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Math Help - Find the limit

  1. #1
    Junior Member
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    Find the limit

    Right now I'm doing some problems that requires you to find the limit using the L'hospital's rule. But our teacher told us not to use the rule for this problem:


    lim (e^x-e^-x-2x/x-sinx)
    x->0

    I'm not sure on how to do this though. The answer I get is 0/0 but I know that's not correct.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dm10 View Post
    I'm sorry but I don't understand what you just said.
    for example,

    e^x = \sum_{n = 0}^\infty \frac {x^n}{n!} = 1 + x + \frac {x^2}{2!} + \frac {x^3}{3!} + \cdots for all x

    with that, you can write the power series for e^{-x}

    Moreover, \sin x = \sum_{n = 0}^\infty (-1)^n \frac {x^{2n + 1}}{(2n + 1)!} = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +- \cdots for all x

    so you can rewrite each expression as power series. see what cancels out and continue that way.
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