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Math Help - Finding Limits Using L'Hopital's Rule

  1. #1
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    Finding Limits Using L'Hopital's Rule

    Okay I'm trying to find this limit:

    lim (ln(x))1/x as x approaches infinity using L'Hopital's rule.



    This is what I've done so far:


    y = (ln(x))1/x

    ln(y) = (ln(ln(x)))/x

    lim ln(y) = lim (ln(ln(x)))/x (Both as x approaches infinity)

    (using L'H) = lim 1/(xln(x)) (as x approaches infinity)



    That is what I have so far and I do not know where to go from there.
    Also, sorry for the bad use of writing the equations and what not; I'm not sure how to write them on here as they should actually look, if that makes sense.
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    Re: Finding Limits Using L'Hopital's Rule

    Both x and ln(x) approach infinity as x approaches infinity. Therefore if you are dividing by them, the reciprocal has to approach 0.

    Edit: typo. The reciprocals, in this case 1/x and 1/ln(x) both have to approach 0, so 1/(x*ln(x)) approaches 0 as well.
    Last edited by SworD; September 3rd 2012 at 09:36 AM.
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    Lightbulb Re: Finding Limits Using L'Hopital's Rule

    \lim_{x\to \infty } \, \sqrt[x]{\log (x)} = 1
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    Re: Finding Limits Using L'Hopital's Rule

    Quote Originally Posted by SworD View Post
    Both x and ln(x) approach infinity as x approaches infinity. Therefore if you are dividing by them, the reciprocal has to approach 0.

    Edit: typo. The reciprocals, in this case 1/x and 1/ln(x) both have to approach 0, so 1/(x*ln(x)) approaches 0 as well.
    True. But that's not the answer. We're supposed to find lim y as x approaches infinity. So far, all I have is what lim ln(y) as x approaches infinity equals. I'm supposed to find lim y as x approaches infinity.
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    Re: Finding Limits Using L'Hopital's Rule

    Quote Originally Posted by MaxJasper View Post
    \lim_{x\to \infty } \, \sqrt[x]{\log (x)} = 1
    I need the steps. And that answer is from wolfram alpha. I know because I already checked hoping to find a solution there first.
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    Re: Finding Limits Using L'Hopital's Rule

    Quote Originally Posted by Preston019 View Post
    I need the steps. And that answer is from wolfram alpha. I know because I already checked hoping to find a solution there first.
    Because {\lim _{x \to \infty }}\frac{{\log (x)}}{x}\mathop  = \limits^H {\lim _{x \to \infty }}\frac{{\frac{1}{x}}}{1} = 0 we know that {\lim _{x \to \infty }}{\left( {\log (x)} \right)^{\frac{1}{x}}} = {e^0} = 1
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    Re: Finding Limits Using L'Hopital's Rule

    True. But that's not the answer. We're supposed to find lim y as x approaches infinity. So far, all I have is what lim ln(y) as x approaches infinity equals. I'm supposed to find lim y as x approaches infinity.
    But it will give you the answer.If limit ln(y) = 0, then lim y has to equal 1, because you know that as the limit tends to infinity, if ln(y) tends to 0, what does y have to tend to? It will = 1, since ln(1) = 0. Basically raise e to both sides. You are allowed to do that because ln() is continuous at that point.
    Last edited by SworD; September 3rd 2012 at 12:18 PM.
    Thanks from Preston019
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