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Math Help - Limit Problem involving ln

  1. #1
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    Limit Problem involving ln

    what is lim as x approaches 1 for ln(ln x)/lnx ?
    Last edited by Jhevon; November 14th 2009 at 09:11 PM.
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  2. #2
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    Quote Originally Posted by mariam View Post
    lim as x approaches 1 for ln(lnx)/lnx
    as x \to 1 , the numerator approaches -\infty , and the denominator approaches 0 ... what does that tell you?
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    i think this means that it does not exist
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    Quote Originally Posted by skeeter View Post
    as x \to 1 , the numerator approaches -\infty , and the denominator approaches 0 ... what does that tell you?
    actually, the numerator only approaches - \infty as x \to 1 from the right...

    Quote Originally Posted by mariam View Post
    i think this means that it does not exist
    you are correct, but probably not for the reason you think you are. the problem is, as x \to 1 from the left, \ln x is negative, and so \ln ( \ln x ) would be undefined. As the logarithm function is only defined when its argument is positive.

    So since the numerator is undefined as we approach 1 from the left, the entire function is undefined there, and so the limit does not exist for that reason.

    Now, \lim_{x \to 1^+} \frac {\ln ( \ln x)}{\ln x} does exist. That should be fun for you to find (but not that difficult--honest)
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    I believe as x approchase 1 from the right, ln(x) equals to negitive infinity.
    By using L Hopital`s rule and taking the derivative of both numerator and denominator. so as x approchase 1 from the right 1/ln(x) = 1/negative infinity = 0
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mariam View Post
    I believe as x approchase 1 from the right, ln(x) equals to negitive infinity.
    By using L Hopital`s rule and taking the derivative of both numerator and denominator. so as x approchase 1 from the right 1/ln(x) = 1/negative infinity = 0
    exactly how would you apply L'Hopital's rule here? you need the form to be \frac 00 or \frac {\infty}{\infty} to use that. you have neither of those forms here...
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    I am not sure if this make sense. now I know that as x approchase 1 from the right ln(x)= negitive infinity.. so as x approchase to 1 from the right ln(ln(x))/ln(x)= ln(negitive infinity)/negitive infinity.. so now the numerator undifined and the denominator equal to negitive infinity
    This all I know
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mariam View Post
    I am not sure if this make sense. now I know that as x approchase 1 from the right ln(x)= negitive infinity.. so as x approchase to 1 from the right ln(ln(x))/ln(x)= ln(negitive infinity)/negitive infinity.. so now the numerator undifined and the denominator equal to negitive infinity
    This all I know
    No, \lim_{x \to 1^+} \ln x = 0

    you have \lim_{x \to 1^+} \frac {\ln \ln x}{\ln x} \to \frac {\ln 0^+}{0^+} \to \frac {- \infty}0...loosely speaking.

    what can you do with that?
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  9. #9
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    Is it this does not exist?
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    Quote Originally Posted by mariam View Post
    Is it this does not exist?
    No, it exists...

    Here's the trick, write it as \lim_{x \to 1^+} \ln \ln x \cdot \frac 1{\ln x}

    Now what?
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  11. #11
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    Trust me I have no clue.
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    Quote Originally Posted by mariam View Post
    Trust me I have no clue.
    \lim_{x \to 1^+} \ln \ln x \cdot \frac 1{\ln x} \to - \infty \cdot \infty

    How about now?
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  13. #13
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    I am just gussing, ok, I think you took the seperate. so you said lim ln lnx . lim 1/lnx as x approachase 1 from the right so you got - infty . infty , which gave you - infty..... as you say 1 * -1=-1
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    Quote Originally Posted by mariam View Post
    I am just gussing, ok, I think you took the seperate. so you said lim ln lnx . lim 1/lnx as x approachase 1 from the right so you got - infty . infty , which gave you - infty..... as you say 1 * -1=-1
    um, yeah. i never thought of -1*1 though... i guess i thought, a negative number times a positive number gives a negative number, and a huge number times a huge number gives a huge number. and so, a huge negative number times a huge positive number will give a huge negative number. or something like that but yeah, your idea seems to work also
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    Hahaha , I like how you said it
    I believe I will never forget it
    Thanks
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