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Need some help with lim

  1. Mar 30, 2005 #1
    Hey every one!

    I have been doing some lim questions and I came across this one.

    Evaluate: lim x-->3 (2x^2-x-15/3x^2-13x+12)

    I have Factorised It and got

    lim x-->3 ( (x-8) (x+7) / (-3x+4) (-x+3) )

    How do I go about the rest?Have I used the right stuff?

    Sorru for the mess. I don't realy know how to use the latex :redface:
     
  2. jcsd
  3. Mar 30, 2005 #2

    matt grime

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    You've not factored the top properly. If that were the correct factorization then the limit wouldn't exist. However, 2x^2 - x -15 evaluates to zero at x=3, so x-3 must be one of the factors., which will cancel with the factor or x-3 on the bottom.
     
  4. Mar 30, 2005 #3
    thanx!


    I see my problem now :)
     
  5. Mar 30, 2005 #4
    After Factorising I would just substitude the x-->3 into the equation wouldnt I?

    Or do I have to use the f(x+h)-f(x)/h? (not too sure about formula too:( I'm feeling kinda sleeeeepppy)
     
  6. Mar 30, 2005 #5

    dextercioby

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    Nope,u'd simplify through the common monom (x-3) and then take the limit...

    Daniel.
     
  7. Mar 30, 2005 #6
    I think I need more explenation.

    After canceling (x-3) from top and bottom do I just change the value of x=3? That is what I currently have done! And got 6+5/9-4=11/5=2 1/5
     
  8. Mar 30, 2005 #7

    dextercioby

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    That's right.The limit in that case means plugging the value "x=3" in the remaining fraction.

    Daniel.
     
  9. Mar 30, 2005 #8
    Thanx.

    Sorry to bother you, But I really apriciate the help.
     
  10. Mar 30, 2005 #9
    Factorising 2X^2-X-15 gives
    (2X+5)(X-3)

    Therefore:
    X=3 or
    X=-5/2
     
  11. Mar 30, 2005 #10

    dextercioby

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    We don't need the second value (-5/2).Jus that "3" was important,because it got simplified with the one in the denominator.

    Daniel.
     
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