$1.$ Let $\{a_n\}$ be a sequence of positive terms such that $\lim_{n \to \infty} a_n = L$ where $L > 0$. Prove that $\lim_{n \to \infty} \sqrt{a_n} = \sqrt{L}$.
Proof of $1:$
\begin{align*}
\forall \epsilon > 0\: \exists N >0\:\: s.t\:\:n>N
&\implies |a_n-L|<\epsilon\\
&\implies|(\sqrt{a_n}-\sqrt{L})(\sqrt{a_n}+\sqrt{L})|<\epsilon\\
&\implies |\sqrt{a_n}-\sqrt{L}||\sqrt{a_n}+\sqrt{L}|<\epsilon\\
&\implies|\sqrt{a_n}-\sqrt{L}|<\epsilon\\
&\therefore \lim_{n \to \infty} \sqrt{a_n} = \sqrt{L}
\end{align*}
Is this proof correct ?
Best Answer
Start with $N$ such that $|a_n-L| <\epsilon \sqrt L$ for $n >N$. Then you get $|\sqrt a_n -{\sqrt L}||\sqrt a_n +{\sqrt L}| <\epsilon \sqrt L$. Hence $|\sqrt a_n -{\sqrt L}| <\frac 1 {\sqrt a_n +{\sqrt L}} \epsilon \sqrt L<\epsilon$ since $\sqrt a_n +{\sqrt L} >{\sqrt L}$.