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Lim x->inf t^(1/t)

  1. Mar 17, 2013 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{t\rightarrow \infty } t^{\frac{1}{t}}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]let \ y = x^{\frac{1}{x}}[/tex]

    [tex]\ln y = {\frac{1}{x}} ln x[/tex]

    [tex]\lim_{x\rightarrow \infty } ln y = \lim_{x\rightarrow \infty } {\frac{ln x}{x}}[/tex]

    L'H
    [tex]\lim_{x\rightarrow \infty } ln y = \lim_{x\rightarrow \infty } {\frac{\frac{1}{x}}{1}} = \frac{0}{1} = 0[/tex]

    Wolfram shows the answer to be 1 and intuition says that a^0 = 1 so i am not sure where my error is.
     
  2. jcsd
  3. Mar 17, 2013 #2

    Ray Vickson

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    Why do you think you have made an error?
     
  4. Mar 17, 2013 #3
  5. Mar 17, 2013 #4

    eumyang

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    It would be better to write it this way:
    Let
    [tex]let \ y = \lim_{x\rightarrow \infty } x^{\frac{1}{x}}[/tex]

    When you take the natural logarithm of both sides, you actually get
    [tex]\ln y = \ln \left( \lim_{x\rightarrow \infty } x^{\frac{1}{x}} \right)[/tex]
    ... but since ln x is continuous, you can rewrite it as
    [tex]\ln y = \lim_{x\rightarrow \infty } \ln \left(x^{\frac{1}{x}} \right) = \lim_{x\rightarrow \infty } {\frac{1}{x}}\ln x[/tex]

    That's because you are not finished. You have one more step to go. What does that 0 represent?
     
  6. Mar 17, 2013 #5
    0 = ln y so e^0 = 1?
     
  7. Mar 17, 2013 #6

    eumyang

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    Yes, ln y = 0, so y = e0 = 1. If you use my corrected definition of y, then you have your answer.
     
  8. Mar 17, 2013 #7

    Ray Vickson

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  9. Mar 18, 2013 #8

    HallsofIvy

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    "a^0= 1" has nothing to do with this. If you just replace x with [itex]\infty[/itex], you get [itex]\infty^0[/itex] which is "undetermined".
     
    Last edited by a moderator: Mar 18, 2013
  10. Mar 18, 2013 #9
    Then are you saying:
    [tex]\lim_{t\rightarrow \infty} t ^ \frac{1}{t} = DNE[/tex]


    After doing some more searching i found this:
    https://www.physicsforums.com/showpost.php?p=3048872&postcount=44
     
    Last edited by a moderator: Mar 18, 2013
  11. Mar 18, 2013 #10

    Dick

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