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$1.$ 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 ?

John M-D94
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Nimantha
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    No, you need to ensure that $|a_{n}^{0.5}+L^{0.5}|\geq 1$ – John M-D94 Jun 09 '20 at 05:32
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    I think you should insert that $\displaystyle |\sqrt{a_n}+\sqrt{L}|\ge \sqrt{L}$ into the third bottom-most line and choose new $\epsilon'=\epsilon/\sqrt{L}$. And follow the bottom comment's as well. – W. Wongcharoenbhorn Jun 09 '20 at 05:36
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    @Nimantha, your factorization of $a_n - L$ into $(\sqrt{a_n} - L)(\sqrt{a_n}+L)$ is wrong. It should be $\pm \sqrt{L}$. – Quotable Jun 09 '20 at 05:36

1 Answers1

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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 $\sqrt a_n +{\sqrt L} >{\sqrt L}$.

Kavi Rama Murthy
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