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Math Help - Epsilon-delta limit proof

  1. #1
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    Epsilon-delta limit proof

    The question:

    Prove that the limit as x goes to infinity is 0 for \frac{x+3}{x^2-3}

    My attempt:

    |\frac{x+3}{x^2-3} - 0| < \varepsilon

    Now I'm stuck. How do I solve this for x? I'm sure this is simple, but for some reason I'm just not seeing it. :/

    Thanks.
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The question:

    Prove that the limit as x goes to infinity is 0 for \frac{x+3}{x^2-3}

    My attempt:

    |\frac{x+3}{x^2-3} - 0| < \varepsilon

    Now I'm stuck. How do I solve this for x? I'm sure this is simple, but for some reason I'm just not seeing it. :/

    Thanks.
    To show a limit at infinity we need to show that for every \epsilon > 0 there exists an N \in \mathbb{R} such that for x > N |f(x)-L|< \epsilon

    Notice that for n > 3

    \displaystyle \bigg|\frac{N+3}{N^2-3} \bigg |< \bigg| \frac{N+3}{N^2-9}\bigg| =\bigg| \frac{N+3}{(N+3)(N-3)}\bigg|=\bigg| \frac{1}{N-3}\bigg|
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  3. #3
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    I'm having trouble following. I think I understand the general proof and conditions, but I'm unsure of why you substituted N for x.
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  4. #4
    Behold, the power of SARDINES!
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    To show that the limit exists we need to show that when x is large enough that the limit is close to 0. So I picked N to be my large x. So now for any x > N the inequality will hold.
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