prove if $(A_n)$ limit is $L$ then $(A_n)^2$ limit is $L^2$

Question

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 :)

Answer 1

$f(x)=x^2$ 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 $$ \forall \varepsilon >0 \exists k \in \mathbb{N} \hbox{ s.t} : \forall n\geq k \implies |f(A_n)-f(L)|<\varepsilon $$

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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 $$\forall \varepsilon>0 \exists h\in\mathbb{N} \text{ s.t. } \forall n\geq h : |A_n^2-L^2| < \varepsilon $$

Answer 2

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.