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Math Help - Limit of a Sum

  1. #1
    Member CalcGeek31's Avatar
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    Limit of a Sum

    How would one work out this limit?

    lim 48(sum(x^n/n! -x -1))/x^2
    x->0


    im sorry for the formatting... thanks in advance
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  2. #2
    MHF Contributor
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    Quote Originally Posted by CalcGeek31 View Post
    How would one work out this limit?

    lim 48(sum(x^n/n! -x -1))/x^2
    x->0


    im sorry for the formatting... thanks in advance
    If you mean  \lim_{x \to 0} \frac{48}{x^2} \left( \sum_{n=0}^{\infty} \frac{x^n}{n!} - 1 - x\right)

    it's  \lim_{x \to 0} \frac{48}{x^2} \left(1 +x + \frac{x^2}{2!} + \frac{x^3}{3!} \cdots - 1 - x\right) = \lim_{x \to 0} \frac{48}{x^2} \left(\frac{x^2}{2!} + \frac{x^3}{3!} \cdots \right) = \lim_{x \to 0} 48 \left(\frac{1}{2!} + \frac{x}{3!} \cdots \right) =24
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  3. #3
    Member CalcGeek31's Avatar
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    everything is over x^2
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  4. #4
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    Quote Originally Posted by CalcGeek31 View Post
    everything is over x^2
    In that case you've been shown the solution.
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