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**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!