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Math Help - limit

  1. #1
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    limit

    trying to solve lim x->1 x^3 - 2x +1 / x-1
    I see this gives 0/0. Now my problem starts, i don't know what to do when i have x^3..
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  2. #2
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    Re: limit

    hi Olden

    for this problem you will need l'Hospital's rule.do you know it?
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  3. #3
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    Re: limit

    I found it in my book but weird since we are not at that chapter yet. I'm at chapter 2.2 and l'Hospital's rule is chapter 5.8... :/
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  4. #4
    Member anonimnystefy's Avatar
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    Re: limit

    oh,i forgot.try dividing the numerator by the denominator.that should give an expression defined for x=1.
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  5. #5
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    Re: limit

    I thought when i did not have lim x->infinity that i had to change the top x^3 - 2x +1 into something else. Like if i had lim x->2 x^2-2x / x^2-3x+2 = 0/0 i would make lim x->2 x(x-2) /(x-2)(x-1) = x/x-1 = 2. And i don't know how to do this when i have x^3. Sorry if i explain badly
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  6. #6
    Member anonimnystefy's Avatar
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    Re: limit

    notice that x^3 - 2x +1 = (x-1)(x^2+x-1)
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  7. #7
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    Re: limit

    Quote Originally Posted by Olden View Post
    I thought when i did not have lim x->infinity that i had to change the top x^3 - 2x +1 into something else. Like if i had lim x->2 x^2-2x / x^2-3x+2 = 0/0 i would make lim x->2 x(x-2) /(x-2)(x-1) = x/x-1 = 2. And i don't know how to do this when i have x^3. Sorry if i explain badly
    x^3-2x+1=(x-1)(x^2+x-1) now divide out x-1
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  8. #8
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    Re: limit

    Note, by the way, that if P(a)= 0 for a polynomial P, then P must have a factor of (x-a). Any time you have a fraction of the form \frac{P(x)}{Q(x)} and both P(a)= 0 and Q(a)= 0, then you can always factor out (x- a) from each and cancel it.
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