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Math Help - Sequence help

  1. #1
    Senior Member Danneedshelp's Avatar
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    Sequence help

    Q: Let x_{n}\geq\\0 for all n\in{N}.

    (a) If (x_{n})\rightarrow\\0, show that (\sqrt{x_{n}})\rightarrow\\0.
    (b) If (x_{n})\rightarrow\\x, show that (\sqrt{x_{n}})\rightarrow\\\sqrt{x}.

    Do I just note that \sqrt{(x_{n})} is less than (x_{n}) so it must also be less than epsilon for part (a)?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    Q: Let x_{n}\geq\\0 for all n\in{N}.

    (a) If (x_{n})\rightarrow\\0, show that (\sqrt{x_{n}})\rightarrow\\0.
    (b) If (x_{n})\rightarrow\\x, show that (\sqrt{x_{n}})\rightarrow\\\sqrt{x}.

    Do I just note that \sqrt{(x_{n})} is less than (x_{n}) so it must also be less than epsilon for part (a)?
    a) Yes you do. Since 0\leq \sqrt{x_n} \leq x_n, \lim 0 \leq \lim\sqrt{x_n}\leq \lim x_n \implies 0\leq\lim\sqrt{x_n}\leq 0 \implies \lim\sqrt{x_n}=0. (This is the Squeeze Theorem.)

    b.) You want to prove that \forall~\epsilon>0, \exists~N such that \forall~ n>N, |\sqrt{x_n}-\sqrt{x}|<\epsilon. Since \{x_n\}\to x, you know that \forall~ n>N, |x_n-x|<\epsilon\sqrt{x}. (Recall that \sqrt{x} is a constant.)

    Thus, for sufficiently large n, |\sqrt{x_n}-\sqrt{x}| = |\sqrt{x_n}-\sqrt{x}|\cdot\frac{\sqrt{x}+\sqrt{x_n}}{\sqrt{x}+  \sqrt{x_n}} = \frac{|x_n-x|}{\sqrt{x}+\sqrt{x_n}}  = \frac{1}{\sqrt{x}}\cdot\frac{|x_n-x|}{1+\sqrt{x_n/x}}<\frac{1}{\sqrt{x}}\cdot\frac{\epsilon\sqrt{x}}  {1+\sqrt{x_n/x}} = \frac{\epsilon}{1+\sqrt{x_n/x}}<\epsilon

    So therefore \{\sqrt{x_n}\}\to \sqrt{x}. \square
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  3. #3
    Super Member Matt Westwood's Avatar
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    Scuse me when 0 < x < 1 we have that \sqrt x > x.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Matt Westwood View Post
    Scuse me when 0 < x < 1 we have that \sqrt x > x.
    Right. Slipped my mind. [Math]\exists~N[/tex] such that n>N \implies |x_n|<\epsilon^2 and I think the rest of a) should follow.
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