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I'm trying to prove that: $$\lim_{n\rightarrow\infty} \sqrt{a_n} = \sqrt{\lim_{n\rightarrow\infty} a_n}$$

Given $a_n > 0$ for all $n$.

My initial idea was to start with the definition of limit (assuming $\lim_{n\rightarrow\infty} a_n = l$):

$$|\sqrt{a_n} - \sqrt{l}| = |\frac{(\sqrt{a_n} - \sqrt{l}) (\sqrt{a_n} + \sqrt{l})}{(\sqrt{a_n} + \sqrt{l})}| = |\frac{a_n - l}{(\sqrt{a_n} + \sqrt{l})}|$$

The problem is that $\sqrt{a_n} + \sqrt{l}$ could be less than $1$. And therefore I can't continue the proof using this approach.

Edit: forgot to mention that $a_n$ converges is another hypothesis.

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  • $\begingroup$ You know that the limit commutes with continuous functions. Can you prove that $\sqrt x$ is continuous on $(0,\infty)$? $\endgroup$ – Ian Coley Mar 1 '14 at 18:25
  • $\begingroup$ I'm following a book called Mathematical analysis by Schröder and continuity has not been defined yet. $\endgroup$ – Jordi Mar 1 '14 at 18:27
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    $\begingroup$ @IanColey : Showing that $\sqrt x$ is continuous is precisely the exercise, if you use the appropriate definition (i.e. if no notion of topology has been introduced). $\endgroup$ – Patrick Da Silva Feb 15 '16 at 17:10
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Hint

Prove this inequality it's useful:

$$|\sqrt{a_n}-\sqrt\ell|\le \sqrt{|a_n-\ell|}$$

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  • $\begingroup$ Assuming $a_n \ge l$ then $(\sqrt{a_n} - \sqrt{l})^2 = (a_n - l) + 2 \sqrt{l}(\sqrt{l} - \sqrt{a_n}) < a_n - l$ (because $\sqrt{l} - \sqrt{a_n} \le 0$) and therefore $|\sqrt{a_n} - \sqrt{l}| \le \sqrt{|a_n - l|}$. Using a similar argument one can prove the inequality when $l \gt a_n$. Something like that? With this inequality proven everything is easy. $\endgroup$ – Jordi Mar 1 '14 at 19:22
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    $\begingroup$ You need not assume $a_n\ge\ell$ or $a_n\le\ell$. Just take the square and use the triangle inequality. $\endgroup$ – user63181 Mar 1 '14 at 19:26
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Hint: Consider two separate cases:

Case I: $a_n\to0$ as $n\to\infty$. In this case, you wish to show that $\sqrt{a_n}\to0$; this can be done pretty easily using the definition of the limit. (Given $\epsilon$, can you see that $\lvert a_n\rvert<\epsilon^2$ for $n$ sufficiently large?)

Case II: $a_n\to\ell>0$. In this case, you can say that for $n$ sufficiently large, $\frac{1}{2}\ell<a_n<\frac{3}{2}\ell$;see if you can use those inequalities on the term in the denominator that you came up with to finish up.

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Suppose that $\frac14l\le a_n\le4l$, which is equivalent to $\frac12\sqrt{l}\le\sqrt{a_n}\le2\sqrt{l}$. Then, $$ \left|\,\sqrt{a_n}-\sqrt{l}\,\right|=\frac{\left|\,a_n-l\,\right|}{\sqrt{a_n}+\sqrt{l}} $$ Therefore, $$ \frac1{3\sqrt{l}}\,\left|\,a_n-l\,\right|\le\left|\,\sqrt{a_n}-\sqrt{l}\,\right|\le\frac2{3\sqrt{l}}\,\left|\,a_n-l\,\right| $$ These two inequalities, along with the definition of a limit, shows that $$ \lim_{n\to\infty}\sqrt{a_n}=\sqrt{\lim_{n\to\infty}a_n} $$


Comment on the Answer in the Question

There is no reason that the coefficient of $\left|\,a_n-l\,\right|$ must be less than $1$. In this case. all you need to show is that for a given $l$, each side is less than a constant multiple of the other.

For example, say we know that $$ \left|\,\sqrt{a_n}-\sqrt{l}\,\right|\le100\left|\,a_n-l\,\right| $$ We can make the left side as small as we want simply by making the right side $\frac1{100}$ the size, and we know we can do that because $$ \lim_{n\to\infty}\left|\,a_n-l\,\right|=0 $$

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It is easy to show that assuming $b_n$ is a convergence sequence, then $b_n^2$ also converges and $$ \lim_{n \to \infty} b_n^2 = \left( \lim_{n \to \infty} b_n \right)^2 $$ This should inspire you the following proof : assume $a_n$ converges to $L^2$ where $L \ge 0$. First assume $L > 0$. Then $$ |\sqrt{a_n} - L| \le |\sqrt{a_n}-L| \frac{\sqrt{a_n}+L}{L} = \frac{|a_n - L^2|}L \underset{n \to \infty}{\longrightarrow} 0. $$ For $L = 0$, one easily sees that $|\sqrt{a_n}| < \varepsilon$ as soon as $|a_n| < \varepsilon^2$, and this eventually happens for all $n \ge N$ large enough since $a_n \to L = 0$.

