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Math Help - Finding a limit

  1. #1
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    Finding a limit

    I'm reading Understand Calculus and having trouble with one of the questions set after the chapter on finding limits.

    The task is to find the limit of f(x) as x approaches 2 when f(x) =
    x^2-4
    -------
    x^2-2x

    I can tell that in this case the limit is 2 by drawing a graph of the equation, but I think there should also be a way of proving it with algebra. I've tried putting the equation into an online limit finder (Wolfram Alpha) and looking at the working there but it uses l'Hospital's rule and as the book I'm working from hasn't covered that yet (or even toucher on differentiation) I think there is a way to prove the limit without using that rule. Any ideas?
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  2. #2
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    Just factor: \dfrac{(x+2)(x-2)}{x(x-2)}.
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  3. #3
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    Thanks, as soon as you point that out I think I have it.

    So if I divide by x^2 I get (x+2)/x

    Which means that as x approaches 2, (x+2)/x approaches 4/2.

    Is that right?

    By the way, how do you make the funky looking equation?
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  4. #4
    Super Member TheChaz's Avatar
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    Quote Originally Posted by djnorris2000 View Post
    Thanks, as soon as you point that out I think I have it.

    So if I divide by x^2 I get (x+2)/x

    Which means that as x approaches 2, (x+2)/x approaches 4/2.

    Is that right?

    By the way, how do you make the funky looking equation?
    1. If you divide by (x - 2), then yes.
    2. Yes. 4/2 = 2
    3. It's called LaTex. You can either "reply with quote" to see what someone typed in, or create your own at...
    Online LaTeX Equation Editor
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  5. #5
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    The answer is 2.
    You can learn to post in symbols. You can use LaTeX tags
    [tex] \left\lfloor x \right\rfloor \;\& \,\left\lceil x \right\rceil [/tex] gives  \left\lfloor x \right\rfloor \;\& \,\left\lceil x \right\rceil
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  6. #6
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    Quote Originally Posted by TheChaz View Post
    1. If you divide by (x - 2), then yes.
    Bah, that's exactly what I meant to say.

    Thanks both of you.
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