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.

  • 2
    $\begingroup$ 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 $|\sqrt{x_n} - \sqrt{L}|$? $\endgroup$ – arkeet Sep 29 '16 at 3:19
  • $\begingroup$ 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, $\sqrt{x_n}<\sqrt{T+\varepsilon}\leq \sqrt{T}+\sqrt{\varepsilon}:=\varepsilon_1\iff \left|\sqrt{x_n}-\sqrt{T}\right|<\varepsilon_1 \implies \lim\limits_{n\to\infty}\sqrt{x_n}=\sqrt{T}$. $\endgroup$ – sequence Sep 29 '16 at 18:08

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 $x_n<\sqrt{A}\epsilon$.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 :)


It's wrong of course.

For $x_n\geq0$ it's true because $f$ is a continuous function, where $f(x)=\sqrt{x}$.

  • 1
    $\begingroup$ What do you mean "it's wrong of course"? $\endgroup$ – Najib Idrissi Sep 29 '16 at 11:46
  • $\begingroup$ @ Najib Idrissi I meant that for $x_n<0$ your statement is wrong, which says that your statement is wrong. $\endgroup$ – Michael Rozenberg Sep 29 '16 at 19:33
  • $\begingroup$ 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... $\endgroup$ – Najib Idrissi Sep 29 '16 at 19:39
  • $\begingroup$ I don't agree with you. You must say that $x_n\geq0$, otherwise your statement is wrong. $\endgroup$ – Michael Rozenberg Sep 29 '16 at 19:45
  • 2
    $\begingroup$ A ton of things are implicitly assumed in math writing. $\endgroup$ – Najib Idrissi Sep 29 '16 at 20:13

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