1. ## 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!

2. Originally Posted by sfspitfire23
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 $\displaystyle a_n\to 3\implies f(a_n)=\sqrt{a_n}\to f(3)=\sqrt{3}$

3. Ah I see!

Just wondering, but is my attempt anywhere close to a correct proof trying to do it with the N-epsilon method?

4. Originally Posted by sfspitfire23
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.