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Math Help - Sequences: Finding limits

  1. #1
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    Sequences: Finding limits

    HI! The problem is: an= The square root of (n+8) over (36n+8). Find the limit as n goes to infinity of an. Thank you!!! I think it converges but I can't find the limit.
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    Quote Originally Posted by elocin View Post
    HI! The problem is: an= The square root of (n+8) over (36n+8). Find the limit as n goes to infinity of an. Thank you!!! I think it converges but I can't find the limit.
    Is this supposed to be
    \lim_{n \to \infty} \frac{\sqrt{n + 8}}{36n + 8}
    or
    \lim_{n \to \infty} \sqrt{\frac{n + 8}{36n + 8}}

    They both exist but are different numbers.

    -Dan
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    Hello, elocin!

    Evaluate: . \lim_{n\to\infty}\sqrt{\frac{n+8}{36n+8}}

    Under the radical, divide top and bottom by n.

    . . \lim_{n\to\infty}\sqrt{\frac{\dfrac{n}{n} + \dfrac{8}{n}}{\dfrac{36n}{n} + \dfrac{8}{n}}} \;\;=\;\;\lim_{n\to\infty}\sqrt{\frac{1 + \dfrac{8}{n}}{36 + \dfrac{8}{n}}}

    . . =\;\;\lim_{n\to\infty}\sqrt{\frac{1+0}{36 + 0}} \;\;=\;\;\sqrt{\frac{1}{36}} \;\;=\;\;\frac{1}{6}

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  4. #4
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    It is the second one! It's the square root over the entire fraction! Thanks!
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