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Limit of x^(1/log(x))

  1. Feb 4, 2016 #1
    I don't understand why the limit of x1/loga(x) as x approaches infinity is a, where a can be any constant for the base.

    Why isn't it 1? The base (x), approaches infinity, while the exponent approaches 0 (1/infinity), so it should be (infinity)0 = 1.
  2. jcsd
  3. Feb 4, 2016 #2


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    Something moving toward infinity raised to an exponent that is moving toward zero is not that simple.
    ## \log_a x ^ {1/\log_a(x)} = \frac{1}{\log_a(x)} \log_a x =1##
  4. Feb 4, 2016 #3


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    ##x^{1/\log_a x}## is a constant function for positive x. Try to find this constant.
  5. Feb 4, 2016 #4
    I'm confused by your hint. Didn't you just prove that it approaches 1?
  6. Feb 4, 2016 #5


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    I showed that if you take the log of your expression, it can be shown to be equivalent to 1.
    What does that tell you about the expression itself? As blue leaf mentioned, it will be a constant.
  7. Feb 4, 2016 #6
    I still don't understand what I'm looking for. What was the purpose of taking the log of the expression? I'm guessing that constant is the base of the logarithm? I don't understand where that comes from though.
  8. Feb 4, 2016 #7


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    Remember that for any y, you can write:
    ## y= a^{\log_a y}##
    Taking the log was just one piece of the puzzle.
  9. Feb 4, 2016 #8


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    Let ## y = x^{\log_a x}## and apply your log rules to simplify it.
  10. Feb 4, 2016 #9


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    The base used in a logarithm has to be positive, and not equal to 1.
    ##\log_a(x) = y \Leftrightarrow x = a^y##, for x > 0. (I'm talking only of the real-valued log function here.)
    If a = 1, the expression ##a^y## = a for any finite value of y.

    Also, ##\infty^0## is one of a number of indeterminate forms. These forms show up in limits. Since they are indeterminate, you can't say in advance what value the limit will be. Further, you can't use ##\infty## in expressions like the one you wrote.
  11. Feb 5, 2016 #10
    The limit can be seen with a little manipulation. Simply switch to the common logarithm.

    ## x^{\frac{1}{log_a(x)}} = x^{\frac{1}{\frac{Ln(x)}{Ln(a)}}} = x^{\frac{Ln(a)}{Ln(x)}} = \Big ( x^{\frac{1}{Ln(x)}} \Big) ^{Ln(a)}##

    The term in parenthesis should look familiar. From there it is straightforward.
  12. Feb 5, 2016 #11
    The original question mentions "(infinity)0 = 1".

    But we don't know what ∞0 is. That is why we have to carefully take the limit.

    When the expression that is going to a limit involves a variable in both the base and exponent, it's often helpful to take the logarithm. So if we set

    S(x) = x1/loga(x)

    then we get

    ln S(x) = ln(x) / loga(x) = ln(x) / (ln(x) / ln(a)) = ln(a).​

    Looks as though we don't have to do much more work.
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