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Math Help - Infinite limits question

  1. #1
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    Infinite limits question

    The question is
    Show that:
    \lim{x \rightarrow\infty}(\surd(x^2+x)-x)=1/2

    I have no idea how I should start. The only possible solution i can come up with is to input different values for x and show that as they get higher the limit gets closer to 1/2...

    Is there a better way to do it? Thanks!!
    Last edited by natashabu; September 30th 2008 at 02:43 PM. Reason: changed a word so the question would make more sense
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  2. #2
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    Hello,
    Quote Originally Posted by natashabu View Post
    The question is
    Show that:
    \lim{x \rightarrow\infty}(\surd(x^2+x)-x)=1/2

    I have no idea how I should start. The only possible solution i can come up with is to input different values for x and show that as they get higher the limit gets closer to 1/2...

    Is there a better way to do it? Thanks!!
    Yes, there is a better way

    \lim_{x \to \infty} \sqrt{x^2+x}-x=\lim_{x \to \infty} ~ x \left(\sqrt{1+\frac 1x}-1\right)

    Multiply top and bottom by the conjugate \sqrt{1+\frac 1x}+1 :

    \lim_{x \to \infty} ~ x \left(\sqrt{1+\frac 1x}-1\right)=\lim_{x \to \infty} ~ x \cdot \frac{(\sqrt{1+\tfrac 1x}-1)(\sqrt{1+\tfrac 1x}+1)}{\sqrt{1+\tfrac 1x}+1}

    =\lim_{x \to \infty} ~ x \cdot \frac{\tfrac 1x}{\sqrt{1+\tfrac 1x}+1}=\lim_{x \to \infty} ~ \frac{1}{\sqrt{1+\tfrac 1x}+1}

    Note that \lim_{x \to \infty} ~ \frac 1x=0 and find the value of the limit.
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  3. #3
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    I get it! Thanks a lot! I wouldn't have been able to figure that out.
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