Proof that the limit of the square root is the square root of the limit 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.
 A: 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,
A: 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.
A: 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
$$
\lim_{n\to\infty}\left|\,a_n-l\,\right|=0
$$
A: Hint
Prove this inequality it's useful:
$$|\sqrt{a_n}-\sqrt\ell|\le \sqrt{|a_n-\ell|}$$
A: 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
$$\lim_{n}a_n^{1/2}= \lim_{n}c^{1/2}(a_n/c)^{1/2}=c^{1/2} \lim_{n}(a_n/c)^{1/2}=c^{1/2} =(\lim_n a_n)^{1/2}$$
A: 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: $$\lim_{n\to\infty}\sqrt{a_n}=A=\sqrt{\lim_{n\to\infty}{a_n}}$$
A: $|\sqrt{a_n} - \sqrt{l}| = |\frac{(\sqrt{a_n} - \sqrt{l}) (\sqrt{a_n} + \sqrt{l})}{(\sqrt{a_n} + \sqrt{l})}| = |\frac{a_n - l}{(\sqrt{a_n} + \sqrt{l})}|=\frac{| a_n-l|}{\sqrt{a_n} + \sqrt{l}}$
$\sqrt{a_n}\geq0\Rightarrow \sqrt{a_n} + \sqrt{l}\geq \sqrt{l}\Rightarrow \frac{| a_n-l|}{\sqrt{a_n} + \sqrt{l}}\leq\frac{| a_n-l|}{\sqrt{l}}$
$\lim(a_n)=l\Rightarrow \exists N\in\mathbb{N}$ 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 $\lim(\sqrt{a_n})=\sqrt{l}$ 
