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Does x->infiity (ln |cosx|)/x^2 exist?

  1. Mar 25, 2005 #1
    Does x->infiity (ln |cosx|)/x^2 exist??

    I can't apply apply L'Hopital here, since the limit of |cosx| doesn't exist..
    Any hints appreciated..
     
  2. jcsd
  3. Mar 25, 2005 #2

    Hurkyl

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    Well, just how badly does the limit of ln |cos x| not exist? How do you think your fraction looks as x grows large?
     
  4. Mar 25, 2005 #3

    Zurtex

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    Not mathematics in anyway at all:

    But wouldn't it be intuitive that this has no limit as as Cos(x) approaches 0, ln(|Cos(x)|) approaches infinity at a faster then exponential rate while 1/(x^2) approaches 0 at a rather slow rate.
     
  5. Mar 25, 2005 #4
    still not get it..

    How can I explain that:
    ln|cosx| grows faster than x^2 if x>X for some X?

     
  6. Mar 25, 2005 #5
    I think you can use the squeeze theorem to find the limit, since:

    0 < |cos x| < 1

    from there, if you apply the log and x^-2 functions, then use l'Hospital, I think you could find the answer.
     
  7. Mar 26, 2005 #6

    dextercioby

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    It's useless,because

    [tex] \lim_{x\searrow 0} \ln x =-\infty [/tex]

    [tex] \lim_{x\rightarrow 1} \ln x =0[/tex]

    Daniel.
     
  8. Mar 26, 2005 #7

    Hurkyl

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    ln|cos x| doesn't grow as x grows... you need to understand how it behaves.
     
  9. Mar 26, 2005 #8

    dextercioby

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    Here's a plot to convince yourself.

    Daniel.
     

    Attached Files:

  10. Mar 26, 2005 #9
    How could there be a limit? What's [itex]\lim_{u\rightarrow -\infty} \frac{u}{x^2}[/itex] for any constant [itex]x[/itex]? This is basically an equivalent problem.
     
  11. Mar 26, 2005 #10

    dextercioby

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    What do u mean "equivalent problem"...?:confused:That fraction has both the numerator & the denominator dependent upon "x"...

    :bugeye:

    Daniel.
     
  12. Mar 26, 2005 #11
    well, [itex]\ln{|\cos{x}|}[/itex] goes to [itex]-\infty[/itex] at every odd multiple of [itex]\frac{\pi}{2}[/itex]. For [itex] x > 0[/itex], which is obviously reasonable in this case, the largest [itex] x^2[/itex] gets on any interval [itex]\left[k\frac{\pi}{2}, \ (k+2)\frac{\pi}{2}][/itex] for an odd integer [itex]k>0[/itex] is just [itex] (k+2)^2\frac{\pi^2}{4}[/itex] since [itex]x^2[/itex] is monotonically increasing there.

    Thus on any such interval

    [tex] \left|\frac{\ln{|\cos{x}|}}{x^2}\right| \geq \left|\frac{\ln|\cos{x}|}{u^2}\right|[/tex]

    for some constant u, and from there it's easy.
     
    Last edited: Mar 26, 2005
  13. Mar 26, 2005 #12

    dextercioby

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    I couldn't follow your logic (it may be my fault),but that limit doesn't exist,because the numerator doesn't have a limit.The denominator goes to [itex] +\infty [/itex] but that still doesn't help.

    Daniel.
     
  14. Mar 26, 2005 #13
    Showing that the fraction goes to [itex]-\infty[/itex] on any interval of the form I posted above is enough (since then you can make x arbitrarily large, and it will still go to [itex]-\infty[/itex] somewhere past that, and in fact (necessarily) infinitely many times).
     
  15. Mar 26, 2005 #14

    dextercioby

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    Nope,it can't be [itex]-\infty[/itex] altogether,because u'd have a [itex] -\frac{\infty}{\infty} [/itex] at certain points (a infinite discrete set,where the "cosine=0") and 0 in the other points....

    Daniel.
     
  16. Mar 26, 2005 #15
    Ooops, somehow the [itex]\infty[/itex] on the bottom of your fraction didn't show up at first.

    Anyways, you never get an indeterminate form like that. Look at the intervals I was examining. They are all finite, ie. x and x^2 are bounded on each of them.
     
  17. Mar 26, 2005 #16

    Hurkyl

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    Not a valid reason. For example, consider (cos x) / x^2
     
  18. Mar 26, 2005 #17

    dextercioby

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    Yes,but that doesn't help.You need to compute the limit [itex] x\rightarrow \pm \infty [/itex],where it doesn't matter whether x^{2} is bounded on an finite interval.

    Daniel.
     
  19. Mar 26, 2005 #18

    dextercioby

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    I agree for the general case.In this context it's valid,though,because the function in the numerator in not bounded.

    Daniel.
     
  20. Mar 26, 2005 #19

    Hurkyl

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    Nope. Consider (x sin x) / x^2
     
  21. Mar 26, 2005 #20
    The definition of a limit at infinity is

    [tex] \lim_{x \rightarrow \infty} f(x) = L \Longleftrightarrow \exists N \ \forall \epsilon > 0 \ \mbox{s.t.} \ x > N \Longrightarrow |f(x) - L| < \epsilon[/tex]

    Now choose any N. I can always show you a point [itex]x^\prime[/itex] with [itex]x^\prime>N[/itex] such that

    [tex] \lim_{x\rightarrow x^\prime} \frac{\ln|\cos{x}|}{x^2} = -\infty[/tex]

    which, from the definition, obviously means that there can't be a finite limit [itex]L[/itex] as [itex]x \rightarrow \infty[/itex].
     
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