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Math Help - Limit problem

  1. #1
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    Limit problem

    I´m having trouble with this limit problem, would appreciate help.

    lim_{x\to\1}\frac{x+x^2+...+x^n-n}{x-1}
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  2. #2
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    Re: Limit problem

    Quote Originally Posted by HK47 View Post
    I´m having trouble with this limit problem, would appreciate help.
    lim_{x\to\1}\frac{x+x^2+...+x^n-n}{x-1}
    Since both numerator and denominator approach zero you can apply L'Hopital's Rule.

    If you are not allowed to use that then the algebra becomes very messy.
    Note for each positive integer k, (x-1) is a factor of x^k-1.
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  3. #3
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    Re: Limit problem

    Yeah, cant use the L´Hopital quite yet. My math textbook sure likes to stump me with these exercises, which can not be solved by provided means. At least I can post them here, so I dont have to rack my brains indefinitely.
    Last edited by HK47; April 30th 2013 at 08:24 AM.
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  4. #4
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    Re: Limit problem

    Quote Originally Posted by HK47 View Post
    Yeah, cant use the L´Hopital quite yet. My math textbook sure likes to stump me with these exercises, which can not be solved by provided means. At least I can post them here, so I dont have to rack my brains indefinitely.
    Take note:
     \begin{align*} \frac{{x + {x^2} +  \cdots  + {x^n} - n}}{{x - 1}} &= \frac{{(x - 1) + ({x^2} - 1) +  \cdots  + ({x^n} - 1)}}{{x - 1}} \\&= \frac{{\sum\limits_{k = 1}^n {({x^k} - 1)} }}{{x - 1}}\\ &= \sum\limits_{k = 1}^n {\left( {\sum\limits_{j = 0}^{k - 1} {{x^j}} } \right)} \end{align*}
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    Re: Limit problem

    So the limit is infinity. Thank you! Didn´t think to factor the n of all things.
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    Re: Limit problem

    Quote Originally Posted by HK47 View Post
    So the limit is infinity.

    No! Actually the limit is \frac{n(n+1)}{2}.
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  7. #7
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    Re: Limit problem

    Oh okay, right, n isnt infinity
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    Re: Limit problem

    Quote Originally Posted by HK47 View Post
    Oh okay, right, n isnt infinity
    n is never infinity, because infinity is not a number.
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