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.