I am trying to prove
if $\left\{a_n\right\}$ and $\left\{b_n\right\}$ are two sequences such that $a_n \to a$ and $b_n \to b$, then $a_nb_n \to ab$
I already know the proof using boundedness. But i tried in this way.
Using the fact that $$a_{n} b_{n}-a b=\left(a_{n}-a\right)\left(b_{n}-b\right)+a\left(b_{n}-b\right)+b\left(a_{n}-a\right)$$ We get $$|a_nb_n-ab|\leq |a_n-a||b_n-b|+|a||b_n-b|+|b||a_n-a|\to (1)$$ Now since $a_n,b_n$ are convergent, we have for every $\epsilon_1 >0$, $\exists n_0 \in \mathbb{N}$ such that $\forall n \geq n_0$ $$|a_n-a|<\epsilon_1 \to (2)$$ Like-wise for every $\epsilon_2>0$, $\exists n_0' \in \mathbb{N}$ such that $\forall n \geq n_0'$ $$|b_n-b|<\epsilon_2 \to (3)$$ Using $(2),(3)$ in $(1)$ with $\Delta$ inequality we get $$|a_nb_n-ab|<\epsilon_1\epsilon_2+|a|\epsilon_2+|b|\epsilon_1$$ whenever $n \geq max(n_0,n_0')$
Now how to conclude?
This is a good time to use a trick from analysis. You have all the epsilons, just have to show they go to zero. This is where you left off: $$|a_nb_n-ab|<\epsilon_1\epsilon_2+|a|\epsilon_2+|b|\epsilon_1$$ whenever $n \geq max(n_0,n_0')$
We have to first define an arbitrary a 'master' epsilon, $\epsilon$. Then we lower $\epsilon_1$ and $\epsilon_2$ as far as we need to get the total under $\epsilon$. (We can lower them as far as we want because they are arbitrarily small.)
For any $\epsilon >0$ we can choose $\epsilon_1 < \min{(1, \ \frac{\epsilon}{3|b|}, \ \epsilon)}$, and $\epsilon_2 < \min{(1, \ \frac{\epsilon}{3|a|}, \ \epsilon)}$. Now $$|a_nb_n-ab|<\epsilon_1\epsilon_2+|a|\epsilon_2+|b|\epsilon_1 < \frac{\epsilon}{3} +\frac{\epsilon}{3}+\frac{\epsilon}{3} = \epsilon$$.
So $|a_nb_n-ab| \rightarrow 0$ and we arrive at $a_nb_n \rightarrow ab$