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Limit of a function

  1. Sep 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Hey guys
    im trying to figure out what lim x->1 ((x^3)-3x+2) / x-1 is.
    I get -1/0 and then when i try factor it i cant get it right..

    2. Relevant equations

    ?

    3. The attempt at a solution

    lim x->1 x^3-3x+2 / x-1 = lim x->1 1^3-3*1+2 / 1-1 = -1/0
     
  2. jcsd
  3. Sep 17, 2012 #2

    Mark44

    Staff: Mentor

    x3 -3x + 2 is divisible by x - 1. Do you know how to do polynomial long division. If not, there's an article on this technique on wikipedia.
    When you write expressions like the above, put parentheses around the entire numerator and the entire denominator, like so:
    (x3 - 3x + 2)/(x - 1)

    Without them, what you wrote is x3 - 3x + (2/x) - 1.
     
  4. Sep 18, 2012 #3
    I do know abit polynomial division, but i cant see my teacher using this method.
    I did mean (x3 - 3x + 2)/(x - 1).

    Im used to doing it like this
    lim x-> 3 (x-3) / (x3 -9) = lim -> 3 (x-3) / ((x-3)(x+3)) = 1/(3+3) = 1/6
    But when i try do this i cant get it right, and i cant cancel the things i dont want 2 have.
     
  5. Sep 18, 2012 #4
    Okay, here's an alternative approach: do you remember that you can write a polynomial into a product form like this
    [tex] p(x) = x^3 + a_2 x^2 + a_1 x + a_0 = (x-r_1)(x-r_2)(x-r_3) [/tex]
    where r are the roots of the polynomial?

    Now you already noticed that one of the roots is +1. Just write out the right hand side and make all of the coefficients equal. You'll get some equations for the remaining roots, and they should be easy to solve.
     
  6. Sep 18, 2012 #5

    Mark44

    Staff: Mentor

    You're the one working this problem, not your teacher. Besides, you are probably underestimating your teacher's abilities. If you know about polynomial long division, why not use it? (x3 - 3x + 2) factors nicely.
     
  7. Sep 18, 2012 #6
    Use x-1 as a factor in the numerator and then cancel it out with the x-1 in the denominator. BTW if you want to factor with x-1 you'll have to do something called synthetic division.
     
  8. Sep 18, 2012 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
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    You do realize that ##1^3-3\times 1 + 2## is equal to 0, not -1, right?
     
  9. Sep 19, 2012 #8
    HAHAHAHA :D no clue what ive been doing. thx!
     
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