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Basic Log Question

  1. Jan 22, 2007 #1

    I come from an engineering background and so have not studied analysis (sadly). I need to figure out the following.

    How does:

    1.) x^y*|ln(1/x)|^m behave for any m given y<0 as x-> infinity

    2.) x^y*|ln(1/x)|^m behave for any m given y>0 as x-> 0

    The way I see it in the first example as x-> infinity the |ln(1/x)|-> infinity
    so effectively you have infinity^y*infinity^m and y is less than 1. So this should explode right?

    However the answer is apparently that the expression->x?

    In the second |ln(1/x)|-> infinity as x tends to 0. So effectivey you have
    However the answer is apparently that the expression ->1?

    Can someone please explain where I am going wrong?
  2. jcsd
  3. Jan 22, 2007 #2

    D H

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    You can simplify this using [itex]|\log(1/x)| = |-\log x| = \log x[/itex].
    Thus, [itex]x^y|\log(1/x)^m| = x^y(\log x)^m[/itex].

    The following only works if m is positive. In both problems, [itex]x^y\to 0[/itex] and [itex](\log x)^m\to\infty[/itex] as [itex]x\to\infty[/itex] (1) or [itex]x\to 0[/itex] (2). The product is indeterminate. Solving it calls for L'Hopital's rule.
  4. Jan 22, 2007 #3

    Sorry problem 1 above should read

    1.) x^y*|ln(1/x)|^m behave for any m given y<1 as x-> infinity

    but I don;t think that changes the nature of your argument.

    In any case thanks will look into L'hopital's rule
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