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Math Help - This equation doesn't have a limit, does it?

  1. #1
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    This equation doesn't have a limit, does it?

    I don't know how to format, sorry...

    "Find the limit, if it exists"

    Limit (as x approaches 2): (x^2 - x +6)/(x-2)

    I used the quadratic formula to try to find values for the numerator, but one ends up with 1 plus or minus the square root of -23 divided by two. Obviously, negative numbers don't have real square roots, so is it safe to say there is no limit to this equation? Thanks...
    Last edited by satx; February 24th 2010 at 03:38 PM.
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  2. #2
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    Quote Originally Posted by satx View Post
    I don't know how to format, sorry...

    "Find the limit, if it exists"

    Limit (as x approaches 2): (x^2 - x +6)/(x-2)

    I used the quadratic formula to try to find values for the numerator, but one ends up with 1 plus or minus the square root of -23 divided by two. Obviously, negative numbers don't have real square roots, so is it safe to say there is no limit to this equation? Thanks...
    Yes satx,

    here is a sketch of it
    Attached Thumbnails Attached Thumbnails This equation doesn't have a limit, does it?-lim-x-2.jpg  
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    Quote Originally Posted by satx View Post
    I don't know how to format, sorry...
    Why not learn to post in symbols? You can use LaTeX tags.
    [tex]\lim _{x \to 2} \frac{{x^2 - x + 6}}{{x - 2}}[/tex] gives \lim _{x \to 2} \frac{{x^2  - x + 6}}{{x - 2}}.
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  4. #4
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    Quote Originally Posted by Archie Meade View Post
    Yes satx,

    here is a sketch of it
    So because 2+ and 2- are converging on opposite infinities, there's no limit? If they were to converge on the same infinity, there would be a limit, right?
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  5. #5
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    Not quite, satx,

    if they were converging on the same infinity,
    then both branches would be shooting off in the same direction,
    once again never meeting.

    In a situation like this, we check to see if the denominator is
    a factor of the numerator.
    If it was, we could say \frac{x-2}{x-2}=1, for all x except 2.

    Then that situation is described as having a "hole" in the graph at x=2.
    If the numerator is a quadratic and the denominator linear,
    the graph would be indistinguishable from a straight line.
    Written as a fraction of course, we'd have to exclude 2 from the domain
    if the denominator contains (x-2).

    In that case the limit may be evaluated as the graph approaches f(2)
    from both sides, since if x is not 2, then

    \frac{x-2}{x-2}=1

    hence

    \frac{x^2-5x+6}{x-2}=\frac{(x-2)(x-3)}{x-2}

    has a "hole" at x=2

    though the graph is indistinguishable from x-3.

    Hence f(2) would be -1 if the graph was x-3,
    therefore -1 is the limit as x approaches 2 for \frac{x^2-5x+6}{x-2}

    We cannot "see" the hole, of course.

    For \frac{x^2-x+6}{x-2}

    the denominator is not a factor of the numerator.

    Another way to express the function is f(x)=(x+1)+\frac{8}{x-2}

    If x is 2, then \frac{8}{x-2} again cannot be evaluated

    since \frac{8}{x-2} approaches infinity as x approaches 2.
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