1. ## 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)?

2. Originally Posted by Danneedshelp
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 $\displaystyle \{\sqrt{x_n}\}\to \sqrt{x}$. $\displaystyle \square$

3. Scuse me when $\displaystyle 0 < x < 1$ we have that $\displaystyle \sqrt x > x$.

4. Originally Posted by Matt Westwood
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.