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Limits question ((x+1)/(x-1)-3)/(x-2)

  1. Mar 15, 2007 #1
    1. The problem statement, all variables and given/known data
    F(x) = [x+1 / x-1] coordinates (2,3)


    2. Relevant equations
    First principle Mtang=(F(x)-F(a))/x-a
    second principle
    finding the derivatives at first glance

    3. The attempt at a solution
    Mtang=(F(x)-F(a))/x-a
    [x+1/x-1]-(3)/(x-2)
    Multiply by (x-1/x-1) to get
    x+1 - 3(x-1)
    ----------- <- over (fraction)
    (x-2)(x-1)
    <<
    x+1-3
    ---------- <- Over (fraction)
    (x-2)
    <<
    x-2
    -----
    x-2

    equals ONE!!
     
    Last edited: Mar 16, 2007
  2. jcsd
  3. Mar 15, 2007 #2
    Are you familiar with l'Hôpital's rule?
     
    Last edited: Mar 15, 2007
  4. Mar 16, 2007 #3
    No I am not
     
  5. Mar 16, 2007 #4

    Dick

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    Check your algebra going from the first 'over (fraction)' to the second one. You somehow made x-1 'disappear' through use of bad algebra. Simplify the numerator in the first expression.
     
  6. Mar 16, 2007 #5
    Thats okay i know what to do know, 3x divided by x is 2x
     
  7. Mar 16, 2007 #6

    Dick

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    3x MINUS x is 2x. 3x divided by x is 3. Are you sure you know what to do?
     
  8. Mar 16, 2007 #7
    well forget about it i got a tutor now
     
  9. Mar 16, 2007 #8

    Dick

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    Good luck!
     
  10. Mar 16, 2007 #9
    Meh, IMO, a tutor can only have a negative impact.
     
  11. Mar 17, 2007 #10

    Gib Z

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    How do we have a better impact than a tutor :S
     
  12. Mar 17, 2007 #11
    We don't but it doesn't exclude the fact that a tutor doesn't, IMO.
     
  13. Mar 17, 2007 #12
    Because we're doing it because we love mathematics and enjoy helping others. The tutor is doing it to make money, or because it's a required work-study job. :tongue2:
     
  14. Mar 17, 2007 #13

    Gib Z

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    Lol Thats a good point, the only reason we would even be helping is because we want to, and because of that we put in our best effort. But id rather be half-assed AND get paid! :P
     
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