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Math Help - Help on limit with log

  1. #1
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    Help on limit with log

    Hello, I have spent several hours stuck on this limit:

    lim(x->0) \frac{ln(1+x)-x}{x^2}

    I managed to solve it by L'hopital, its limit is -0.5 but I was wondering if anyone can come up with a different method, as I am not supposed to use L'hopital at this stage.

    I tried

    lim\hspace{0.1cm} ln(1+x)^{\frac{1}{x^2} } - lim \frac{1}{x} = ln \hspace{0.1cm} lim(1+x)^{\frac{1 1}{x x}} - lim \frac{1}{x} = ln e^{1/x} - lim \frac{1}{x}

    But it did not work.

    Any idea? Thanks!
    Last edited by bizan; December 30th 2012 at 07:24 AM.
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  2. #2
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    Re: Help on limit with log

    \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5} - \cdots

    \ln(1+x)-x=-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5} - \cdots

    \frac{\ln(1+x)-x}{x^2}=-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+\frac{x^3}{5} - \cdots

    Easier to see now?
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  3. #3
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    Re: Help on limit with log

    Wow!, now it makes sense. Thank you very much .
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  4. #4
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    Re: Help on limit with log

    That would work if you were allowed to use series. But that is equivalent to L'Hopital's rule":

    \lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x\to 0} \frac{f(0)+f'(0)x+c_1x^2+\dots}{g(0)+g'(0)x+c_2x^2  +\dots}

    and assuming f(0)=g(0)=0,

    \lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x\to 0} \frac{f'(0)+c_1x+\dots}{g'(0)+c_2x+\dots} = \frac{f'(0)}{g'(0)}

    - Hollywood
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  5. #5
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    Re: Help on limit with log

    Thanks for the reply.

    I am allowed to use L'hopital's rule. But I am also asked to do it with a different method.

    Given that in another exercise I had to expand ln(1+x) using a taylor series (which gives as a result the series abender posted) I think that is the extra way they want me to do it.
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  6. #6
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    Re: Help on limit with log

    I see. You're right - abender's calculation is undoubtedly what they are looking for.

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