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Lim x→∞ (x - ln x)?

  1. Jun 3, 2009 #1
    Can somebody show a step by step process of how to get this answer?

    I keep getting -∞ when it should be ∞

    heres my work:

    lim x→∞ (x - ln x) = lim x→∞ ((1 - 1/x ln x) / (1/x)

    then apply l'hospitals rule

    lim x→∞ (x^(-2) ln x - x^(-2)) / (-x^(-2))

    then I cancel all the x^(-2) and im left with -ln x which would equal to -∞
     
  2. jcsd
  3. Jun 3, 2009 #2

    Mark44

    Staff: Mentor

    You can apply L'Hopital's rule only when you have an indeterminate form of [0/0] or [+/-infinity/infinity].

    In this expression -- ((1 - 1/x ln x) / (1/x) -- 1/x approaches zero, but ln x approaches infinity, which makes it another indeterminate form.

    I don't think that there's anything you can do with L'Hopital's rule on this one. You might try an approach that looks at the derivative of f(x) = x - ln x. It's easy to show that the derivative approaches 1 as x gets large, which suggests that x - ln x gets larger and larger.
     
  4. Jun 3, 2009 #3

    Dick

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    Why don't you apply l'Hopital to the ln(x)/x part of your expression and use the result to simplify it?
     
  5. Jun 3, 2009 #4

    George Jones

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    (At the risk of confusing things) or take the first x into the ln(x).
     
  6. Jun 3, 2009 #5
    Remember that [tex] ln(x) = \int_1^x \frac{1}{u} du [/tex] and use the MVT
     
  7. Jun 4, 2009 #6
    where does the ln(x)/x come from?
     
  8. Jun 4, 2009 #7
    This is what I would do:

    [tex]\stackrel{lim}{x\rightarrow\infty}(x-ln(x))\times\frac{(x+ln(x))}{(x+ln(x))}[/tex]

    This would give:


    [tex]\stackrel{lim}{x\rightarrow\infty}\frac{(x^{2}-ln^{2}(x))}{(x+ln(x))}[/tex]

    You could work it out from there, split the limit up by [tex]x^{2}[/tex] and [tex]ln^{2}x[/tex]

    Hope this helped, and sorry about the poor formatting, this is probably the first time im doing latex.

    By the way, i'm sure the value is infinity.
     
  9. Jun 4, 2009 #8

    zcd

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    You could do lim x→∞ (ln e^x - ln x) = lim x→∞ ln (e^x /x) and apply l'hopital's rule inside; the domain change of x to ln e^x is irrelevant because your limit exists within the domain of log.
     
  10. Jun 5, 2009 #9
    I think I see it, but just to make sure, does ln e equal 1? so that can be plugged into any equation?
     
  11. Jun 5, 2009 #10

    HallsofIvy

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    Yes, ln e= 1.
     
  12. Jun 5, 2009 #11

    Dick

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    Write it (as you basically did) as x*(1-ln(x)/x).
     
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