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Lim inf/sup question

  1. Dec 18, 2008 #1
    i need to find the sup and the inf of

    [itex]
    f(x)=1/(1+(ln x)^2 )

    [/itex]

    i can find the limit for the function for + infinity
    and - infinity

    but what to do next?
    the sup is the least upper bound
    the limit for +infinity is not SUP
    ??
     
  2. jcsd
  3. Dec 18, 2008 #2

    HallsofIvy

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    I assume you mean sup and inf of the range of that function. You seem to be thinking that the limit as x goes to infinity SHOULD be the sup and there is no reason for that. As x goes to either infinity or negative infinity f(x) goes to 0. Since f is never negative, 0 is the inf (greatest lower bound) of the range of f. Since the denominator is always larger than or equal to the numerator, f is never larger than 1- but f(1)= 1 so sup (least upper bound) of the range of f is 1.
     
    Last edited: Dec 18, 2008
  4. Dec 18, 2008 #3
    thanks
     
  5. Dec 18, 2008 #4

    statdad

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    Halls, I think you have a typo here:
    because of the [tex] \ln x [/tex] term, [tex] f(0) [/tex] is undefined. I believe you meant to
    type

    [tex]
    f(1) = 1
    [/tex]
     
  6. Dec 18, 2008 #5

    HallsofIvy

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    Thanks statdad. I have edited so I can pretend I never made that mistake!
     
  7. Dec 18, 2008 #6

    statdad

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    Nothing to see here. Move along. :smile:
     
  8. Jan 21, 2009 #7
    ....ln(x) is not defined for x<=0.
    The domain of f(x) is {x:x>0}.
    Cheers.
     
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