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Math Help - another Convergence Q

  1. #1
    Senior Member sfspitfire23's Avatar
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    another Convergence Q

    Suppose a_n>0 for all n and a_n\rightarrow 3 show that \sqrt{a_n}\rightarrow \sqrt{3}.

    Attempt-

    Well, we know that |a_n-3|<\epsilon and therefore 3-\epsilon<a_n<3+\epsilon and we know then that there is some N such that this is true for all n>N But then there will be a smaller N (because \sqrt{3}<3 such that \sqrt{3}-\epsilon<\sqrt{a_n}<\sqrt{3}+\epsilon will be true for all n>N....

    What have I shown here? am I close?


    Thanks!
    Last edited by sfspitfire23; March 1st 2010 at 08:18 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Suppose a_n>0 for all n and a_n\leftarrow 3 show that \sqrt{a_n}\leftarrow \sqrt{3}.

    Attempt-

    Well, we know that |a_n-3|<\epsilon and therefore 3-\epsilon<a_n<3+\epsilon and we know then that there is some N such that this is true for all n>N But then there will be a smaller N (because \sqrt{3}<3 such that \sqrt{3}-\epsilon<\sqrt{a_n}<\sqrt{3}+\epsilon will be true for all n>N....

    What have I shown here? am I close?


    Thanks!
    f(x)=\sqrt{x} is continuous everywhere on \mathbb{R} and so a_n\to 3\implies f(a_n)=\sqrt{a_n}\to f(3)=\sqrt{3}
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Ah I see!

    Just wondering, but is my attempt anywhere close to a correct proof trying to do it with the N-epsilon method?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Ah I see!

    Just wondering, but is my attempt anywhere close to a correct proof trying to do it with the N-epsilon method?
    Hmm. I think you have a little risky business going on there. I'm not too sure about the step where the answer magically appears.
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