Suppose a sequence ${x_n}$ converges. Prove that $\displaystyle\lim_{n\rightarrow\infty} \sqrt{|x_n|} = \sqrt{|\lim_{n\rightarrow\infty} x_n|}$

Question

I have been manipulating the triangle inequality with the definition of a limit but cannot get a solid proof.

Answer 1

It is already answered somewhere. It is just because the function $$f(x)=\sqrt{|x|}$$ is continuous everywhere.

Answer 2

Let $a_{m,n} = \sqrt{|(x_n)|}$ for all $m$. And use information on this other site inside math stack exchange: When can you switch the order of limits?

Answer 3

Let's prove it by definition. Let $x\in\mathbb{R}$ such that $\lim x_n=x$. We want to show that given $\epsilon>0$ there exists $n_0\in\mathbb{N}$ such that $$n>n_0\Rightarrow|\sqrt{|x_n|}-\sqrt {|x|}|<\epsilon.$$

If $x=0$, since $\lim x_n=0$, then $\lim|x_n|=0$ and therefore there exists $n_0\mathbb{N}$ such that $$n>n_0\Rightarrow|x_n|<\epsilon^2,$$ and thus $$n>n_0\Rightarrow|\sqrt{|x_n|}|=\sqrt{|x_n|}<\epsilon.$$

Now, if $x\neq 0$, since $$||x_n|-|x||=|\sqrt {|x_n|}-\sqrt {|x|}||\sqrt {|x_n|}+\sqrt {|x|}|,$$ we have $$|\sqrt {|x_n|}-\sqrt {|x|}|=\frac{||x_n|-|x||}{|\sqrt {|x_n|}+\sqrt {|x|}|}.$$ Since $\lim x_n=x$, there exists $n_1\in\mathbb{N}$ such that $$n>n_1\Rightarrow|x_n-x|<\frac{|x|}{2},$$ and so, by triangle inequality, $$n>n_1\Rightarrow \frac{|x|}{2}=|x|-\frac{|x|}{2}<|x_n|.$$Hence $$n>n_1\Rightarrow \frac{1}{\sqrt{|x_n|}}<\sqrt{\frac{2}{x}}$$

Now, note that $\lim x_n=x$ implies $\lim|x_n|=|x|$ (the converse isn't true!), so there exists also $n_2\in\mathbb{N}$ such that $$n>n_2\Rightarrow||x_n|-|x||<\epsilon(\sqrt{\frac{2}{x}}+\sqrt{|x|}).$$ Finally, take $n_3=\max\{n_1,n_2\}$. We conclude that $$n>n_3\Rightarrow|\sqrt{|x_n|}-\sqrt {|x|}|<\epsilon.$$