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Find the limit algebraically lim ((x-1)/(x^2-1)) x ->

  1. Aug 19, 2004 #1
    Find the limit algebraically lim ((x-1)/(x^2-1)) x -->

    Find the limit algebreically:

    lim ((x-1)/(x^2-1))
    x --> 1

    Thanks
     
  2. jcsd
  3. Aug 19, 2004 #2
    Hello.

    To find your limit we cannot just set x equal to 1. This would make the expression indeterminate, i.e. 0/0. Hence we have to use L ' Hopital's Rule.

    lim f ' (x) / g ' (x) = 1 / 2x. Setting x = 1 we get 1/2 for the limit.
    x-->1

    If you want to solve it algebraically:

    lim (x-1)/((x^2 - 1) = lim (x-1)/(x-1)(x+1) = lim 1 / (x+1) = 1/2
    x-->1 x-->1 x-->1
     
    Last edited: Aug 19, 2004
  4. Aug 19, 2004 #3

    Hurkyl

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    That's not doing it algebraically. The questions wants you to do something algebraic with (x-1)/(x^2-1)
     
  5. Aug 19, 2004 #4
    Algebra is the real key here. Factor the denominator

    [tex]x^2-1=(x-1)(x+1)[/tex]

    Then we have

    [tex]\lim\limits_{x \to 1} \frac {x-1}{x^2-1} = \lim\limits_{x \to 1} \frac {x-1}{(x-1)(x+1)} = \lim\limits_{x \to 1} \frac {1}{x+1} = \frac {1}{2} [/tex]
     
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