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Math Help - Continous Question

  1. #1
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    Continous Question

    I need help on this problem:



    Also please explain how to do it. I started out by graphing the first function but not sure how to do the others/what its asking for.
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  2. #2
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    This looks like an assignment of some sort so I won't give direct answers.

    By definition of conitnuity, limit x-> a f(x) = f(a)

    The key here is to find a,b such that the limit from the left of 2 and the limit from the right of 2 correseponds to f(2).
    Basically you'll need to find this:

     \displaystyle \lim_{x\rightarrow 2} \frac{x^2-4}{x-2}
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  3. #3
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    Alright, im still confused...

    the limit of that is 4. I dont get how that relates with a and b
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  4. #4
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    \displaytype\lim_{x\to a} f(x) exists if and only if \displaytype\lim_{x\to a^+} f(x) and \displaytype\lim_{x\to a^-} f(x) both exist and are the same.

    Yes, \displaytype\lim_{x\to 2^-} f(x)= \displaytype\lim_{x\to 2^-}\frac{x^2- 4}{x- 2}= 4. That is " x\to 2^-, the limit from below, because that fraction form is true only for x< 2. In order that the limit at 2 (from both directions) exist, we must get the same limit "from above": \displaytype\lim_{x\to 2^+} f(x)= \displaytype\lim_{x\to 2^+} ax^2- bx+ 3= 4x- 2b+ 3 must equal 4. That gives you one equation.

    Do the same with \displaytype\lim_{x\to 3^-} ax^2- bx+ 3 and \displaytype\lim_{x\to 3^+} 3x- a+ b to get a second equation.
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