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Limit question

  1. Aug 13, 2007 #1

    EnumaElish

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    [SOLVED] Limit question

    1. The problem statement, all variables and given/known data

    Show that the limit of [itex]f'(x)[/itex] as x --> 1 is [itex]-4/\pi[/itex]:

    2. Relevant equations

    [tex]f(x) = 1 - 4 \arccos\left[\frac 1 2 \left(x+\sqrt{2-x^2}\right)\right]/ \pi[/tex]

    3. The attempt at a solution
    [tex]f'(x)=\frac{2\sqrt 2\left(1-x/\sqrt{2-x^2}\right)}
    {\left(\sqrt{1-x\sqrt{2-x^2}}\right)\pi}[/tex]

    Both the numerator and the denominator --> 0 as x-->1. I tried l'Hopital's rule. The derivative of the numerator is [itex]-4\sqrt 2/(2-x^2)^{3/2}[/itex], which evaluates to [itex]-4\sqrt 2[/itex] at x = 1. To get the stated answer, the derivative of the denominator should be [itex]\pi\sqrt 2[/itex] at x=1. But it is actually
    [tex]
    \frac{-(1-x^2)\pi}
    {\sqrt{2-x^2}\sqrt{1-x\sqrt{2-x^2}}},
    [/tex]

    which is [itex]0/0[/itex] at x=1. :uhh:
     
    Last edited: Aug 14, 2007
  2. jcsd
  3. Aug 13, 2007 #2

    learningphysics

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    L'Hopital seems to go on and on when you try it with f'(x)...

    Find the limit of (f'(x))^2 at x=1, then taking the square root of that limit...
     
  4. Aug 13, 2007 #3

    EnumaElish

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    (f'(x))^2 works nicely. Thanks!
     
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