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Homework Help: A tricky limit

  1. Aug 24, 2007 #1

    daniel_i_l

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    Gold Member

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

    Find the limit:
    [tex]lim_{x \rightarrow 1} (2-x)^{tan \frac{\pi x}{2}}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I first tried to find the limit of ln of the function inorder to turn the power into a multiplication and got:
    [tex]lim_{x \rightarrow 1} \frac{ ln(2-x) sin \frac{\pi x}{2}}{cos \frac{\pi x}{2}}[/tex]
    Then I used L'hopitals rule and got:
    [tex]lim_{x \rightarrow 1} ln( (2-x)^{tan \frac{\pi x}{2}} ) = \pi / 2[/tex]
    That means that [tex]lim_{x \rightarrow 1} (2-x)^{tan \frac{\pi x}{2}} = e^{\pi / 2}[/tex]
    Is that right? I tried putting in values to my calc and it looks like the answer should be 1?
    What did I do wrong?
    Thanks.
     
  2. jcsd
  3. Aug 24, 2007 #2

    Dick

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    Homework Helper

    Check your l'Hopital's rule again. I get 2/pi, not pi/2.
     
  4. Aug 24, 2007 #3

    Gib Z

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    Homework Helper

    I also get [itex]2/\pi[/itex], and when I put in x=0.9999 on my calculator, the limit and [itex]e^{2/\pi}[/itex] agree to reasonable accuracy, around 1.89, not 1.
     
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