# If $\lim_{n \rightarrow \infty} a_n = L > 0$. Prove. $\lim_{n \rightarrow \infty} \sqrt{a_n} = \sqrt{L}$ converges

If $$\lim_{n \rightarrow \infty} a_n = L > 0$$. Prove. $$\lim_{n \rightarrow \infty} \sqrt{a_n} = \sqrt{L}$$ converges.

Proof Attempt: $$\Bigg|\sqrt{a_n} - \sqrt{L}\Bigg|= \Bigg|\frac{a_n - L}{\sqrt{a_n} + \sqrt{L}}\Bigg|$$

Note:

$$\Bigg|\sqrt{a_n} - \sqrt{L}\Bigg| \leq \Bigg|\sqrt{a_n} + \sqrt{L}\Bigg| \leq \Bigg| a_{n} + L \Bigg|$$

If I take the limit of the last inequality this would mean: $$\Bigg| a_{n} + L \Bigg| \leq 2L$$

Therefore take $$\epsilon = 2L$$ and this would satisfy the expression.

I feel it should've been a little simpler than this. Where did I go wrong?

EDIT/ Possible Solution Redone:

By the Archimedian Property of real numbers there is a big enough value of $$n$$ such that $$a_n > \frac{L}{4}$$ (Other fractions could also be used assuming truth of property).

$\therefore \ $$a_n > \frac{L}{4} \Rightarrow \sqrt{a_n} > \frac{\sqrt{L}}{2}$$ Then: $$\Bigg|\sqrt{a_n}+\sqrt{L}\Bigg| > \Bigg| \frac{\sqrt{L}}{2} + \sqrt{L} \Bigg|$$ Which implies: $$|\frac{1}{|\frac{3 \sqrt{L}}{2}|} > \frac{1} {|\sqrt{a_n} + \sqrt{L}|}$$ Looking at the expression again and manipulation: $$\Bigg|\frac{a_n - L}{\sqrt{a_n} + \sqrt{L}}\Bigg| \leq \frac{|a_n - L|}{|\sqrt{a_n} + \sqrt{L}|} < \frac{2 \Bigg| a_n - L \Bigg|}{3 \sqrt{L}} < \frac{2 \epsilon}{3 \sqrt{L}} < \epsilon$$ Therefore $$|\sqrt{a_n} - \sqrt{L}|$$ will converge if we take the same $$N$$ that works for our assumption. • Where did$|\sqrt{a_n}-\sqrt{L}|\leq|\sqrt{a_n}+\sqrt{L}|\leq|a_n+L|$come from? Oct 6, 2018 at 17:10 • Where did you bring the inequality$\frac{|a_n-L|}{|\sqrt{a_n}+\sqrt{L}|}<|a_n-L|$from? The denominator might be smaller than$1$and then it is not true. – Mark Oct 6, 2018 at 17:16 ## 2 Answers You cannot take $$\varepsilon=2L$$. Asserting that $$\lim_{n\to\infty}\sqrt{a_n}=\sqrt L$$ means that for every $$\varepsilon>0$$, there is a natural $$N$$ such that$$n\geqslant N\implies\left\lvert\sqrt{a_n}-\sqrt L\right\rvert<\varepsilon.$$ Concerning the edited version: why do you think that$$\frac{|a_n - L|}{\left|\sqrt{a_n} + \sqrt L\right|} < \lvert a_n - L \rvert?$$ In order to prove what you wish to prove, you can use the fact that$$(\forall n\in\mathbb{N}):\left\lvert\sqrt{a_n}+L\right\rvert=\sqrt{a_n}+L\geqslant L$$and that therefore$$\frac{\lvert a_n-L\rvert}{\left\lvert\sqrt{a_n}+\sqrt L\right\rvert}\leqslant\frac1{\sqrt L}\lvert a_n-L\rvert.$$ • @dc3rd I have already made an edition to my answer concerning the edited version of your question. Oct 6, 2018 at 17:16 • Hint: for big values of$n$you have$a_n>\frac{L}{4}$. Try to think how can it effect the denominator in your expression. – Mark Oct 6, 2018 at 17:18 • Yes, your solution looks good. – Mark Oct 6, 2018 at 18:08 • @Mark Thank you, but note that it has to be changed a little if$L=0$. Oct 6, 2018 at 18:10 • No, you are not. The existence of such a$n\$ follows from the definition of convergent sequence. It has nothing to do with the Archimedes principle. Oct 6, 2018 at 18:17

$$\Bigg|\frac{a_n - L}{\sqrt{a_n} + \sqrt{L}}\Bigg| \leq \frac{|a_n - L|}{|\sqrt{a_n} + \sqrt{L}|} \le \frac{| a_n - L |}{\sqrt{L}}$$

Hence, given $$\epsilon>0$$.

Find $$N>0$$ such that $$n>N$$ implies $$|a_n-L| < \sqrt{L}\epsilon$$