# Proof that $\lim\limits_{n\to \infty} \sqrt{x_n} = \sqrt{\lim\limits_{n\to \infty} x_n}$

It is asked to prove that $\lim\limits_{n\to \infty} \sqrt{x_n} = \sqrt{\lim\limits_{n\to \infty} x_n}$, and suggested to use the following two inequalities:

$$a+b\leq a+ 2\sqrt{a}\sqrt{b}+b$$ $$\sqrt{b}\sqrt{b}\leq \sqrt{a}\sqrt{b}$$

The second inequality holds iff $a\geq b \geq 0$.

I've tried different possibilities, but couldn't figure out how to either take the limit sign out of the square root, or take the limit sign into the square root. Would appreciate some hints, but not an entire solution please.

• Don't fiddle around with limit signs. If $x_n \to L$, you need to show that $\sqrt{x_n} \to \sqrt{L}$. Given a bound for $|x_n - L|$, can you get a bound for $|\sqrt{x_n} - \sqrt{L}|$? Commented Sep 29, 2016 at 3:19
• I'm not sure where to use the above inequalities. Please let me know what you think about my proof: $\lim\limits_{n\to\infty}x_n = T \implies \sqrt{\lim\limits_{n\to\infty}x_n} =\sqrt{T}$. Since $\lim\limits_{n\to\infty}x_n = T$, $\forall\varepsilon>0,\exists N>0$ such that $\left| x_n-T \right|<\varepsilon$ for $n>N$. Thus $T-\varepsilon <x_n<T+\varepsilon \implies \sqrt{x_n}<\sqrt{T+\varepsilon}$. Now, $\sqrt{x_n}<\sqrt{T+\varepsilon}\leq \sqrt{T}+\sqrt{\varepsilon}:=\varepsilon_1\iff \left|\sqrt{x_n}-\sqrt{T}\right|<\varepsilon_1 \implies \lim\limits_{n\to\infty}\sqrt{x_n}=\sqrt{T}$. Commented Sep 29, 2016 at 18:08
• I think you missed one step. Why $\sqrt{x_n}-\sqrt{T} < \sqrt{\varepsilon}$ implies $|\sqrt{x_n}-\sqrt{T}| < \sqrt{\varepsilon}$? Commented Mar 20, 2021 at 19:59
• for the second comment, you can choose $|T-x_n|\lt \epsilon^2$ and show with the same proof you wrote that $|sqrt(T)-sqrt(x_n)|\lt \epsilon . for the third, by the [triangle inequality](youtu.be/051WWg9twO0?si=NNCX9umOGFnZ1mNA) and by$\sqrt{a}>0 \sqrt{x_n}-\sqrt{T}=|\sqrt{x_n}+(-|\sqrt{T}|)\ge |\sqrt{x_n}-\sqrt{T}|$and thus$ |\sqrt{x_n}-\sqrt{T}|\le |\sqrt{x_n}-\sqrt{T}|<\sqrt{\epsilon}$Commented Feb 24 at 23:10 ## 2 Answers You have to make sure that $$\forall n\in\mathbb{N}^*$$, $$x_n\geq 0$$. Otherwise the limit doesn't exist because the sequence $$\{\sqrt{x_n}\}_{n=1}^{\infty}$$ isn't even defined(well, in $$\mathbb{R}$$). Suppose the statement above is true. Let $$A=\lim\limits_{x\to\infty}x_n$$ Then there are 2 cases: (1)$$A=0$$. $$\forall\epsilon>0$$,$$\exists N\in\mathbb{N}^*$$ so that $$\forall n\geq N$$ there is $$x_n<\epsilon^2$$,that is, $$\sqrt{x_n}<\epsilon$$, so that $$\lim\limits_{x\to\infty}\sqrt{x_n}=0=\sqrt{\lim\limits_{n\to\infty}x_n}$$. (2)Otherwise,$$A>0$$. $$\forall\epsilon>0$$,$$\exists N\in\mathbb{N}^*$$ so that $$\forall n\geq N$$ there is $$|x_n-A|<\sqrt{A}\epsilon$$.This means that $$\epsilon>\frac{x_n}{\sqrt{A}}\geq|\frac{x_n-A}{\sqrt{x_n}+\sqrt{A}}|=|\sqrt{x_n}-\sqrt{A}|$$, therefore $$\lim\limits_{n\to\infty}\sqrt{x_n}=\sqrt{A}=\sqrt{\lim\limits_{n\to\infty}x_n}$$. This proof uses solely the definition of the limit of a sequence :) • I don't understand where "...there is$x_n < \sqrt{A}\epsilon$..." comes from?$\lim_{n \to \infty} x_n = A$. Hence this should be$|x_n - A|<\sqrt{A}\epsilon$. Commented Mar 19, 2021 at 20:16 • @alexander.myltsev Sorry for the mistake. I have fixed it now. Commented Mar 21, 2021 at 4:35 • Ok, thanks. It isn't clear now why "...This means that$\varepsilon > \frac{x_n}{\sqrt{A}}$..." is true? Commented Mar 21, 2021 at 9:19 • I think the step is pretty clear:$\varepsilon > \frac{|x_n - A|}{\sqrt{A}} = \frac{|x_n - A|}{|\sqrt{A}|} = |\frac{x_n - A}{\sqrt{x_n} + \sqrt{A}}|=|\sqrt{x_n}-\sqrt{A}| $. Please, correct me if I'm wrong. Commented Apr 1, 2021 at 20:03 • The second equality should be$\geq$instead Commented Apr 8, 2021 at 7:26 It's wrong of course. For$x_n\geq0$it's true because$f$is a continuous function, where$f(x)=\sqrt{x}$. • What do you mean "it's wrong of course"? Commented Sep 29, 2016 at 11:46 • @ Najib Idrissi I meant that for$x_n<0$your statement is wrong, which says that your statement is wrong. Commented Sep 29, 2016 at 19:33 • There has to be some confusion here. There is no "my statement", the only thing I did in this thread is ask you a question. And OP's question completely obviously assumes that$x_n \ge 0$, if only because they're taking the square root of$x_n$without saying anything... Commented Sep 29, 2016 at 19:39 • I don't agree with you. You must say that$x_n\geq0\$, otherwise your statement is wrong. Commented Sep 29, 2016 at 19:45
• A ton of things are implicitly assumed in math writing. Commented Sep 29, 2016 at 20:13