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Math Help - Natural log limit

  1. #1
    Member nautica17's Avatar
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    Natural log limit

    lim x->0 from the right of ln(4/x^2)

    I think it is infinity but I'm not sure how to prove that? I know that if you divide by zero that will give you infinity and I can see that from the graph. However that logarithm being there is making things a bit confusing. I feel like I might be missing a step because of that. Can I just pull out that 4 and take the limit of the rest of the function and then that will give zero?
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  2. #2
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    Quote Originally Posted by nautica17 View Post
    lim x->0 from the right of ln(4/x^2)

    I think it is infinity but I'm not sure how to prove that? I know that if you divide by zero that will give you infinity and I can see that from the graph. However that logarithm being there is making things a bit confusing. I feel like I might be missing a step because of that. Can I just pull out that 4 and take the limit of the rest of the function and then that will give zero?
    Just pull it apart using the properties of logs.

    ln\left(\frac{4}{x^2}\right)=ln(4)-ln(x^2)

    =ln(4)-2ln(x)

    So now the limit 2\lim_{x->0^+}ln(x) is -\infty right? If you look at a graph of ln(x), you should be able to see that. So the limit is:

    \lim_{n->0^+}ln(4)-2\lim_{n->0^+}ln(x)=\infty
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  3. #3
    Super Member General's Avatar
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    Quote Originally Posted by nautica17 View Post
    lim x->0 from the right of ln(4/x^2)

    I think it is infinity but I'm not sure how to prove that? I know that if you divide by zero that will give you infinity and I can see that from the graph. However that logarithm being there is making things a bit confusing. I feel like I might be missing a step because of that. Can I just pull out that 4 and take the limit of the rest of the function and then that will give zero?
    No, The result is not infinity always.
    \frac{a}{0}=\infty if a > 0.

    \frac{a}{0}=-\infty if a < 0.

    \frac{a}{0} undefined if a=0.

    What do you mean by "pull out that 4" ??!
    In general : ln (\frac{a}{b}) \neq a ln (\frac{1}{b}) !!

     ln (\frac{4}{x^2})=ln(4) - 2ln(x)
    Now take the limit, And clearly it infinity.
    If you see the graph of f(x)=ln(\frac{4}{x^2}), you will see that y-axis is a vertical asymptote.
    which prove what do you want.
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