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anonymous
 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.
anonymous
 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.

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you need to find \(\lim\limits_{n\to\infty}~a_n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a _{1}=1,a _{n+1}=12\frac{ 1 }{ a _{n} }, n \ge1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would I go abouts doing that? @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1use below \(\lim\limits_{n\to\infty}~a_{n+1} = \lim\limits_{n\to\infty}~a_n\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1let \(\lim\limits_{n\to\infty}~a_n =\lim\limits_{n\to\infty}~a_{n+1} =x\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so you want to solve \(x\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we 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 \(x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x=6\pm \sqrt{35}\] But how did you manage to get that from the question?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1ahh i have already told you :) you need to be a lil more specific haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How did you know that you had to equate the two limits together to figure out the question and stuff? :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1okay 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...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahh, fair enough then, well thanks for the assistance as always! :)
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