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Math Help - crazy limit

  1. #1
    Senior Member polymerase's Avatar
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    crazy limit

    Find the value of \displaystyle\lim_{x\to\infty}({\frac{\pi}{2}-\arctan\:x})^{\frac{1}{ln\:x}}.

    Thanks!
    Last edited by polymerase; December 7th 2008 at 09:40 AM.
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by polymerase View Post
    Find the value of \displaystyle\lim_{x\to\infty}({\frac{\pi}{2}-arctan\:x})^{\frac{1}{ln\:x}}.

    Thanks!
    It goes to 0^0, so it is either 0 or 1.
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by colby2152 View Post
    It goes to 0^0, so it is either 0 or 1.
    NO IT DOESN'T
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by polymerase View Post
    NO IT DOESN'T
    Well, the arctan(x) will go to \frac{/pi}{2} as x goes to infinity, because the range is -\frac{\pi}{2}, \frac{\pi}{2}

    So the inside goes to \frac{\pi}{2}-\frac{\pi}{2}=0

    And the power will go to 0, because ln(\infty)=\infty and \frac{1}{\infty} =0.

    So, yes, it does go to 0^0
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by angel.white View Post
    Well, the arctan(x) will go to \frac{/pi}{2} as x goes to infinity, because the range is -\frac{\pi}{2}, \frac{\pi}{2}

    So the inside goes to \frac{\pi}{2}-\frac{\pi}{2}=0

    And the power will go to 0, because ln(\infty)=\infty and \frac{1}{\infty} =0.

    So, yes, it does go to 0^0
    OMG you noobs...no it doesn't ok! The answer should be \frac{1}{e}. trust me this is challenge question my prof asked...and if you knew him you would know his challenge questions are never easy. I may know the answer but i don't know how to get it...and either do you guys.
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  6. #6
    Super Member angel.white's Avatar
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    Quote Originally Posted by polymerase View Post
    OMG you noobs...no it doesn't ok! The answer should be \frac{1}{e}. trust me this is challenge question my prof asked...and if you knew him you would know his challenge questions are never easy. I may know the answer but i don't know how to get it...and either do you guys.
    Prove it
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  7. #7
    Senior Member polymerase's Avatar
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    Quote Originally Posted by angel.white View Post
    Prove it
    How about you graph it and see....
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  8. #8
    Super Member angel.white's Avatar
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    On a graph, yes, I get 1/e. The equation does go to 0^0, but that form isn't differentiable, so it is wrong to say it must be zero or 1.
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  9. #9
    Senior Member polymerase's Avatar
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    Quote Originally Posted by angel.white View Post
    On a graph, yes, I get 1/e. The equation does go to 0^0, but that form isn't differentiable, so it is wrong to say it must be zero or 1.
    you learn something everyday don't ya....

    see if you find or talk to someone who can figure out the answer....i really would like to know.
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  10. #10
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    Quote Originally Posted by polymerase View Post
    Find the value of \displaystyle\lim_{x\to\infty}({\frac{\pi}{2}-arctan\:x})^{\frac{1}{ln\:x}}.

    Thanks!
    \left( \frac{\pi}{2} - \tan^{-1} x \right) ^{ \frac{1}{\log x} } = \exp \left( \frac{\log (\pi/2 - \tan^{-1} x)}{\log x} \right)
    But,
    \frac{\log (\pi/2 - \tan^{-1} x)}{\log x} \to - 1.
    Thus the limit is,
    e^{-1}.
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  11. #11
    Super Member angel.white's Avatar
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    \frac{\log (\pi/2 - \tan^{-1} x)}{\log x} \to - 1

    TPH: can you show this? I messed with it a while, I kept getting zero (maybe I made an early mistake that kept interfering with my answer)
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  12. #12
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    log 0 and log x are both infinite so we can use L'hopital's rule.

     \frac {d}{dx} (log(\pi/2-arctan x)) = \frac {1/(x^2+1)}{\pi/2-arctan x}

     \frac {d}{dx} (log x) = 1/x

    so we have <br />
\frac {\frac {1/(x^2+1)}{\pi/2-arctan x}}{1/x}

    which simplifies to <br />
\frac {x/(x^2+1)}{\pi/2-arctan x}

    since this is 0/0 we can use l'Hopital's rule again":
    <br />
\frac {d}{dx}(\frac {x}{x^2+1} = \frac {x^2+1-2x^2}{(x^2+1)^2}<br />
=\frac {-x^2+1}{(x^2+1)^2}

    Edit: an angel found my mistake here and I have now fixed it
    \frac {d}{dx} (\pi/2-arctan x) = \frac {-1}{x^2+1}

    so we have
    \frac {\frac {-x^2+1}{(x^2+1)^2}}{\frac{-1}{x^2+1}}

    which simplifies to

    \frac {x^2-1}{x^2+1} -> 1
    Last edited by badgerigar; December 6th 2007 at 02:05 AM. Reason: fixed mistake kindly pointed out by an angel :)
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  13. #13
    Super Member angel.white's Avatar
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    Thankyou badgerigar, when I followed along, though, I got

    \frac {d}{dx} (\pi/2-arctan x) = -\frac {1}{x^2+1}

    as opposed to positive 1/(x^2+1), so my answer was positive one.
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  14. #14
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    Thanks for pointing that out angel-white. I guess there must be some other mistake in there also to make the answer come out right.

    I'm new here, so could someone tell me for future reference: If somebody finds a mistake you made is it good or bad to edit your post and fix it?
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  15. #15
    Super Member angel.white's Avatar
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    Hmm, I don't know. For me, if the mistake invalidates the entire post, I just put a disclaimer at the top, but leave it there so that the responses to the post make sense. If the mistake is something simple I can fix, then I usually do.
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