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Math Help - L'Hopital problem

  1. #1
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    L'Hopital problem

    The first appearance in print of LíHąopitalís Rule was in the book Analyse des Infiniment Petits published by the Marquis de LíHopital in 1696. This was the first calculus textbook ever published and the example that the Marquis used in that book to illustrate his rule was to find the limit of the function

    x =

    (√(2a3x−x4)) − (a(3√(aax)))
    a - (4√(ax3))


    as x approaches a, where a > 0. (At that time it was common to write aa instead of a2.) Solve this problem

    N.B. : √ = route
    Last edited by asifrahman1988; March 17th 2013 at 02:12 AM.
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  2. #2
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    Re: L'Hopital problem

    is it 0? the numerator becomes \sqrt{(2a^{3}a-a^4)-(a(\sqrt[3]{aaa})) as x\rightarrow{a}
    = \sqrt{a^4}-a\sqrt[3]{a^3}
    = a^2-aa
    =0
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  3. #3
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    Re: L'Hopital problem

    Quote Originally Posted by iMaths View Post
    is it 0? the numerator becomes \sqrt{(2a^{3}a-a^4)-(a(\sqrt[3]{aaa})) as x\rightarrow{a}
    = \sqrt{a^4}-a\sqrt[3]{a^3}
    = a^2-aa
    =0
    No, because the denominator also goes to 0. This is the reason we want to use L'hopital's rule.
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  4. #4
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    Re: L'Hopital problem

    oh right i forgot that. i keep think in limit you just ignore the denominator. so you differentiate the numerator and the denominator?

    assuming you equation is: \frac{\sqrt{2a^{3}x-x^4}-a\sqrt[3]{aax}}{a-\sqrt[4]{ax^3}}

    using L'hopital's rule you differentiate both top and bottom? Chain rule?
    = \frac{0.5(2a^{3}x-x^4)^{-1/2}(2a^3-4x^3)-(1/3)a^{5/3}x^{-2/3}}{-(3/4)a^{1/4}x^{-1/4}}

    as x\rightarrow{a} we get:
    = \frac{0.5(2a^{3}a-a^4)^{-1/2}(2a^3-4a^3)-(1/3)a^{5/3}a^{-2/3}}{-(3/4)a^{1/4}a^{-1/4}}

    so
    = \frac{-a^{3}a^{-2}-a/3}{-{3/4}a}

    = \frac{-a-a/3}{-(3/4)a}

    = \frac{-(4/3)a}{-(3/4)a}

    =(4/3)/(3/4)

    =16/9
    Last edited by iMaths; March 17th 2013 at 03:11 AM.
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  5. #5
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    Re: L'Hopital problem

    Quote Originally Posted by iMaths View Post
    using L'hopital's rule you differentiate both top and bottom?
    Yes. Then you take the limit. If it exists, it is the same limit as the original expression.
    im sorry i cant remember how to differentiate the two terms under the square root sign.
    You can always put brackets around them and think of the two terms as a single term. If you don't like to do this for some reason, let s,t be your two terms under the square root sign. Then \frac{d}{dx}\sqrt{s+t}=\frac{d}{dx}(s+t)^{\frac{1}  {2}}= \frac{1}{2}(s+t)^{-\frac{1}{2}}\frac{d}{dx}(s+t)=\frac{1}{2}(s+t)^{-\frac{1}{2}}\left(\frac{ds}{dx}+\frac{dt}{dx} \right)
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  6. #6
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    Re: L'Hopital problem

    i edited #4, can you check if its correct?
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  7. #7
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    Re: L'Hopital problem

    Yep that looks about right.
    Thanks from iMaths
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  8. #8
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    Re: L'Hopital problem

    Quote Originally Posted by iMaths View Post
    oh right i forgot that. i keep think in limit you just ignore the denominator. so you differentiate the numerator and the denominator?

    assuming you equation is: \frac{\sqrt{2a^{3}x-x^4}-a\sqrt[3]{aax}}{a-\sqrt[4]{ax^3}}

    using L'hopital's rule you differentiate both top and bottom? Chain rule?
    = \frac{0.5(2a^{3}x-x^4)^{-1/2}(2a^3-4x^3)-(1/3)a^{5/3}x^{-2/3}}{-(3/4)a^{1/4}x^{-1/4}}

    as x\rightarrow{a} we get:
    = \frac{0.5(2a^{3}a-a^4)^{-1/2}(2a^3-4a^3)-(1/3)a^{5/3}a^{-2/3}}{-(3/4)a^{1/4}a^{-1/4}}

    so
    = \frac{-a^{3}a^{-2}-a/3}{-{3/4}a}

    = \frac{-a-a/3}{-(3/4)a}

    = \frac{-(4/3)a}{-(3/4)a}

    =(4/3)/(3/4)

    =16/9
    It actually ends up as 16a/9, because you are dividing by an extra 'a' because you have earlier a^(1/4)*a^(-1/4)=a, really it equals one, because you will get a^0, so the final answer is (16/9)*a
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