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

Question

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.

Answer 1

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$$n\geqslant N\implies\left\lvert\sqrt{a_n}-\sqrt L\right\rvert<\varepsilon.$$

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Concerning the edited version: why do you think that$$\frac{|a_n - L|}{\left|\sqrt{a_n} + \sqrt L\right|} < \lvert a_n - L \rvert?$$

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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$$\frac{\lvert a_n-L\rvert}{\left\lvert\sqrt{a_n}+\sqrt L\right\rvert}\leqslant\frac1{\sqrt L}\lvert a_n-L\rvert.$$

Answer 2

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

Hence, given $\epsilon>0$.

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