1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook