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Math Help - Infiniate Limit of a Natural Log

  1. #1
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    Infiniate Limit of a Natural Log

    How do I find the limit as x goes to +infinity for (ln 2x)/(ln 3x)?

    Also, how do I find the limit as x approaches 1 from the negative direction for ln (1-x)?

    Thank You.

    Jeannine
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  2. #2
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    For the first question, writing that \ln(2x)=\ln 2 + \ln x= \left(\frac{\ln 2}{\ln x} + 1\right)\ln x may help you.

    For the second one, notice that if x tends to 1 from the left, then 1-x tends to 0 from the right, so that the limit you are looking for is in fact the limit of \ln y when y tends to 0 from the right.

    Laurent.
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  3. #3
    Super Member 11rdc11's Avatar
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    Try using l'hopital rule for the 1st problem
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    Infinite Limit of a Natural Log, Part Two

    Thanks. I get the first one.

    For the second, I'm still unclear. I notice that:

    ln(1-x) = ln 1 - ln x = 0 - ln x = -ln x.

    Does this help? How do I find the limit as x approaches 1 from the negative direction for -ln x?

    Sorry for my clouded mind!

    Jeannine
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  5. #5
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    Quote Originally Posted by jbecker007 View Post
    I notice that:
    ln(1-x) = ln 1 - ln x = 0 - ln x = -ln x.
    No you don't. You have \ln (ab)=\ln a+\ln b... Logarithms turns multiplications into additions (by the way, log was first introduced to simplify computations, since additions are easier).

    What I suggested in my first post may be called "composition of limits". The limit of 1-x, as x tends to 1 from the right, is 0^+ (0, reached by upper values). And the limit of \ln(y) when y tends to 0 from the right is -\infty. So, composing these results, the limit of \ln(1-x) when x tends to 1 from the left is -\infty.

    Is it clearer now?

    Laurent.
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