another Convergence Q

**sfspitfire23** · March 1st 2010, 07:27 PM

Suppose $a_n>0$ for all $n$ and $a_n\rightarrow 3$ show that $\sqrt{a_n}\rightarrow \sqrt{3}$ .

Attempt-

Well, we know that $|a_n-3|<\epsilon$ and therefore $3-\epsilon<a_n<3+\epsilon$ and we know then that there is some $N$ such that this is true for all $n>N$ But then there will be a smaller N (because $\sqrt{3}<3$ such that $\sqrt{3}-\epsilon<\sqrt{a_n}<\sqrt{3}+\epsilon$ will be true for all n>N....

What have I shown here? am I close?

Thanks!

**Drexel28** · March 1st 2010, 07:31 PM

Originally Posted by sfspitfire23

Suppose $a_n>0$ for all $n$ and $a_n\leftarrow 3$ show that $\sqrt{a_n}\leftarrow \sqrt{3}$ .

Attempt-

Well, we know that $|a_n-3|<\epsilon$ and therefore $3-\epsilon<a_n<3+\epsilon$ and we know then that there is some $N$ such that this is true for all $n>N$ But then there will be a smaller N (because $\sqrt{3}<3$ such that $\sqrt{3}-\epsilon<\sqrt{a_n}<\sqrt{3}+\epsilon$ will be true for all n>N....

What have I shown here? am I close?

Thanks!

$f(x)=\sqrt{x}$ is continuous everywhere on $\mathbb{R}$ and so $a_n\to 3\implies f(a_n)=\sqrt{a_n}\to f(3)=\sqrt{3}$

**sfspitfire23** · March 1st 2010, 08:21 PM

Ah I see!

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

**Drexel28** · March 1st 2010, 08:28 PM

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.

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