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C2 sequences: the sequence converges to L; find L watch

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    If xn = 2 - (2/3) ^ n, find x1 , x2 , x3 and x4.
    The sequence converges to L. Find L.

    Done the first part but stuck on the method for the second part; the book says the answer is 2, and i think you have to find the limit as n approaches infinity by putting L= f(L).

    My working so far shows (as I have replaced any n's with L)
    L= 2- (2/3) ^ L
    But then i get stuck, i have a feeling you're meant to replace the whole of (2/3)^n with just L, but I don't understand why?
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    (Original post by Bluebell2525)
    If xn = 2 - (2/3) ^ n, find x1 , x2 , x3 and x4.
    The sequence converges to L. Find L.

    Done the first part but stuck on the method for the second part; the book says the answer is 2, and i think you have to find the limit as n approaches infinity by putting L= f(L).

    My working so far shows (as I have replaced any n's with L)
    L= 2- (2/3) ^ L
    But then i get stuck, i have a feeling you're meant to replace the whole of (2/3)^n with just L, but I don't understand why?
    Hehe no, the L=f(L) thing only applies when you have x(n+1)=f(x(n)).

    With this one you simply need to note that (2/3)^n gets smaller and smaller as n gets bigger and bigger, ie. (2/3)^n -> 0 as n->infinity, ie. x(n)->2-0=0, as n->infinity
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    (Original post by Haddock3)
    Hehe no, the L=f(L) thing only applies when you have x(n+1)=f(x(n)).

    With this one you simply need to note that (2/3)^n gets smaller and smaller as n gets bigger and bigger, ie. (2/3)^n -> 0 as n->infinity, ie. x(n)->2-0=0, as n->infinity
    Thankyou!
 
 
 
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