I want to prove that if:
$$\lim_{n \to \infty}s_n = L_1, \lim_{n \to \infty}t_n = L_2$$
then $$\lim_{n \to \infty}(s_n t_n) = L_1L_2$$
Wrong (?) proof:
Fix $\epsilon >0$. By definition, there are integers $N_1,N_2$ such that:
$$n>N_1 \implies |s_n-L_1|< \frac{\epsilon}{|s_n|+|L_2|}$$
$$n>N_2 \implies |t_n-L_2|< \frac{\epsilon}{|s_n|+|L_2|}$$
Hence, for $n > \max\{N_1,N_2\}$, we have:
$$|s_nt_n – L_1L_2| = |s_n(t_n – L_2) + s_nL_2 – L_1L_2|$$
$$\leq |s_n||t_n – L_2| + |L_2||s_n – L_1|$$
$$< \frac{\epsilon}{|s_n|+|L_2|} (|s_n| + |L_2|) = \epsilon$$
I was taught that the $\epsilon$ can't depend on $n$, but I can't see why. What goes wrong?
EDIT: I know how to fix the proof, I made a post on this one: Limit of product of sequences is the product of the limits of the sequences
Best Answer
I always wonder why people insist in cutting epsilons in pieces.
Just remember that eventually proving $\forall n>N, |u_n-\ell|<K\varepsilon$ is sufficient as long as $K$ is constant.
Trying to cut epsilons is maybe aesthetically nice but, I think it hurts the understanding at basic level.
Instead with the straightforward proof you get to : $\begin{cases} \forall n>N_1, |s_n-L_1|<\varepsilon\\ \forall n>N_2, |t_n-L_2|<\varepsilon\\ \end{cases}$
So for $N>\max(N_1,N_2)$
we have $|s_nt_n-L_1L_2|\le|s_n||t_n-L_2|+|L_2||s_n-L_1|\le\left(|s_n|+|L_2|\right)\varepsilon$
Now you see that you do not have a constant before $\varepsilon$, and get to think about why $(s_n)_n$ should be bounded.
And indeed, any convergent sequence is bounded, thus $|s_n|<M$ independently of $n$.
You arrive to $|s_nt_n-L_1L_2|<\underbrace{(M+|L_2|)}_{\text{a constant }K}\varepsilon$
And you should be happy with that, it is not mandatory to get to a bare $\varepsilon$ in the end, $0.0003\,\varepsilon,\ 210734\,\varepsilon,\ 10^{513}\,\varepsilon$ or $K\varepsilon$ are all the same.