Where I made mistake When I was proving the product rule of sequence limits, I find a fact that it seems to not allow us to do the substitution on $\epsilon$. (sorry my English is not good) 
$$\lim_{n\to \infty}a_n=a, \lim_{n\to \infty }b_n =b$$
Then we have $|a_n-a|<\epsilon$ when $n>N_1$
$|b_n-b|<\epsilon$ when $n>N_2$
$N=max(N_1,N_2)$
Thus we can have $|a_n|<\epsilon+|a|,|b_n|<\epsilon+|b|$ when $n>N$
Then $|a_n||b_n|<\epsilon^2+|a|\epsilon+|b|\epsilon+|a||b|$
For $|a_n|-\epsilon<|a|,$ and $|b_n|-\epsilon<|b|$
$\epsilon^2+|a|\epsilon+|b|\epsilon+|a||b|<\epsilon^2+(|a_n|-\epsilon)\epsilon+|b|(|a_n|-|a|)+|a||b|=|a_n|\epsilon+|b||a_n|=|a_n||\epsilon+b|$
Since the limit of $a_n$ exists, then we have $|a_n|<M_1$ where $M_1$ is a constant.
As the  same result, $|b_n|<M_2$.
If we take the substitution $\epsilon_1=M_1(\epsilon+M_2)$
Thus the $|a_nb_n-0|<\epsilon_1$
which doesn't always to hold.
Where I made mistake?
 A: You are using inequalities with wrong directions $$\epsilon^2+|a|\epsilon+|b|\epsilon+|a||b|<\epsilon^2+(|a_n|-\epsilon)\epsilon+|b|(|a_n|-|a|)+|a||b|=|a_n|\epsilon+|b||a_n|=|a_n||\epsilon+b|$$
How did you get $$|a|<(|a_n|-\epsilon )$$
A: To show that $a_nb_n$ converges to $ab$, we need to estimate the magnitude of $$|a_nb_n-ab|$$
Now $$|a_nb_n-ab|=|(a_nb_n-a_nb)+(a_nb-ab)|\le |a_nb_n-a_nb|+|a_nb-ab|$$
$$|a_nb_n-ab|\le |a_n||b_n-b|+|b||a_n-a|$$
Since $\{a_n\}$ is a convergent sequence, so it is bounded and therefore $\quad\exists\quad$ a real number $B_1\gt 0$ such that $$|a_n|\le B_1\qquad \forall ~n\in \mathbb{N}.$$
Let $M=sup\{B_1,|b|\}$, then $$|a_nb_n-ab|\le M|b_n-b|+M|a_n-a|$$
Since$\{a_n\}$ and $\{b_n\}$ converges to $\quad a\quad$ and $\quad b\quad$ respectively, so for $\quad \epsilon \gt 0,\quad \exists\quad$ natural numbers $\quad k_1,\quad \text{and}\quad k_2\quad $ such that $$|a_n-a|\lt \frac{\epsilon}{2M}\qquad \forall \quad n\ge k_1$$and $$|b_n-b|\lt \frac{\epsilon}{2M}\qquad \forall \quad n\ge k_2$$
Let $\quad k=sup\{k_1,k_2\}\quad$ and then for $\quad n\ge k\quad$
$$|a_nb_n-ab|\le M \frac{\epsilon}{2M}+M \frac{\epsilon}{2M}=\epsilon$$
Since $\quad \epsilon \gt 0\quad$be arbitrary, so $\quad a_nb_n\quad$ converges to $\quad ab\quad$
