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Math Help - finding the limit

  1. #1
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    finding the limit

    hey, i have a question that i cannot figure out.

    Given that lim(n->infinity)=an exists when a1=1 and an+1=sqrt[1+(6/an)] for n>1, find the limit.

    how do i go about solving this?

    cheers!
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  2. #2
    Senior Member pankaj's Avatar
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    a_{n+1}=\sqrt{1+\frac{6}{a_n}}

    \lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}\sqrt{1+\frac{6}{a_n}}


    Let \lim_{n\to \infty}a_{n}=a,then we also have \lim_{n\to \infty}a_{n+1}=a

    Therefore, a=\sqrt{1+\frac{6}{a}}

    a^2=1+\frac{6}{a}

    a^2-a-6=0

     <br />
a=3<br />
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  3. #3
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    thank you, i was getting different answers every time i did that question.

    i have also stumbled across another ones i am having problems with.

    lim(n->infinity) (1+1/2n)^n-1

    lim(n->infinity) ln(n+cos(n))/ln(n^2)

    any help would be appreciated
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  4. #4
    Senior Member pankaj's Avatar
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     <br />
\lim_{n\to \infty}((1+\frac{1}{2n})^{2n})^{\frac{n-1}{2n}}<br />

    =  <br />
\lim_{n\to \infty}((1+\frac{1}{2n})^{2n})^{\frac{1}{2}-\frac{1}{2n}}<br />

    = \sqrt{e}

    Apply L'Hospital for the second limit
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