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Math Help - Finding the limit

  1. #1
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    Finding the limit

    I am to find the limit as n approaches infinity of (sqrt(n+) + sqrt(n-1) - 2sqrt(n).

    I separated the last term and then grouped. I did this so I could multiply by a conjugate.

    When I simplified I ended up with this:

    Lim as n approaches infinity of [(sqrt(n+1) - sqrt(n))^2 - (sqrt(n-1) - sqrt(n))^2]/(sqrt(n+1) - sqrt(n-1))

    I don't think I did anything algebracially wrong. But any ways when I try to take the limit, I get a 0 in the denominator.

    Can you help?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Nichelle14
    I am to find the limit as n approaches infinity of (sqrt(n+) + sqrt(n-1) - 2sqrt(n).

    I separated the last term and then grouped. I did this so I could multiply by a conjugate.

    When I simplified I ended up with this:

    Lim as n approaches infinity of [(sqrt(n+1) - sqrt(n))^2 - (sqrt(n-1) - sqrt(n))^2]/(sqrt(n+1) - sqrt(n-1))

    I don't think I did anything algebracially wrong. But any ways when I try to take the limit, I get a 0 in the denominator.

    Can you help?
    What follows hindges on \sqrt{x}>x when x<1, and \sqrt{x}<x when x>1.

    Now:

    <br />
\sqrt{n+1} + \sqrt{n-1} - 2\sqrt{n}= \sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]<br />

    Now as 1+1/n>1 when n>1:

    <br />
1<\sqrt{1+1/n}<1+1/n<br />

    Similarly as 1-1/n<1 when n>1:

    <br />
1-1/n<\sqrt{1-1/n}<1<br />

    So:

    <br />
\sqrt{n}[1+1-1/n-2]< \sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]<\sqrt{n}[1+1/n+1-2]<br />

    <br />
-1/\sqrt{n}< \sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]<1/\sqrt{n}<br />

    Hence:

    <br />
\lim_{n \to \infty}}\ \sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]=0<br />

    so:

    <br />
\lim_{n \to \infty}}\ \sqrt{n+1} + \sqrt{n-1} - 2\sqrt{n}=0<br />

    RonL
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