Let $\{x_n\}_{n\ge1}$ be a sequence of positive real number and if $\{x^2_n\}_{n\ge1}$ is convergent, then is $\{x_n\}_{n\ge1}$ is convergent?
$\{x_n\}_{n\ge1}$ it's convergent because $\{x^2_n\}_{n\ge1}$ is convergent means $\lim_{n\to \infty}\{x^2_n\}_{n\ge1}=l(\text{for some finite l})$ that means $\lim_{n\to \infty}\{x_n\}_{n\ge1}=\sqrt l$ that indicates it's convergent. I want some counter example. please help. Thanks in advance.
The statement is true and you can prove it via a Cauchy sequence argument, using that for positive sequences $x_n+x_m = |x_n+x_m|\geq |x_n-x_m|$. This implies that
$$|x_m^2-x_n^2|=|x_m+x_n|\cdot |x_m-x_n| \geq |x_m-x_n|^2.$$
Since $(x_n^2)_n$ converges there is for all $\varepsilon > 0$ a $N\in\mathbb{N}$ such that for all $n,m\geq N$ we have $$\varepsilon > |x_m^2-x_n^2|\geq |x_m-x_n|^2.$$
Let $\tilde{\varepsilon}\in(0,1)$ and let $\varepsilon =\tilde{\varepsilon}^2$. Then,
$$\tilde\varepsilon =\sqrt{\varepsilon} > \sqrt{|x_m^2-x_n^2|}\geq |x_m-x_n|$$
for sufficiently large $n,m$. Hence, $(x_n)_n$ is a Cauchy sequence as well and, therefore, convergent.
Yes the theorem is true .
Your proof is correct but you should specify why you can take the function $\sqrt x$ inside the limit (One way to do this is to use the fact that the continuous function of a limit is the limit of a continuous function )
You should also prove that if $x_{n}$ is a convergent sequence of positive numbers then it’s limit is $\ge 0$ otherwise the square root function will be undefined this is why the proof fails in the negative case
Claim: If $x_n^2 \to L$,then $x_n \to \sqrt{L}$.
There are $2$ possibilities:
Case 1: $ L > 0$. For any $\epsilon > 0, \exists N$ such that if $n \ge N \implies |x_n^2- L| < \epsilon\sqrt{L}\implies |x_n - \sqrt{L}|= \dfrac{|x_n^2-L|}{x_n+\sqrt{L}}\le \dfrac{1}{\sqrt{L}}|x_n^2-L|< \dfrac{1}{\sqrt{L}}(\epsilon\cdot \sqrt{L})= \epsilon$.
Case 2: $L = 0$. For any $\epsilon > 0, \exists N$ such that if $n \ge N \implies |x_n^2| <\epsilon^2\implies |x_n| < \epsilon.$
In both cases, we've shown $x_n \to \sqrt{L}$.