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Lim of integral

  1. Oct 18, 2004 #1
    [tex]\lim_{x \rightarrow 0} \frac{1}{x} \int_0^x {(1-tan2t)}^{\frac{1}{t}} dt[/tex]

    and nevermind the { in front of the 0, i couldnt figure out how to take it out, its my first time posting

    I can't figure out how to start attacking this problem, do I have to intregrate by parts? if so what do I use as u and v.
    I tried thinking of any identities that could help but nothing came to mind.

    Any help will be greatly appreciated

    Thanks in advance.

    Evaluating it with maple gives [tex] e^{-2} [/tex]

    Admin note: pesky { removed. Click the latex image to see its code.
     
    Last edited by a moderator: Oct 18, 2004
  2. jcsd
  3. Oct 18, 2004 #2

    NateTG

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    Well, you could re-write it as:
    [tex]\lim_{x\rightarrow 0} \frac{ \int_0^x {(1-tan2t)}^{\frac{1}{t}} dt}{x}[/tex]
    And then apply l'Hopital's rule.

    No clue if that's the right way to go, and I'm too lazy to check.
     
  4. Oct 18, 2004 #3
    you would apply lhopital, take the limit and then diifferentiate?
     
  5. Oct 18, 2004 #4
    oh damn just as i was writing this message i figured out how to apply lhopital , so simple but it escaped by grasp, its just a differentiation of an integral, thx so much
     
  6. Oct 19, 2004 #5

    NateTG

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    [tex]\lim_{x\rightarrow 0} \frac{ \int_0^x {(1-tan2t)}^{\frac{1}{t}} dt}{x}[/tex]
    Take the derivatives of the top and bottom:
    [tex]\lim_{x \rightarrow 0} \frac{(1-tan(2x))^{\frac{1}{x}}}{1}[/tex]
    [tex]\lim_{x \rightarrow 0} (1 - \tan (2x))^{\frac{1}{x}}[/tex]

    But now, the expression inside the parens goes to 1 since [tex]\tan(0)=0[/tex] and the exponent should also tend to bring things to 1, so the limit should be 1.
     
  7. Oct 19, 2004 #6

    arildno

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    This is wrong, Nate G!
    You've got the indeterminate form [tex]1^{\infty}[/tex]
    we have:
    [tex]\lim_{x\to0}(1-tan(2x))^{\frac{1}{x}}=e^{\lim_{x\to0}\frac{ln(1-tan(2x))}{x}}[/tex]
    [tex]=e^{\lim_{x\to0}\frac{-2x}{x}}=e^{-2}[/tex]
     
  8. Oct 19, 2004 #7

    NateTG

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    Stupid brain....for some strange reason I thought it was going to [tex]1^0[/tex]. <Puts on pointy hat with donkey ears.>
     
  9. Oct 19, 2004 #8
    just take the integral first then it will be real easy to do this question
     
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