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Math Help - Three similar limit problems

  1. #1
    Junior Member simplysparklers's Avatar
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    Three similar limit problems

    Hey guys!
    I have 3 similar problems, if someone could please show me how to do one of them that would be great!
    They all involve finding \lim_{n\rightarrow\infty} of a_{n}, when a_{n} isn't given.

    1.) a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3} (n\geq{1})
    Find \lim_{n\rightarrow\infty}a_{n}

    2.) a_{1}=2, a_{n+1}=\frac{2a_{n}+3}{a_{n}+2} (n\geq{1})
    Find \lim_{n\rightarrow\infty}a_{n}

    3.) Determine the limit of (a_{n}), where
    a_{1}=2,a_{n+1}=\sqrt{2+2a_{n}} (n\geq{1})

    If someone could show me how to do one of them I'd really appreciate it!!

    Thanks,
    Jo
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by simplysparklers View Post
    Hey guys!
    I have 3 similar problems, if someone could please show me how to do one of them that would be great!
    They all involve finding \lim_{n\rightarrow\infty} of a_{n}, when a_{n} isn't given.

    1.) a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3} (n\geq{1})
    Find \lim_{n\rightarrow\infty}a_{n}

    2.) a_{1}=2, a_{n+1}=\frac{2a_{n}+3}{a_{n}+2} (n\geq{1})
    Find \lim_{n\rightarrow\infty}a_{n}

    3.) Determine the limit of (a_{n}), where
    a_{1}=2,a_{n+1}=\sqrt{2+2a_{n}} (n\geq{1})

    If someone could show me how to do one of them I'd really appreciate it!!

    Thanks,
    Jo
    In every case, assume that L is the limit.

    Then you can notice that \lim_{n \to \infty} a_n=\lim_{n \to \infty} a_{n+1}=L

    For the second one, as an example :

    L=\frac{2L+3}{L+2}

    ---> L^2=3

    Then, see if a_n is negative or positive in order to chose the negative or the positive solution
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  3. #3
    Junior Member simplysparklers's Avatar
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    Thank you so much Moo!!You rock!!!
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  4. #4
    Eater of Worlds
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    #3, you could try this way.

    \lim_{n\rightarrow{\infty}}\sqrt{2+2a_{n}}

    L=\sqrt{2+2L}

    L^{2}-2L-2=0

    L=\sqrt{3}+1, \;\ L=1-\sqrt{3}

    Reject the negative one because the terms in the sequence are positive.

    \lim_{n\rightarrow{\infty}}a_{n}=1+\sqrt{3}
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  5. #5
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by simplysparklers View Post
    Hey guys!
    I have 3 similar problems, if someone could please show me how to do one of them that would be great!
    They all involve finding \lim_{n\rightarrow\infty} of a_{n}, when a_{n} isn't given.

    1.) a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3} (n\geq{1})
    Find \lim_{n\rightarrow\infty}a_{n}

    2.) a_{1}=2, a_{n+1}=\frac{2a_{n}+3}{a_{n}+2} (n\geq{1})
    Find \lim_{n\rightarrow\infty}a_{n}

    3.) Determine the limit of (a_{n}), where
    a_{1}=2,a_{n+1}=\sqrt{2+2a_{n}} (n\geq{1})

    If someone could show me how to do one of them I'd really appreciate it!!

    Thanks,
    Jo
    All extremely similar,

    For the first one we have already proved in earlier threads that a_n is monotonically increasing. So let the limit be L, then

    \lim_{n \to \infty} a_{n+1}=\lim_{n \to \infty} \frac{3a_{n}+1}{a_{n}+3} \Rightarrow L = \frac{3L+1}{L+3} \Rightarrow L = \pm1

    But the negative solution is not possible since a_n is monotonic increasing and a_1 = 0.

    So L = 1
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    For the second one, do a similar thing:

    First show by induction 2 \leq a_n < \sqrt 3
    Then show its monotonically increasing. Then finally since the limit exists, do the same ol' trick of letting \lim_{n \to \infty}a_n = \lim_{n \to \infty} = a_{n+1} = L again
    Then finally choose the positive solution because 2 \leq a_n < \sqrt 3


    EDIT: I am sooo slow
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  6. #6
    Junior Member simplysparklers's Avatar
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    Isomorphism got there before me Moo!!But I do get it now, it's actually fairly simple!
    Thank you guys so much for yer help!!Much love and hugs to ye all!!
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