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Math Help - A Couple Of Limits

  1. #1
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    A Couple Of Limits

    Im having trouble solving the following limits, any help would be appreciated Ive tried and for the second one I kept on getting that it DNE (Does not exist) but I've been told the answer is 1.


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  2. #2
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    <br />
2)\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {x + 1} \right)}}<br />
{x}\underbrace  = _{LH}\mathop {\lim }\limits_{x \to 0} \frac{1}<br />
{{x + 1}} = 1<br />
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  3. #3
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    Thanks! It seems I was doing the right thing but made a mistake when taking the derivative of ln(x+1) no idea how I made such a silly mistake
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  4. #4
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    Here's a way to look at the second one if you like.

    Use the series for \frac{ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\frac{x^{4}}{5}-\frac{x^{5}}{6}+......

    \lim_{x\to 0}\left[1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\frac{x^{4}}{5}-\frac{x^{5}}{6}+......\right]

    See now where the 1 comes in?. As all the x terms drop due to the 0, all that's left is 1.
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  5. #5
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    Thanks I got #2 now, its just #1 and #3 thats bugging me.
    Last edited by trigoon; December 5th 2008 at 09:59 AM.
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