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Math Help - limit of a function!

  1. #1
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    limit of a function!

    find a \delta such that for the given value of L and \varepsilon in the statement \midf(x)-L\mid\le\varepsilon whenever 0\le\midx-a\mid\le\delta
    given that f(x)=(x^2-9)/(x+3) a=3, L=-6 and \varepsilon=0.005

    pls am sorry i can't get the formatting of the question right, kindly help on solving this. thanks!
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  2. #2
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    Re: limit of a function!

    Quote Originally Posted by lawochekel View Post
    find a \delta such that for the given value of L and \varepsilon in the statement |f(x)-L|\le\varepsilon holds whenever 0\le|x-a|\le\delta
    given that f(x)=(x^2-9)/(x+3), a=3, L=-6 and \varepsilon=0.005.
    You probably mean a = -3 because f(x) is close to 0, not -6, in the neighborhood of x = 3. Note that f(x) = x - 3 everywhere except for x = -3. I recommend drawing a graph of f(x).

    This question belongs in the Pre-Calculus or Calculus forum.
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