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Clarence

  • one year ago

What is this question asking me to do? Assuming that the sequence {an} defined below is convergent, find its limit. a1=1,an+1=12−1an,n≥1.

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  1. ganeshie8
    • one year ago
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    you need to find \(\lim\limits_{n\to\infty}~a_n\)

  2. clarence
    • one year ago
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    \[a _{1}=1,a _{n+1}=12-\frac{ 1 }{ a _{n} }, n \ge1\]

  3. clarence
    • one year ago
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    How would I go abouts doing that? @ganeshie8

  4. ganeshie8
    • one year ago
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    use below \(\lim\limits_{n\to\infty}~a_{n+1} = \lim\limits_{n\to\infty}~a_n\)

  5. ganeshie8
    • one year ago
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    let \(\lim\limits_{n\to\infty}~a_n =\lim\limits_{n\to\infty}~a_{n+1} =x\)

  6. ganeshie8
    • one year ago
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    so you want to solve \(x\)

  7. ganeshie8
    • one year ago
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    we have \(a _{n+1}=12-\frac{ 1 }{ a _{n} }\) take limit \(\lim\limits_{n\to\infty} a _{n+1}=\lim\limits_{n\to\infty} 12-\frac{ 1 }{ a _{n} }\\~\\x = 12-\frac{1}{x}\) you can solve \(x\)

  8. clarence
    • one year ago
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    \[x=6\pm \sqrt{35}\] But how did you manage to get that from the question?

  9. ganeshie8
    • one year ago
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    get what ?

  10. clarence
    • one year ago
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    Get what to do :p

  11. ganeshie8
    • one year ago
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    ahh i have already told you :) you need to be a lil more specific haha

  12. clarence
    • one year ago
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    How did you know that you had to equate the two limits together to figure out the question and stuff? :P

  13. ganeshie8
    • one year ago
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    okay honestly i didn't invent that trick... i don't think its some thing that occurs easily to anyone... it needs to be shown by others...

  14. clarence
    • one year ago
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    Ahh, fair enough then, well thanks for the assistance as always! :)

  15. ganeshie8
    • one year ago
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    np :)

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