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Math Help - Limit convergence?

  1. #1
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    Limit convergence?

    I need to show that

    \lim_{n}(\sqrt{n+1}-\sqrt{n})=0

    and

    \lim_{n}(\sqrt{n^2-n}-n)=-\frac{1}{2}

    Help would be greatly appreciated. Thanks.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by miatchguy View Post
    I need to show that

    \lim_{n}(\sqrt{n+1}-\sqrt{n})=0
    The standard trick here is to multiply with \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}:

    \lim_{n}(\sqrt{n+1}-\sqrt{n})=\lim_{n\to\infty}\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\lim_{n\to\infty}\frac{1}{  \sqrt{n+1}+\sqrt{n}}=0


    and

    \lim_{n}(\sqrt{n^2-n}-n)=-\frac{1}{2}
    Well, let's try the same basic trick once more

    \lim_{n\to\infty}(\sqrt{n^2-n}-n)=\lim_{n\to\infty}\frac{n^2-n-n^2}{\sqrt{n^2-n}+n}=\lim_{n\to\infty}\frac{-n}{\sqrt{n^2-n}+n}
    =\lim_{n\to\infty}\frac{-1}{\sqrt{1-\frac{1}{n}}+1}=-\frac{1}{2}
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  3. #3
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    Talking

    Quote Originally Posted by Failure View Post
    The standard trick here is to multiply with \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}:

    \lim_{n}(\sqrt{n+1}-\sqrt{n})=\lim_{n\to\infty}\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\lim_{n\to\infty}\frac{1}{  \sqrt{n+1}+\sqrt{n}}=0



    Well, let's try the same basic trick once more

    \lim_{n\to\infty}(\sqrt{n^2-n}-n)=\lim_{n\to\infty}\frac{n^2-n-n^2}{\sqrt{n^2-n}+n}=\lim_{n\to\infty}\frac{-n}{\sqrt{n^2-n}+n}
    =\lim_{n\to\infty}\frac{-1}{\sqrt{1-\frac{1}{n}}+1}=-\frac{1}{2}
    Thanks alot for the help there. My professor was showing me a method based on graphing, where I'd use the quotient the slopes to bound the difference of the values. It got to be very tricky. I don't know why I didn't see this. Thanks again.
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