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.

Question

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.

ganeshie8 · Answer

you need to find $\lim\limits_{n\to\infty}~a_n$

clarence · Answer

$$a _{1}=1,a _{n+1}=12-\frac{ 1 }{ a _{n} }, n \ge1$$

clarence · Answer

How would I go abouts doing that? @ganeshie8

ganeshie8 · Answer

use below
 $\lim\limits_{n\to\infty}~a_{n+1} = \lim\limits_{n\to\infty}~a_n$

ganeshie8 · Answer

let $\lim\limits_{n\to\infty}~a_n =\lim\limits_{n\to\infty}~a_{n+1} =x$

ganeshie8 · Answer

so you want to solve $x$

ganeshie8 · Answer

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 $x$

clarence · Answer

$$x=6\pm \sqrt{35}$$
But how did you manage to get that from the question?

ganeshie8 · Answer

get what ?

clarence · Answer

Get what to do :p

ganeshie8 · Answer

ahh i have already told you :)
you need to be a lil more specific haha

clarence · Answer

How did you know that you had to equate the two limits together to figure out the question and stuff? :P

ganeshie8 · Answer

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...

clarence · Answer

Ahh, fair enough then, well thanks for the assistance as always! :)

ganeshie8 · Answer

np :)