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Lim x->0+ and l'Hospital Rule

  1. Jul 10, 2010 #1
    1. The problem statement, all variables and given/known data

    lim(x→0^+)(lnx/x)


    3. The attempt at a solution
    First, what does x->0+ mean? Is it positive infinity?
    Why lim(x→0^+) (lnx)=-∞?
    Why the answer is

    = -∞ * (1/0+)
    = -∞* ∞
    = -∞?
    Why cannot I use l'Hospital Rule?
     
  2. jcsd
  3. Jul 10, 2010 #2

    Mark44

    Staff: Mentor

    It means that x approaches zero from the right (positive values close to zero).
    Look at the graph of y = ln(x). The domain is {x | x > 0}. As x gets closer to zero, y gets more and more negative.
    What you have above is mostly incorrect. 1/0 is not a number, so you can't use it in calculations. [tex]\lim_{x \to 0^+} \frac{ln(x)}{x} = \lim_{x \to 0^+} \frac{1}{x}ln(x) [/tex]
    The first factor gets larger and larger without bound; the second factor gets more and more negative without bound. As a result the product's limit is -∞.
    Read the fine print in L'Hopital's Rule. There are certain conditions that must be satisfied before you can use it.
     
  4. Jul 10, 2010 #3
    That's a great explanation. Thank you a lot.
     
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