1. ## 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)?

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

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

4. Originally Posted by Matt Westwood
Scuse me when $0 < x < 1$ we have that $\sqrt x > x$.
Right. Slipped my mind. $$\exists~N$$ such that $n>N \implies |x_n|<\epsilon^2$ and I think the rest of a) should follow.