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I have to prove the following:

$$\lim a_n = L \implies \lim a_n^2 = L^2$$

I know that $\lim a_n = L \implies \forall\epsilon>0 \ \exists n_0$ such that $n>n_o\implies |a_n-L|<\epsilon$

I need to prove that there exists another $n_o$ such that:

$$|a_n^2 -L^2|<\epsilon$$

What I tried:

$$|a_n^2 -L^2| = |a_n-L||a_n+L|<\epsilon|a_n+L|$$

but if $|a_n-L|<\epsilon$ by hypotesis, then $-\epsilon-L<a_n<L+\epsilon\implies a_n+L<2L+\epsilon$

I thought this would help me but this $2L$ is not good, I cannot make $|a_n^2-L^2|$ as close as I want to $0$, but wait... maybe I can? Because I have:

$$|a_n^2 -L^2| = |a_n-L||a_n+L|<\epsilon|a_n+L|<\epsilon(2L+\epsilon)$$

if I choose $\epsilon_1 = \frac{1}{2L+\epsilon}$ then it's ok, right?

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$$\lim a_n = L \implies \lim a_n-L=0 \implies \forall\epsilon>0, \;\exists N_{\epsilon} \in \mathbb{N} : |a_n-L|\leq \epsilon,\; \forall n>N_{\epsilon}$$ $$\implies (a_n-L)^2<\epsilon^2 \implies a_n^2-2a_nL+L^2 < \epsilon^2,\; \forall n>N_{\epsilon}$$

So,

$$\lim a_n^2-2a_nL+L^2 = 0$$

However,

$$\lim a_nL = L^2$$

as limits are preserved under scalar multiplication. The overall limit becomes:

$$\lim a_n^2-L^2=0 \implies \lim a_n^2 = L^2$$

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You are close.

For a start, you can assume that $\epsilon_1 < 1$.

This means that $|a_n^2 -L^2| <\epsilon_1|a_n+L| <\epsilon_1(2L+\epsilon_1) <\epsilon_1(2L+1) $.

Then, to make $\epsilon_1(2L+1) < \epsilon $, just choose $\epsilon_1 <\dfrac{\epsilon}{2L+1} $.

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    $\begingroup$ I know this inequality $|a_n + L| < 2L + \epsilon_1$ is taken from the OP, but is it correct? Seems like it should be $|a_n + L| < 2|L| + \epsilon_1$. ($L$ might be negative.) $\endgroup$ – Bungo Apr 28 '16 at 1:04
  • $\begingroup$ No biggy. Just use |L| as you say. $\endgroup$ – marty cohen Apr 28 '16 at 2:04
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You can use, more generally, the fact that continuous functions and limits commute. That is, if $f$ is continuous, and $(a_{n})_{n}$ converges to $L$, $f(a_{n})\rightarrow f(L)$. (take $f(a)=a^2$). Proof here.

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