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Help with Calc problem, limits

  1. Jun 13, 2005 #1
    This shouldn't be too hard but I'm stuck, brain is getting fried.

    lim [(1+2x-x^3)-(1+2(1)-(1)^3)]/(x-1)
    x->1

    have gotten this far and I'm stuck:

    lim (-x^3+2x-1)/(x-1)
    x->1

    Any help is appreciated, thx.

    hk

    ETA: fixed typo
     
    Last edited: Jun 13, 2005
  2. jcsd
  3. Jun 13, 2005 #2

    OlderDan

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    -x^3 + 2x - 1 = x - x^3 + x - 1 = x(1 - x^2) + x - 1

    continue
     
  4. Jun 13, 2005 #3
    working...

    hk
     
  5. Jun 13, 2005 #4
    I really really want to just cancel x-1 from top and bottom! But knowing I can't, do I need to multiply top and bottom by something to change the denominator?

    hk
     
  6. Jun 13, 2005 #5

    OlderDan

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    The numerator has what you need. I just gave you one step in the right direction. Look for another step.
     
  7. Jun 13, 2005 #6
    Well, I gave up and solved it with the [f(a+h)-f(a)]/h formula instead of
    [f(x)-f(a)]/(x-a). I was able to get the correct solution of -1 (according to the back of the book), but I'm still curious where I was going wrong with my algebra on the first try. Thanks for your help Dan, I appreciate it.

    hk
     
  8. Jun 14, 2005 #7

    dextercioby

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    [tex] -x^3 + 2x - 1 = x - x^3 + x - 1 = x(1 - x^2) + x - 1=x(1+x)(1-x)-(1-x)=(1-x)[x(1+x)-1] [/tex]

    And you can see that the (1-x) has been factored.Simplify with the denominator and then take the limit.

    Daniel.
     
  9. Jun 14, 2005 #8
    Doh! Difference of two squares, I tend to miss a lot of those when one of the squares is 1. Thanks Daniel.

    hk
     
  10. Jun 14, 2005 #9

    saltydog

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    I just wish to clear up something for the record:

    When taking the limit, we are considering values of x close to 1 but NOT equal to 1. Thus it is possible to divide by (x-1). Not trying to upstage anyone here. Anyway, by synthetic division:

    [tex]\mathop\lim\limits_{x\to 1}\frac{-(x^3-2x+1)}{x-1}=\mathop\lim\limits_{x\to 1}\frac{-(x^2+x-1)(x-1)}{x-1}=\mathop\lim\limits_{x\to 1}-(x^2+x-1)=-1[/tex]

    I know, I'm such a Johnny-come-lately.
     
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