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Math Help - Limits at a point

  1. #1
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    Limits at a point

    Hey guys,

    I'm trying to understand how to determine the left and right side of this limit:

    limit as x -> 2

    |x-2|/(x-2)

    Do I have to graph it, or is there an algebraic way to do this? Cheers.
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  2. #2
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    Quote Originally Posted by Glitch View Post
    Hey guys,

    I'm trying to understand how to determine the left and right side of this limit:

    limit as x -> 2

    |x-2|/(x-2)

    Do I have to graph it, or is there an algebraic way to do this? Cheers.
    Recall that

    |x - 2| = \begin{cases}\phantom{-(}x-2\phantom{)}\textrm{ if }x-2\geq 0\\-(x - 2)\textrm{ if }x-2<0\end{cases}

    |x - 2| = \begin{cases}x-2\textrm{ if  }x\geq 2\\2-x\textrm{ if }x<2\end{cases}


    When you are are approaching x = 2 from the left, x < 2.

    So \lim_{x \to 2^{-}}\frac{|x-2|}{x-2} = \lim_{x\to 2^{-}}\frac{2-x}{x-2}

    = \lim_{x\to  2^{-}}\frac{-(x-2)}{x-2}

    = \lim_{x\to 2^{-}}(-1)

     = -1.



    When you are approaching x=2 from the right, x>2.

    So \lim_{x \to 2^{+}}\frac{|x-2|}{x-2} = \lim_{x\to 2^{+}}\frac{x-2}{x-2}

    =\lim_{x \to 2^{+}}(1)

     = 1.



    Clearly, since the left hand limit is not the same as the right hand limit, the limit does not exist.
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  3. #3
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    Ahh, I see. I was thinking that that'd be 0/0 (when you sub in 2) which is undefined. Thanks.
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