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?
Best Answer
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} $.