Proving the simple limit theorems

Question

If {$a_n$} coverges to A and c is a real number then the sequence

{$ca_n$} converges to cA

Case 1: c=0, therefore cA = 0 and the statement is true (is it that simple?)

Case 2: c > 0 or c < 0 then let $\epsilon > 0$ such that

$\lvert a_n - A \rvert < \epsilon$ for some n $\ge$ N

so we know that $A-\epsilon < a_n <A +\epsilon $

Could I just multiply this inequality by c so that

$c(A-\epsilon)< ca_n < c(A+\epsilon)$

I also don't think I'm allowed to use the product rule but that would be,

$\lim \limits_{n\to \infty} c$ {$a_n$} = $(\lim \limits_{n\to \infty} c)(\lim \limits_{n\to \infty}) = c(\lim \limits_{n\to \infty}a_n) = cA $

This problem seems harder than I'm assuming it to be

Answer 1

For case 1 it is that easy because $\{0,0,0,0,...\}$ converges to $0$. Try this for case 2: Fix $\varepsilon>0$. We know there exists $N$ such that if $n\geq N$, $|a_n-A|<\varepsilon/|c|.$ Then $|c||a_n-A|=|ca_n-Ac|<\varepsilon$, so $\{ca_n\}\to cA.$