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Math Help - L'Hospital's Rule

  1. #1
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    L'Hospital's Rule

    I need help finding the limits of these expression using L'Hospital's rule.


    1. lim x --> 1 {x^a - ax + a -1/(x -1)^2}

    2. lim x --> 0+ {sqrt(x) lnx}

    3. lim x --> 0 {cot2x sin6x}

    4. lim x --> infinity {x^3e^-x^2}

    5. lim --> infinity {xe^1/x - x}

    6. lim --> infinity {x - lnx}

    5. lim --> 0+ {x^x^2}
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  2. #2
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    Quote Originally Posted by zachb View Post
    I need help finding the limits of these expression using L'Hospital's rule.


    1. lim x --> 1 {x^a - ax + a -1/(x -1)^2}
    Derivative,
    (ax^{a-1}-a)/(2*(x-1))
    Take derivative again,
    (a(a-1)x^{a-2})/(2)
    Now substitute,
    (a)(a-1)/2
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  3. #3
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    Quote Originally Posted by zachb View Post
    2. lim x --> 0+ {sqrt(x) lnx}
    This is in the form 0*(infinity), so do the following:

    lim(x-->0+){sqrt(x) lnx} = lim(x -->0+){ln(x)/[1/sqrt(x)]}

    Now we have an (infinity)/(infinity) format for the limit, so we can take L'Hopital's rule:
    lim(x-->0+){sqrt(x) lnx} = lim(x-->0+){[1/x]/[(-1/2)/sqrt(x^3)]}

    = (-2)lim(x-->0+){sqrt(x^3)/x}

    = (-2)lim(x-->0+){sqrt(x)} = 0

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zachb View Post
    6. lim x --> infinity {x - lnx}
    Think of x as ln(e^x). Then
    x - ln(x) = ln(e^x) - ln(x) = ln[e^x/x]

    Now, if lim(x-->infinity){e^x/x} is finite, then lim(x-->infinity){ln[e^x/x]} is finite. (If lim(x-->infinity){e^x/x} = 0 then lim(x-->infinity){ln[e^x/x]} is negative infinity.) If lim(x-->infinity){e^x/x} is infinite, then lim(x-->infinity){ln[e^x/x]} is infinite.

    So let's take a look at: lim(x-->infinity){e^x/x}. This is of the form (infinity)/(infinity) so use L'Hopital's rule:
    lim(x-->infinity){e^x/x} = lim(x-->infinity){e^x/1} = (infinity)

    Thus
    lim(x -->infinity){x - lnx} --> (infinity)

    -Dan

    PS Note that I would need to be VERY careful with this argument if lim(x-->infinity){e^x/x} had been finite. In general it is not a good idea to switch the order of the limit and ln operators (or the limit and any other operator.) As the limit turned out to be infinite this argument didn't matter.
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  5. #5
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    Quote Originally Posted by zachb View Post
    I need help finding the limits of these expression using L'Hospital's rule.




    3. lim x --> 0 {cot2x sin6x}

    lim x --> 0 {cot2x sin6x}

    = lim x --> 0 { sin6x / tan2x} to make it 0/0 form

    = lim x --> 0 { 6cos6x / 2(sec2x)^2}

    = 6/2 = 3
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  6. #6
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    lim x --> 0 {cot2x sin6x}

    = lim x --> 0 { sin6x / tan2x} to make it 0/0 form

    = lim x --> 0 { 6cos6x / 2(sec2x)^2}

    = 6/2 = 3
    how did you get from lim x --> 0 {cot2x sin6x} to lim x --> 0 { sin6x / tan2x} ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bobnine View Post
    how did you get from lim x --> 0 {cot2x sin6x} to lim x --> 0 { sin6x / tan2x} ?
    cot(2x) = 1/tan(2x)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zachb View Post

    4. lim x --> infinity {x^3e^-x^2}
    lim x --> infinity {x^3e^-x^2}
    = lim x --> infinity {x^3/e^x^2} ......infinity/infinity
    Apply L'Hopital's
    = lim x --> infinity {3x^2/2x(e^x^2)}
    = lim x --> infinity {(3/2) x/(e^x^2)} ........(3/2)*infinity/infinity
    Apply L'Hopital's again
    = lim x --> infinity {(3/2)1/2x(e^x^2)}
    = 0
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