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The limit as x approaches 1 of x / ln (x)

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Hi. I'm having problems with the limit of x --> 1 of ( x / ln(x) ).

    2. Relevant equations

    L'Hopital's Rule
    Algebraic Manipulation

    3. The attempt at a solution

    I understand that the the limit will give me a semi-indeterminate form (that is, it's answer is 1 / 0).

    What I don't understand is how I can manipulate ln (x) so it's not in the denominator. I tried thinking of ways to use e^x.... but realized that multiplying by e^x does nothing. I'm presuming that we can raise both the numerator and the denominator by e^x but was not sure if it was legitimate.

    Thanks in advance.
     
  2. jcsd
  3. Sep 6, 2008 #2

    Dick

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    Homework Helper

    If you are getting the 'semi-indefinite form' 1/0, you can forget about manipulating it further. The limit doesn't exist.
     
  4. Sep 6, 2008 #3

    HallsofIvy

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    Any time you have a ratio of two continuous functions, and just putting the value gives 1/0 (or more generally any non-zero value over 0) the limit does not exist. As x gets closer and closer to the "target value" the numerator stays close to 1 while the denominator gets close to 0. I.e. the fraction gets larger and larger. There is no limit.
     
  5. Sep 6, 2008 #4
    Thanks, both. I understand it now. I made a numerical table to estimate the limit.... and as you two pointed out... the limit does not exist (infinite discontinuities).
     
  6. Apr 9, 2011 #5
    well i don't think so maybe I'm wrong but let's see

    lim as x->1 (x/lnx) now me remove the natural lag

    lim as x->1 ( e^x/ x) so as X approaches 1 we get lim x->1 (e^1/1)=e :)

    and yes it's legit :)
     
    Last edited: Apr 9, 2011
  7. Apr 9, 2011 #6

    gb7nash

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    To add to this, the limit could be infinity, -infinity, or not exist. If you have something like [tex]\lim_{x\to 0}\frac{1}{x}[/tex], x could approach 0 from the left, from the right, or from both directions. In this case, we can't put a definitive answer.

    Note: This only works if you include the extended real line. If not, then it wouldn't exist at all.
     
    Last edited: Apr 9, 2011
  8. Apr 9, 2011 #7

    Mark44

    Staff: Mentor

    What makes you think that you can remove the natural "lag" (sic) in this way?

    As was already said, this limit does not exist. The two one-sided limits are as far apart as they could possibly be.
    [tex]\lim_{x \to 1^+}\frac{x}{ln(x)} = \infty[/tex]
    while
    [tex]\lim_{x \to 1^-}\frac{x}{ln(x)} = -\infty[/tex]

     
  9. Apr 10, 2011 #8
    Actually ex/lnx cannot be simplified unlike elnx = x. For your second expression you'll get:

    [tex]
    \lim_{x \to 1^+}e^(x/lnx) = \infty
    [/tex]
    [tex]
    \lim_{x \to 1^-}e^(x/lnx) = 0
    [/tex]
     
  10. May 1, 2011 #9

    sorry yes I'm wrong... but you don't have to be rude and I know that you have to write log instead of "LAG" but please it's not the end of world.
     
  11. May 2, 2011 #10

    HallsofIvy

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    There was nothing rude about Mark44's response.
     
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