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Math Help - Algebra after using l'Hopital's Rule

  1. #1
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    Algebra after using l'Hopital's Rule

    I would like to ask another question.
    I had to calculate a limit using l'Hospital's rule.
    I know that I got the correct numerical answer but unfortunately my answer has a different form from that of the professor.

    Professor's answer: \frac{3^\frac{5}{3}}{2^\frac{3}{2}}

    My answer:

    \frac{\frac{-1}{2}*2^\frac{-1}{2}}{\frac{-1}{3}*3^\frac{-2}{3}}

    How can I change my form in the better one which he uses?
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  2. #2
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    Quote Originally Posted by Sebastian de Vries View Post
    I would like to ask another question.
    I had to calculate a limit using l'Hospital's rule.
    I know that I got the correct numerical answer but unfortunately my answer has a different form from that of the professor.

    Professor's answer: \frac{3^\frac{5}{3}}{2^\frac{3}{2}}

    My answer:

    \frac{\frac{-1}{2}*2^\frac{-1}{2}}{\frac{-1}{3}*3^\frac{-2}{3}}

    How can I change my form in the better one which he uses?
    Multiply top and bottom by -1 to get rid of the minus signs, so now you have:

    \frac{\frac{1}{2}\times 2^\frac{-1}{2}}{\frac{1}{3}\times 3^\frac{-2}{3}}

    Then observe that 1/2=2^{-1} and 1/3=3^{-1} to get:

    \frac{2^{-1}\times 2^\frac{-1}{2}}{3^{-1}\times 3^\frac{-2}{3}}

    Now laws of indices give:

    \frac{ 2^\frac{-1-1}{2}}{ 3^\frac{-1-2}{3}} =\frac{ 2^\frac{-3}{2}}{ 3^\frac{-5}{3}}

    Now flip the whole thing upside down to get rid of the negative powers to get the Prof's answer.

    .
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  3. #3
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    \lim_{ x\to1}\frac{(x+1)^\frac{1}{2}-(2x)^\frac{1}{2}}{(x+2)^\frac{1}{3}-(2x+1)^\frac{1}{3}}

    I attempted to type the asked limit but as you can see the software can't execute the code (my mistake of course).

    Constatine, thanks for the answer.


    Mr F edit: You would see that I had fixed the latex if you stopped re-editing and hence undoing my editing!
    Sebastian: sorry.
    Last edited by Sebastian de Vries; January 5th 2009 at 05:28 AM. Reason: Fixed the latex. Again!
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  4. #4
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    Quote Originally Posted by Sebastian de Vries View Post
    \lim_{ x\to1}\frac{(x+1)^\frac{1}{2}-(2x)^\frac{1}{2}}{(x+2)^\frac{1}{3}-(2x+1)^\frac{1}{3}}

    I attempted to type the asked limit but as you can see the software can't execute the code (my mistake of course).

    Constatine, thanks for the answer.


    Mr F edit: You would see that I had fixed the latex if you stopped re-editing and hence undoing my editing!
    If x approaches 1 your quotient will become \dfrac00. Therefore you should use de l'Hopitals rule:

    \lim_{ x\to1}\dfrac{(x+1)^\frac{1}{2}-(2x)^\frac{1}{2}}{(x+2)^\frac{1}{3}-(2x+1)^\frac{1}{3}} = \lim_{x\to1}\dfrac{\dfrac{\sqrt{x}-\sqrt{2} \sqrt{x+1}}{2\sqrt{x}\sqrt{x+1}}}{\dfrac{(2x+1)^{\  frac23} - 2(x+2)^{\frac23}}{3(x+2)^{\frac23}(2x+1)^{\frac23}  }}

    Now plug in x = 1 and you'll get:

    \lim_{ x\to1}\dfrac{(x+1)^\frac{1}{2}-(2x)^\frac{1}{2}}{(x+2)^\frac{1}{3}-(2x+1)^\frac{1}{3}} = \dfrac34 \cdot \sqrt{2} \cdot \sqrt[3]{9}\approx 2.206
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