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Looking for a function

  1. Jun 13, 2005 #1
    Does anyone know a function t(x) satisfying the following conditions.

    as x -> 0 we have

    (1) t(x)-> oo

    (2) x t(x) -> 0

    (3) x t(x) exp(t(x)) -> 0

    (4) x t(x)^2 exp(t(x)) -> oo

    I have "two extremes" as

    t(x) = sqrt(-log x) satisfies (1-3) but x t(x)^2 exp(t(x)) -> 0

    t(x) = -log( -x log(x))

    satisfies (1-2) and (4), but x t(x) exp(t(x)) -> c >0

    I have been searching desperately.
     
  2. jcsd
  3. Jun 13, 2005 #2
    What level course is this?
     
  4. Jun 13, 2005 #3
    hello

    if your really desperate you should try ploting it through mathematica or some similar program and watch what happens as x goes to infinity it shouldnt be too hard, a few guesses and you should be able to get your solution
     
  5. Jun 13, 2005 #4
    This is in the reals, right? And am I misreading this:
    or are (1) and (2) mutually exclusive?
     
  6. Jun 13, 2005 #5
    I want to agree with you but first I want to verify that the only functions that tend to infinity as x-> 0 are ones where the denominator has a quicker tendency to infinity than the numerator.
     
  7. Jun 13, 2005 #6

    [tex]t(x) = \frac{1}{\sqrt{\vert x \vert}}[/tex]

    Am I missing something?
     
  8. Jun 13, 2005 #7
    [tex]\lim_{x\rightarrow 0} \frac{x}{\sqrt{|x|}} \neq 0 [/tex]
     
  9. Jun 13, 2005 #8
    edit: I misread the OP, ignore this post.
     
    Last edited by a moderator: Jun 13, 2005
  10. Jun 13, 2005 #9

    shmoe

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    It's zero. When x>0 it's sqrt(x). It appears he's only looking at right handed limts (hence his log's), but for this function the two-sided limit is also zero, it's -sqrt(x) for x<0.

    These limits are as x->0, none of them exclude any others.


    t(x) = -log( -x log(x)) is hopeful. In condition 3, you're looking at [-log(x)-log(-log(x))]/(-log(x)). You'd like to speed up the growth of the denominator without affecting the numerator signifigantly. Try fiddling with the inside log of t(x)
     
  11. Jun 13, 2005 #10
    I thought of this last night, does it have to be continuous?
    And also, for a limit to exist, the function must approach the value from both sides, however since the log function can't take negative values, there is a whole in his domain for t(x) = -log( -x log(x)), correct?
     
  12. Jun 13, 2005 #11

    shmoe

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    The log(x) is what made me think he was considering right hand limits. Finding a function satisfying those limits from the right is enough in any case, just put |x| where you see x and you've got your two sided limits as well. There's nothing about continuity as he's written it, just a limit. In fact the first condition rules on continuity at 0. The t(x) = -log( -x log(x)) will work, with some minor alterations.
     
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