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Thread: Finding the limit

  1. #1
    Junior Member
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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 $\displaystyle \sqrt{x}>x$ when $\displaystyle x<1$, and $\displaystyle \sqrt{x}<x$ when $\displaystyle x>1$.

    Now:

    $\displaystyle
    \sqrt{n+1} + \sqrt{n-1} - 2\sqrt{n}=$$\displaystyle \sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]
    $

    Now as $\displaystyle 1+1/n>1$ when $\displaystyle n>1$:

    $\displaystyle
    1<\sqrt{1+1/n}<1+1/n
    $

    Similarly as $\displaystyle 1-1/n<1$ when $\displaystyle n>1$:

    $\displaystyle
    1-1/n<\sqrt{1-1/n}<1
    $

    So:

    $\displaystyle
    \sqrt{n}[1+1-1/n-2]<$$\displaystyle \sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]<\sqrt{n}[1+1/n+1-2]
    $

    $\displaystyle
    -1/\sqrt{n}<$$\displaystyle \sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]<1/\sqrt{n}
    $

    Hence:

    $\displaystyle
    \lim_{n \to \infty}}\ \sqrt{n}\ [\sqrt{1+1/n}+\sqrt{1-1/n}-2]=0
    $

    so:

    $\displaystyle
    \lim_{n \to \infty}}\ \sqrt{n+1} + \sqrt{n-1} - 2\sqrt{n}=0
    $

    RonL
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