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I A limit as t-->0 of log(t) / SQRT(t)

  1. May 6, 2017 #1
    I am trying to find the following limit ##\displaystyle \lim_{x \to 0^{+}} \frac{\log (t)}{\sqrt{t}}##, however, I don't see how. Obviously the answer is negative infinity, but I don't see how to get that. L'hospital's rule doesn't seem to work.
     
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  3. May 6, 2017 #2

    andrewkirk

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    Use an epsilon-delta* argument, and take care with signs.

    * Actually, where the limit is infinite, its an M-delta argument for big M.
     
  4. May 6, 2017 #3

    Orodruin

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    Why would you use l'Hôpital? The numerator goes to minus infinity and the denominator to zero ...
     
  5. May 6, 2017 #4
    Just to be clear, would that be the only way?
     
  6. May 6, 2017 #5

    Orodruin

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    Ultimately, any way is based on the definition as proofs of all theorems regarding limits must be based on the definition of a limit. In your case thoguh, you have a product of a function whose limit is infinity and another whose limit is minus infinity.
     
  7. May 6, 2017 #6
    Can we rigorously say that infinity times negative infinity is negative infinity?
     
  8. May 6, 2017 #7

    pwsnafu

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    Of course not, infinity is not a real number.

    However you can rigorously prove: Let ##f(x)## diverge to infinity and ##g(x)## diverge to negative infinity as ##x \to \infty##. Then ##(fg)(x)## diverges to negative infinity.
    If you have done epsilon proofs, then you should be able to do this.
     
  9. May 25, 2017 #8
    I don't see why not, you can write -∞ as a product of ∞ and -1 so you get -(∞2)=-∞
     
  10. May 25, 2017 #9
    ##\infty = \infty + 1 \iff \infty - \infty = 1 \iff 0 = 1##

    Now ?
     
  11. May 25, 2017 #10

    Mark44

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    @Buffu, your example is unrelated to what @phasacs said, which by the way has an error.
    If ##\lim_{x \to a}f(x) = -\infty## and ##\lim_{x \to a} g(x) = -1##, then ##\lim_{x \to a} f(x) g(x) = -\infty##. It wouldn't take much to prove this rigorously.
     
    Last edited: May 25, 2017
  12. May 25, 2017 #11

    Mark44

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    It doesn't apply in this problem.
    L'Hopital's Rule is applicable for limits of the form ##\frac 0 0## or the form ##\frac{\pm \infty}\infty##.
     
  13. May 25, 2017 #12

    jack action

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    Wouldn't the proof be the following:
    $$\lim_{t \to 0^{+}} \frac{\log (t)}{\sqrt{t}} = \lim_{t \to 0^{+}} \log (t)\frac{1}{\sqrt{t}}= \left( \lim_{t \to 0^{+}} \log (t)\right) \left(\lim_{t \to 0^{+}} \frac{1}{\sqrt{t}}\right) = (-\infty)(\infty) = -\infty$$
     
  14. May 25, 2017 #13
    My example was to suggest that playing with infinity like a real number is not valid. Isn't that his error ?
     
  15. May 25, 2017 #14
    I think somebody might argue with your proof as to how you know that "-ve infinity" times "+ve infinity" is "-ve infinity" ?
     
  16. May 25, 2017 #15

    Mark44

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    No, it isn't, although I don't know who you're referring to with "his error."
    If ##\lim_{x \to a} f(x) = \infty## and ##\lim_{x \to a} g(x) = \infty##, then it's fairly easy to rigorously show that
    1. ##\lim_{x \to a} f(x) \cdot (\pm g(x)) = \pm \infty##, and
    2. ##\lim_{x \to a} f(x) + g(x) = \infty##.
    The limits you can't show are the ones that are indeterminate: ##\lim_{x \to a} \frac{f(x)}{g(x)}## and ##\lim_{x \to a} f(x) - g(x)##.
     
  17. May 25, 2017 #16

    jack action

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    If I replace ##\infty## with a large number, then ##-\infty \times \infty## will get me a very large negative number (even if both ##\infty## are different values).

    From what I know, infinity multiplication (with any real number or infinity) is not an indeterminate form. If you can find some reference stating otherwise, I would be glad to learn something new.
     
  18. May 25, 2017 #17

    Mark44

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    Right.
     
  19. May 25, 2017 #18
    :sorry::sorry:. You are correct. :smile::smile:
     
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