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Limits Questions

  1. Mar 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Evaluate: lim x(squared) - 2x - 8
    x-->4 _________________
    x (squareroot of x) -8
    2. Relevant equations



    3. The attempt at a solution

    subbing 4 in as x resulted in an indterminant.
    Ive tried using the conjugate but I get stuck when I can not simplify or cancel.
    The main problem I'm having is trying to get rid of the squareroot.
     
  2. jcsd
  3. Mar 20, 2007 #2

    Dick

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    What 'conjugate' did you multiply by and what did you get? Show more of your work.
     
  4. Mar 20, 2007 #3
    mulitplied both num/den by x(squared) + 2x + 8

    which resulted x(to the power of 4) - 4x(squared) - 64 on top
    but x (squareroot of x) -8 * x(squared) + 2x + 8

    can the denominator be simplified more?
     
  5. Mar 20, 2007 #4

    Dick

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    Try multiplying top and bottom by x*sqrt(x)+8.
     
  6. Mar 20, 2007 #5
    if i got you right, the limit you wrote goes like this

    lim (x^2-2x-8)/(x*x^1/2 -8), x-->4 right??
    if it is like this than you have to multibly both numerator and denominator by

    x*x^1/2+8, what do you get? try this first, than you will get further instructions.

    after that try to factorize x^2-2x-8, find x1,x2, what do you get, than on the denominator you get x^3-64, try to factorize this too, and the problem is solved!!

    i hope this helps
     
    Last edited: Mar 20, 2007
  7. Mar 20, 2007 #6
    ok i got x(cubed)-2x-64/x(squared)*(x)-64
     
  8. Mar 20, 2007 #7

    Dick

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    Not really right. You should still have a sqrt(x) in the numerator. You don't have to multiply the numerator out. The denominator is ok though, factor it.
     
  9. Mar 20, 2007 #8
    x(cubed)-2x(sqrt(x))-64

    is the denominator correct?
     
  10. Mar 20, 2007 #9

    Dick

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    If that's supposed to be the numerator, no it's not. It's a binomial times a trinomial - if you do it right you'll get six terms. But I would still encourage you not to do it. Just leave it factored. The denominator IS x^3-64. Have you factored it yet?
     
  11. Mar 20, 2007 #10
    x(sqrd)-2x-8 * x((sqrt)x) + 8
    _________________________

    (x-4)((x(sqrd) + 4x + 16)
     
  12. Mar 20, 2007 #11

    Dick

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    Ok so far. I'd use more parentheses in the numerator. What do you think you should do now? Hint: the (x-4) term is the source of your problems.
     
  13. Mar 20, 2007 #12
    I assume theres somewhere that i can reduce but I dont see where.
     
  14. Mar 20, 2007 #13
    Is the answer 1/24?
     
  15. Mar 20, 2007 #14

    Dick

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    I doubt it. There's a hidden factor of (x-4) in the numerator. Where could it be?
     
  16. Mar 20, 2007 #15
    (x-4)(x+2){x(sqrtofx)+8}
    _____________________
    (x-4){(xsqrd)+4x+16}

    the x-4's cancel and then I can sub x = 4?

    I got 1/8
     
  17. Mar 20, 2007 #16

    Dick

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    Good work! But sub again. Carefully this time.
     
  18. Mar 20, 2007 #17
    haha whoops 96/48 = 2

    Thanks for the help.. much appreciated
     
  19. Mar 21, 2007 #18
    yes after you cancel out the (x-4) you can sub the x for 4, but i still think you got the wrong answer. It should be 2.
     
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