prove if $(A_n)$ limit is $L$ then $(A_n)^2$ limit is $L^2$ Hello,
What I need to prove is: if $(A_n)$ limit is $L$ then $(A_n)^2$ limit is $L^2$.
I've added my attempt to prove it. I got stuck so I'm guessing I'm missing something here.  

help will be appreciated :)
 A: $f(x)=x^2$ is continuous, which by definition in equivalent to $$\forall \varepsilon>0 \exists\delta>0 \hbox{ s.t.} : |x-y|<\delta \implies |f(x)-f(y)|<\varepsilon.$$
By the assumption that $(A_n)$ converges to $L$ you got that  $\forall \varepsilon_1>0 \exists k\in \mathbb{N} $ s.t. : $\forall n\geq k \implies |A_n -L|<\varepsilon_1.$
Now you are so use these two statements in order to show that
$$
\forall \varepsilon >0  \exists k \in \mathbb{N} \hbox{ s.t} : \forall n\geq k \implies |f(A_n)-f(L)|<\varepsilon
$$

If you cant use that $f(x)=x^2$ is continuous, it goes as follows:
fix $\varepsilon>0$. 
$$
|A_n^2-L^2|=|(A_n-L)(A_n+L)|=|A_n-L||A_n+L|=|A_n-L||A_n-L+2L|$$
$$
\leq |A_n-L|(|A_n-L|+2|L|) 
$$
Now fix $k_1$ s.t. $\forall n\geq k_1 $ : $|A_{n}-L|< \varepsilon_1$,  and  $k_2$ s.t. $\forall n\geq k_2$  : $|A_{n}-L|< 1.$ (note that this can be done because of the assumption, that $A_n$ converges to $L$)
Now for all $n\geq \max\{k_1,k_2\}=h$ we got that
$$
|A_n^2-L^2| \leq |A_n-L|(|A_n-L|+2|L|) < \varepsilon_1|(1+2|L|)
$$ 
If we  started to set $\varepsilon_1=\varepsilon / (1+2|L|)$ we would get that $|A_n^2-L^2|<\varepsilon$.
This was done for an arbitrary $\varepsilon>0$, thus it must hold that 
$$\forall \varepsilon>0 \exists h\in\mathbb{N} \text{ s.t. } \forall n\geq h : |A_n^2-L^2| < \varepsilon
$$
A: You want to show that $|A_n^2-L^2|$ becomes small. You have already reached $|(A_n-L)(A_n+L)|$, which is good. Now, $|A_n-L|$ becomes small, so you should not throw it away, it is what helps you. $|A_n+L|$ does not become small, but fortunately not too big either. 
