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Thread: another Convergence Q

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

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

    Attempt-

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

    What have I shown here? am I close?


    Thanks!
    Last edited by sfspitfire23; Mar 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 $\displaystyle a_n>0$ for all $\displaystyle n$ and $\displaystyle a_n\leftarrow 3$ show that $\displaystyle \sqrt{a_n}\leftarrow \sqrt{3}$.

    Attempt-

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

    What have I shown here? am I close?


    Thanks!
    $\displaystyle f(x)=\sqrt{x}$ is continuous everywhere on $\displaystyle \mathbb{R}$ and so $\displaystyle a_n\to 3\implies f(a_n)=\sqrt{a_n}\to f(3)=\sqrt{3}$
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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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