Hope that helps,

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1) Suppose that $(a_n) \to 1$ and $a_n\in \mathbb{R}^{>0}$ we shall show that $(a_n)^{1/2}\to1$. Given $\varepsilon>0$ there is a $K$ such that $(1+1/k)$ and $(1-1/k) $ are $\varepsilon$-close to $1$ whenever $k\ge K$. Since $a_n \to 1$ so there is some $N$ such that $|a_n -1|\le 1/K$ whenever $n\ge N$ we have. Thus

$$1-1/K\le a_n\le 1 + 1/K$$ $$(1-1/K)^{1/2}\le a_n^{1/2}\le (1 + 1/K)^{1/2}$$

Now since $1 + 1/K>1$, then $(1+1/K)>(1+1/K)^{1/2}$ also we have that $1-1/K<1$ so $(1-1/K)^{1/2}>(1-1/K)$. Hence

$$(1-1/K)< a_n^{1/2}< (1 + 1/K)$$

Since both are $\varepsilon$-close to $1$, we're done. Then $a_n^{1/2}\to 1$.

2) Now suppose that $(a_n) \to c$, $a_n\in \mathbb{R}^{>0}$ and $c\not=1$. Then by the limit laws we can conclude that $a_n/c \to 1$ so $(a_n/c)^{1/2}\to 1$. Thus

$$\lim_{n}a_n^{1/2}= \lim_{n}c^{1/2}(a_n/c)^{1/2}=c^{1/2} \lim_{n}(a_n/c)^{1/2}=c^{1/2} =(\lim_n a_n)^{1/2}$$

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  • $\begingroup$ What if $a_n \to 0$, as in the case of $1/n$ for instance? $\endgroup$ – Patrick Da Silva Sep 30 '15 at 23:33
  • $\begingroup$ Given $\epsilon>0$, for a sufficiently large $N$, we have $0<a_n<\epsilon^2$ for all $n\ge N$... @PatrickDaSilva $\endgroup$ – Jose Antonio Oct 1 '15 at 1:57
  • $\begingroup$ I know (see my answer), my point is it wasn't in your answer. $\endgroup$ – Patrick Da Silva Oct 1 '15 at 1:59
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If you assuming that the sequence $\sqrt{a_n}$ converges, so you can write: $$\lim_{n\to\infty} \sqrt{a_n} =A$$

Then:

$$\lim_{n\to\infty}\sqrt{a_n}*\lim_{n\to\infty}\sqrt{a_n}=A*A$$

Then, as both limits exists:

$$\lim_{n\to\infty}(\sqrt{a_n}*\sqrt{a_n})=A^{2}$$

$$\Rightarrow\lim_{n\to\infty}(\sqrt{a_n})^{2}=A^{2}$$ $$\Rightarrow\lim_{n\to\infty}{a_n}=A^{2}$$

Using this result, you write: $$\sqrt{\lim_{n\to\infty}{a_n}}=\sqrt{A^{2}}$$ $$\sqrt{\lim_{n\to\infty}{a_n}}=A$$

Then you have: $$\lim_{n\to\infty}\sqrt{a_n}=A=\sqrt{\lim_{n\to\infty}{a_n}}$$

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  • $\begingroup$ You can't assume that sequencce converges $\endgroup$ – Daniel Segal Nov 19 '19 at 22:59
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$|\sqrt{a_n} - \sqrt{l}| = |\frac{(\sqrt{a_n} - \sqrt{l}) (\sqrt{a_n} + \sqrt{l})}{(\sqrt{a_n} + \sqrt{l})}| = |\frac{a_n - l}{(\sqrt{a_n} + \sqrt{l})}|=\frac{| a_n-l|}{\sqrt{a_n} + \sqrt{l}}$

$\sqrt{a_n}\geq0\Rightarrow \sqrt{a_n} + \sqrt{l}\geq \sqrt{l}\Rightarrow \frac{| a_n-l|}{\sqrt{a_n} + \sqrt{l}}\leq\frac{| a_n-l|}{\sqrt{l}}$

$\lim(a_n)=l\Rightarrow \exists N\in\mathbb{N}$ such that $\forall n\geq N, |a_n-l|<\sqrt{l}\ \epsilon.$

Thus, $\forall n\geq N, |\sqrt{a_n} - \sqrt{l}|\leq\frac{| a_n-l|}{\sqrt{l}}<\frac{\sqrt{l}\ \epsilon}{\sqrt{l}}=\epsilon$ which is exactly the definition that $\lim(\sqrt{a_n})=\sqrt{l}$

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