln x

    x x x x

    1

    11lim

    1ln

    x

    x x

    x x

    1

    1lim 1 1 ln x x

    LHpital

    again.

    1

    2

    1

    1lim

    1 ln x

    x

    x x x

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    Cautionary Example

    1

    1 1

    1

    lnExample: Evaluate lim

    Step 1. Check to see if L'Hopital/s rule applies:

    lim ln ln1 0 but lim 1 0

    L'Hopital's Rule does not apply.

    Use the quotient property for limits instead:limln

    lim

    x

    x x

    x

    x

    x x

    x x

    x x

    1

    1

    ln 00

    lim 1

    Using L'Hopital's Rule would have given us

    an incorrect result.

    x

    x

    x

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    Indeterminate Forms: 1 00 0

    Evaluating these forms requires a mathematical trick tochange the expression into a fraction.

    ln lnnu n u

    When we take the log of an exponential function,the exponent can be moved out front.

    ln1u

    n

    We can then write theexpression as a fraction,which allows us to useLHpitals

    rule.

    lim x a

    f x ln lim

    x a f x

    e

    lim ln x a

    f x

    e

    We can take the log of the function as long

    as we exponentiate

    at the same time.

    Then move thelimit notationoutside of the log.

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    Indeterminate Forms: 1 00 0

    1/lim x x

    x

    1/lim ln x x xe

    1lim ln x

    x xe

    lnlim

    x

    x

    x

    e1

    lim 1 x x

    e0

    e1

    0

    LHpitalapplied

    Example: