Is this a valid proof that $(a_n) \rightarrow l$ implies $(\sqrt{a_n}) \rightarrow \sqrt{l}$?

Question

Let $(a_n)$ be a sequence of non-negative real numbers converging to $l>0$. Show that $\lim_{n\rightarrow\infty}\sqrt{a_n} = \sqrt{l}$.

Fix $\epsilon>0$.

$\exists N\in \mathbb{N}$ s.t. $n\geq N \implies |a_n - l|<\epsilon$

But $|a_n-l|=|\sqrt{a_n} - \sqrt{l}||\sqrt{a_n} + \sqrt{l}| \leq |\sqrt{a_n} - \sqrt{l}|(|\sqrt{a_n}|+|\sqrt{l}|) \ \ \ \ \ (*)$

But $(a_n) \rightarrow l$ implies it's bounded, so, by completeness, we can talk of $sup$.

Let $M_1 = sup \{\sqrt{a_n}:n \in \mathbb{N}\}$ and $M=\{M_1, |\sqrt{l}|, 1\}$

Note: I've included 1 to protect against the case where the maximum of $M_1$ and $\sqrt{l}$ is $0$.

Then $(*) < |\sqrt{a_n} - \sqrt{l}|2M \leq \epsilon$

In particular $|\sqrt{a_n} - \sqrt{l}|2M \leq \epsilon \implies |\sqrt{a_n} - \sqrt{l}| \leq \epsilon/2M$.

Done.

Answer 1

No, your argument is not correct. (*) only gives $|\sqrt {a_n} -\sqrt l| \geq \frac {|a_n-l|} {|\sqrt {a_n}+|\sqrt l|}$ and it does not help you in proving the result. Here is a proof: $|\sqrt {a_n} -\sqrt l|=\frac {|a_n-l|} {\sqrt {a_n} +\sqrt l} $. Now use the fact that $\sqrt {a_n} +\sqrt l >\sqrt l$ which gives $|\sqrt {a_n} -\sqrt l| \leq \frac {|a_n-l|} {\sqrt l} $. Can you complete the proof now?

Answer 2

One can see there is something wrong ni your proof because your last statement that the distance in square toots is less than $\epsilon/2M$ is not always true.

For example, let $$ a_n = 1+ \frac{10000}{n} $$ Then $M = M_1 = 10001$.

Now take $N$ large enough that $a_n - 1 < \epsilon = 0.01$; the first $N$ satisfying this is $1,000,001$. At that point, $$ \sqrt{1+10000/n}-1 \approx 0.005 > \frac{\epsilon}{2M} \approx 0.5\cdot 10^{-9} $$