Proving $\lim_{n\rightarrow\infty}a_n=l\Rightarrow \lim_{n\rightarrow\infty}a^2_n=l^2$ $$a_n\ge0, l\ge0$$If $$\lim_{n\rightarrow\infty}a_n=l$$
Then $$\lim_{n\rightarrow\infty}a^2_n=l^2$$
Here is a trick, which will help you solve this problem. 

$$a_n^2 - l^2 = (a_n - l)^2 + 2l(a_n - l)$$
using this you should be able to show that $a_n^2 - l^2 \rightarrow 0$. 

EDIT: To elaborate further, by the triangle inequality we now know that 
$$\lvert a_n^2 - l^2\rvert \leq \lvert (a_n - l)^2 \rvert + \lvert 2l(a_n-l)\rvert \hspace{10mm} (1)$$
So we just need to show that each of the two terms on the right get small as $n$ gets large. You should be able to make both of them small by using the fact that $a_n \rightarrow l$. 

FINAL EDIT: To see a complete proof. Fix $\epsilon &gt; 0$. Choose $N_0$ so that for $n\geq N_0$, $$\lvert a_n - l\rvert \leq \sqrt{\epsilon};$$ and $N_1$ large so that  for $n\geq N_1$, $$\lvert a_n - l \rvert \leq \frac{1}{\lvert 2 l \rvert} \epsilon.$$Then if $n\geq\max(N_0,N_1)$, the inequality (1) above gives us $$\lvert a_n^2 - l^2 \rvert \leq \epsilon + \epsilon = 2 \epsilon$$
and since $\epsilon$ was arbitrary we conclude that $\lvert a_n^2 - l^2\rvert \rightarrow 0$.
Suppose: $$\lim_{n\rightarrow\infty}a_n^2\not\rightarrow l^2$$
Exists an $\epsilon&gt;0$ such that: $$|n^2-l^2|\ge\epsilon \Leftrightarrow l^2-\epsilon\ge n^2 \ge l^2 + \epsilon$$

Left side of the equation is false, because if it were true, $\epsilon&lt;0$. Which is false.

Right side is false because:$$n^2-l^2\ge\epsilon \Leftrightarrow -l^2 \ge \epsilon-n^2 \ge 0$$

Which is again false, because $n,l\ge0$

EDIT: If this is wrong, please let me know why.