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Math Help - Limit to infinity question

  1. #1
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    Limit to infinity question

    I'm really having trouble with this question...
    When I solve it I find the answer to just be zero but apparently the answer is two or -2.

    Find lim 1/(((x^2+x)^1/2)+x)
    x-> - infinity

    Could somebody please explain to me how to do this type of question. I understand what to do when there are variables on the top, but since there is only a one I'm very confused.
    Last edited by poipoipoi10; October 12th 2011 at 09:35 AM.
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    Re: Limit to infinity question

    Quote Originally Posted by poipoipoi10 View Post
    When I solve it I find the answer to just be infinity but apparently the answer is two or -2.
    Find lim 1/(((x^2+x)^1/2)+x)
    x-> infinity
    If the question is \lim _{x \to \infty } \frac{1}{{\sqrt {x^2  + x}  + x}} then the limit is 0.
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    Re: Limit to infinity question

    Oh man, sorry I meant to type negative infinity. But wouldnt that still make it 0?
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    Re: Limit to infinity question

    Quote Originally Posted by poipoipoi10 View Post
    Oh man, sorry I meant to type negative infinity. But wouldnt that still make it 0?
    No, that makes a real difference.
    In that case because if x<0 then \frac{1}{{\sqrt {x^2  + x}  + x}} =  - \sqrt {1 + \frac{1}{x}}  - 1
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    Re: Limit to infinity question

    Would that not still make the numerator 0? Because How I learned to do it was divide by the highest power denominator, which in this case would be x^2
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    Re: Limit to infinity question

    Quote Originally Posted by poipoipoi10 View Post
    Would that not still make the numerator 0? Because How I learned to do it was divide by the highest power denominator, which in this case would be x^2
    The answer is -2.
    The working is just algebra. And using |x|=\sqrt{x^2} because x<0.
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    Re: Limit to infinity question

    I'm fairly new to this website so I dont know how to format all my algebra the way that you guys do =(. But this is my train of thought when I do the question:

    1. Divide by highest power x in denominator.

    (1/x^2)/-(((1/x^4)^1/2)((x^2+x)^1/2)+x/x^2)

    2. Continue algebra (divide into radical).

    (1/x^2)/-(((x^2/x^4 + x/x^4)^1/2) + x/x^2)

    3. Simplify

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

    4. Plug infinity's in

    0/-((0+0)^2)+0)

    So basically I end up with 0/0.

    I'm sorry that it's so hard to understand, I wish I could format it the way you do, but I can't figure it out.
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    Re: Limit to infinity question

    Quote Originally Posted by poipoipoi10 View Post
    1. Divide by highest power x in denominator.
    Forget about dividing. We don't use it until the last step.

    \begin{gathered}  \frac{1}{{\sqrt {x^2  + x}  + x}}\frac{{\sqrt {x^2  + x}  - x}}{{\sqrt {x^2  + x}  - x}} \hfill \\   = \frac{{\sqrt {x^2  + x}  - x}}{x} \hfill \\   = \frac{{\left| x \right|\sqrt {1 + \frac{1}{x}}  - x}}{x} ,~~x<0\hfill \\   =  - \sqrt {1 + \frac{1}{x}}  - 1 \hfill \\ \end{gathered}
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    Re: Limit to infinity question

    Ah, that cleared it up for me, thanks =)

    If you can still see this I have one quick question:

    How in your third step did you get the 1/x when you took out the x^2?
    Last edited by poipoipoi10; October 12th 2011 at 11:28 AM.
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    Re: Limit to infinity question

    Quote Originally Posted by poipoipoi10 View Post
    Ah, that cleared it up for me, thanks =)

    If you can still see this I have one quick question:

    How in your third step did you get the 1/x when you took out the x^2?
    First, since you are making \displaystyle x \to -\infty, you need to know that \displaystyle \sqrt{x^2} = -x if \displaystyle x < 0. So

    \displaystyle \begin{align*} \sqrt{x^2 + x} &= \sqrt{x^2\left(1 + \frac{1}{x}\right)} \\ &= \sqrt{x^2}\sqrt{1 + \frac{1}{x}} \\ &= -x\sqrt{1 + \frac{1}{x}} \end{align*}
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