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Did I do anything wrong in this limit evaluation?

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    A problem in my book asks the reader to evaluate lim x→0 xln(2x) using L'Hôpital.

    2. Relevant equations

    None


    3. The attempt at a solution

    I think I got it right but I simply wanted to make sure I did no illegal manipulations (I'm self-studying this stuff).

    I started by substituting u = 1/x to get an undeterminate form suitable for L'Hôpital. By doing so, lim x→0 becomes lim u→∞

    lim x→0 xln(2x) = lim u→∞ ln(2/u) / u

    Applying L'Hôpital:

    = lim u→∞ (-u)/2u2

    = lim u→∞ (-1)/2u

    = 0


    I know that's the right answer, but did I make any mistakes? Also, was I allowed to switch lim x→0 by lim u→∞? Thanks in advance!
     
  2. jcsd
  3. Mar 8, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure, it's right. You didn't even really need a substitution. Write the original limit as ln(2x)/(1/x) which is equally indeterminant. Then l'Hopital gives (1/x)/(-1/x^2)=(-x). That's still 0.
     
  4. Mar 8, 2012 #3
    Oh right, that sure is simpler. Thanks again!
     
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