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Math Help - L'Hopital's rule question

  1. #1
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    L'Hopital's rule question

    if we consider 1/exp(sinx) and take the limit as x--> infinity i think it's intuitively clear that the limit doesn't exist because -1<=sinx<=1. But if we multiply top and bottom by x (which doesn't change 1/exp(sinx)) and apply L'Hopital's rule we can argue that the limit is 0. Try differentiating top and bottom and look at the result. there is a factor of x in the denominator which will take the denominator to infinity while the numerator is 1.

    can someone resolve this inconsistency?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kkoutsothodoros View Post
    if we consider 1/exp(sinx) and take the limit as x--> infinity i think it's intuitively clear that the limit doesn't exist because -1<=sinx<=1. But if we multiply top and bottom by x (which doesn't change 1/exp(sinx)) and apply L'Hopital's rule we can argue that the limit is 0. Try differentiating top and bottom and look at the result. there is a factor of x in the denominator which will take the denominator to infinity while the numerator is 1.

    can someone resolve this inconsistency?
    the denominator still does not go to infinity, it oscillates as well. so the limit does not exist. the fraction is negative if the cos(x) is in the negative range and positive if the cos(x) is in the positive range, so the denominator can go to +/- infinity depending on where the cosine is. this is inconsistent
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    i agree that it oscillates but it oscillates at a higher rate for each oscillation. if you use some graphing software you can see that xcosx gets bigger and bigger as x--> infinity. am i still missing something?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kkoutsothodoros View Post
    i agree that it oscillates but it oscillates at a higher rate for each oscillation. if you use some graphing software you can see that xcosx gets bigger and bigger as x--> infinity. am i still missing something?
    that doesn't matter, the mere fact that it oscillates means it does not have a limit. in this respect, you can think of the limit as a settling point. the y-value for which the function settles or is asymptotic to, will be the limit. but there is no such y-value because we're oscillating
    Last edited by Jhevon; July 7th 2007 at 08:26 AM.
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    i see what i was missing. thanks!!
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    when i was looking at the graph i was just looking to see it increase indefinitely. i didn't think to look that it was also DECREASING indefinitely!!

    thanks again!
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    The function f(x) = e^{\sin x} has no limit as x\to \infty.

    Because if we approach \infty as \pi, 2\pi , 3\pi ,... then \sin x =0 and so e^0 = 1

    But if we approach \mbox{Me} as \frac{\pi}{2}, \frac{3\pi}{2} , ... we get a different result, in fact, I think the limit just fails to exist.
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    Quote Originally Posted by ThePerfectHacker View Post
    The function f(x) = e^{\sin x} has no limit as x\to \infty.
    In fact the function f(x)=e^(sinx) is a periodic function.
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    Quote Originally Posted by curvature View Post
    In fact the function f(x)=e^(sinx) is a periodic function.
    Yes, but not all periodic functions fails to have a limit as x\to \infty.
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    Quote Originally Posted by curvature View Post
    In fact the function f(x)=e^(sinx) is a periodic function.
    As shown
    Attached Thumbnails Attached Thumbnails L'Hopital's rule question-july04.gif  
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    Quote Originally Posted by ThePerfectHacker View Post
    Yes, but not all periodic functions fails to have a limit as x\to \infty.
    Unless it is constant ?
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    Quote Originally Posted by kkoutsothodoros View Post
    if we consider 1/exp(sinx) and take the limit as x--> infinity i think it's intuitively clear that the limit doesn't exist because -1<=sinx<=1.
    Since f(x)=1/exp(sin(x)) is a periodic function (as shown by the graph below), the limit (as x to infinity) does not exist.
    Attached Thumbnails Attached Thumbnails L'Hopital's rule question-july05.jpg  
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  13. #13
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    i think what's important here is to be careful when quickly assuming the expression goes to 0 because the denominator goes to infinity while the numerator is a fixed constant. i made the mistake of not noticing that the denominator oscillates between infinity and - infinity. so in effect the limit doesn't exist and thus invoking L'Hopital's rule a mistake.
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  14. #14
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    originally when i posted this thread i knew 1/e^sinx had no limit as x-->infinity. what i did was multiply top and bottom by x. i then used L'Hopital's rule and incorrectly showed it's limit was 0. i couldn't resolve the inconsistency. i assumed after taking derivative of top and bottom of x/xe^sinx that the derivative of the bottom (e^sinx+xcosxe^sinx) goes to infinity. it doesn't because it oscillates between positive and negative infininty. i was only seeing the positive!!
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  15. #15
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    [quote=kkoutsothodoros;59530] i made the mistake of not noticing that the denominator oscillates between infinity and - infinity.
    [quote]

    Note that the denominator only oscillates between e and 1/e. See the first graph above.
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