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Math Help - lim f(x)

  1. #1
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    lim f(x)

    Hi, i was just wondering why f(2) is not defined in the expression below:

    f(x) = (x^2 - x - 2) / (x - 2)

    I factorised (x^2 - x - 2) to get (x - 2)(x + 1). I then cancelled the (x - 2) in the numerator and the denominator out. The final f(x) would be (x + 1).

    However, my calculus textbook says that the function will still be discontinuous at x = 2.

    May i ask the principle behind this? Hasn't (x - 2) already been removed from the denominator?
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  2. #2
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    Quote Originally Posted by Cplusplusdummy View Post
    Hi, i was just wondering why f(2) is not defined in the expression below:

    f(x) = (x^2 - x - 2) / (x - 2)

    I factorised (x^2 - x - 2) to get (x - 2)(x + 1). I then cancelled the (x - 2) in the numerator and the denominator out. The final f(x) would be (x + 1).

    However, my calculus textbook says that the function will still be discontinuous at x = 2.

    May i ask the principle behind this? Hasn't (x - 2) already been removed from the denominator?
    This is because you are assuming that,
    \frac{x^2 - x - 2}{x-2} = (x+1).

    This is not true. Because the LHS is not defined at x=2.
    Remember for two functions to be equal they need to have the same domain.

    The correct thing to do is to define,
    f(x) = x+1 for x\not = 2 i.e. leave it undefined.

    Then f(x) = \frac{x^2 - x  - 2}{x-2} would be correct.
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  3. #3
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    Thanks a lot for the explanation! I get it now!
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