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Math Help - limit problem

  1. #1
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    limit problem

    2) Evaluate: lim_{x \to 0} (\frac{1}{ln(x+1)} - \frac{1}{x})
    I don't know how to do this. Please show me a step-by-step solution
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  2. #2
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    The only solution I can think of right now involves Maclaurin series. Have you learnt that yet?

    \frac{1}{\ln(x+1)}-\frac{1}{x}=\frac{x-\ln(x+1)}{x\ln(x+1)}


    Then replace \ln(x+1) with its Maclaurin series:


    \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O(x^5)
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  3. #3
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    Replacing ln(x+1) with the McLaurin series, you wll still have a problem with both numerator and denominator going to 0.

    So use L'Hopital's rule.
    \frac{d(x-ln(x+1))}{dx}= 1- \frac{1}{x+1}
    \frac{d(x ln(x+1))}{dx}= ln(x+1)+ \frac{x}{x+1}

    Now the limit, as x goes to 0, of each of those, is 0 so do it again:
    \frac{d(1- \frac{1}{x+1})}{dx}= \frac{1}{(x+1)^2}
    \frac{d(ln(x+1)+ \frac{x}{x+1})}{dx}= \frac{1}{x+1}+ \frac{1}{x+1}- \frac{x}{(x+1)^2}

    The limit, as x goes to 0, of the numerator is 1 while the limit of the denominator is 2. The original limit is 1/2.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Replacing ln(x+1) with the McLaurin series, you wll still have a problem with both numerator and denominator going to 0.
    You can factor out x^2 to get something like \frac{\frac{1}{2}+O(x)}{1+O(x)}
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  5. #5
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    Okay, good point.
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