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Basic limit question

  1. Feb 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the limit:

    [itex]lim_{x -> \infty}[n*(x^{\frac{1}{n}}-1)-ln(x)][/itex], for any n.

    2. Relevant equations
    L'Hopitals rule.

    3. The attempt at a solution
    I know that the ratio of the first expression over the last goes to zero, by L'Hopital, but unfortunately I now have a difference and not a quotient. Is it possible to transform it in some way to use L'Hopital?
  2. jcsd
  3. Feb 25, 2014 #2


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    Staff: Mentor

    a-b = (ab-b^2)/b = (a/b - 1)/(1/b)
    Not sure if one of those helps.
  4. Feb 25, 2014 #3
    [itex]lim_{x -> \infty}[n*(x^{\frac{1}{n}}-1)-ln(x)][/itex], for any n.

    ##y = \displaystyle \lim_{x \to \infty}(ln[e^{n*(x^{\frac{1}{n}}-1)}]-ln(x))##
    ##y = \displaystyle \lim_{x \to \infty}\displaystyle ln\left[\frac{e^{n*(x^{\frac{1}{n}}-1)}}{x}\right]##
    ##e^y = \displaystyle \lim_{x \to \infty}\displaystyle \left[\frac{e^{n*(x^{\frac{1}{n}}-1)}}{x}\right]##

    That gives it in fraction form. I'm not sure if it makes it easier to evaluate or not, I haven't tried the problem myself.
  5. Feb 27, 2014 #4


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    If you apply l'Hopital's rule on that fraction, you end up with the same fraction again (plus some irrelevant part you can ignore).
  6. Feb 27, 2014 #5
    You're right.

    I believe factoring out ln(x) so you have ##\displaystyle\lim_{x\to\infty}\ln{(x)} \cdot \left(\frac{n(x^{1/n}-1)}{\ln{(x)}} - 1\right)## might work. You can use l'Hospital's on the inside fraction which works better than what I suggested above.
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