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Thread: Sequence help

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

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

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

    Do I just note that $\displaystyle \sqrt{(x_{n})}$ is less than $\displaystyle (x_{n})$ 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 $\displaystyle x_{n}\geq\\0$ for all $\displaystyle n\in{N}$.

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

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

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

    Thus, for sufficiently large $\displaystyle n$, $\displaystyle |\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}}$$\displaystyle = \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 $\displaystyle \{\sqrt{x_n}\}\to \sqrt{x}$. $\displaystyle \square$
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  3. #3
    MHF Contributor Matt Westwood's Avatar
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    Scuse me when $\displaystyle 0 < x < 1$ we have that $\displaystyle \sqrt x > x$.
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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 $\displaystyle 0 < x < 1$ we have that $\displaystyle \sqrt x > x$.
    Right. Slipped my mind. [Math]\exists~N[/tex] such that $\displaystyle n>N \implies |x_n|<\epsilon^2$ and I think the rest of a) should follow.
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