Proof that the limit of the square root is the square root of the limit

Hint

Prove this inequality it's useful:

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

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,

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.