If $\{a_n\}$ and $\{b_n\}$ converge, then $\{a_nb_n\}$ converges. My solution:
Case 1: $b \neq 0$
WTS:
(1) $\exists a \in \mathbb R, \forall \epsilon > 0, \exists N_1 > 0$, such that for all $n \in \mathbb N$,
if $n > N_1$, then $|a_n - a| < \dfrac{\epsilon}{2|b|}$
(2) $\exists b \in \mathbb R, \forall \epsilon > 0, \exists N_2 > 0$, such that for all $n \in \mathbb N$,
$$\text{Let } M := max(|a-\epsilon|, |a + \epsilon|, |a_n| \text{ for n} < N)$$
if $n>N_2$, then $|b_n-b|<\dfrac{\epsilon}{2M}$
Let $\epsilon > 0$ be arbitrary. Choose N = $\max(N_1, N_2)>0$
Suppose $n > N$, then
\begin{align*}
|a_nb_n - ab| &= |a_nb_n - a_nb + a_nb - ab|\\
&=|a_n (b_n-b)+b(a_n-a)|\qquad\text{(by algebra)}\\
&\leq |a_n||b_n - b| + |b| |a_n - a|\qquad\text{(by triangle inequality)}\\
&< M\dfrac{\epsilon}{2M} + |b|\dfrac{\epsilon}{2|b|}\\
&=\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon
\end{align*}
 A: You can use the fact that there exists a real constant $M := \max\{|a+\epsilon|, |a-\epsilon|, |a_n|\ \mathrm{for}\ n < N\}$.
Now $|a_n||b_n−b|+|b||a_n−a| ≤ M|b_n−b|+|b||a_n−a|$. You can do the rest right?
Also, at the end of the proof, you want it to say $< \epsilon$. If you start off by saying "for all $\epsilon$ there exists ... such that $|a_n-a|<\frac{\epsilon}{2|b|}$", the last line will simplify to $M|b_n−b|+|b||a_n−a| < M|b_n−b| + \frac{\epsilon}{2}$. In the same way you can make the first term simplify to $\frac{\epsilon}{2}$ and end up with a clean proof. But to do this you need to know that |b| is non-zero, so this will require a separate case.
A: Here is another way let $$\lim _{ n\rightarrow \infty  }{ { a }_{ n }=a } ,\lim _{ n\rightarrow \infty  }{ { b }_{ n }=b } $$then $${ a }_{ n }=a+{ \alpha  }_{ n },{ b }_{ n }=b+{ \beta  }_{ n }\quad ,\quad n=1,2,...$$ where $\lim _{ n\rightarrow \infty  }{ { \alpha  }_{ n }=\lim _{ n\rightarrow \infty  }{ { \beta  }_{ n }=0 }  } $
$${ a }_{ n }{ b }_{ n }=\left( a+{ \alpha  }_{ n } \right) \left( b+{ \beta  }_{ n } \right) =ab+\left( { \alpha  }_{ n }b+{ \beta  }_{ n }a+{ \alpha  }_{ n }{ \beta  }_{ n } \right) \\ \lim _{ n\rightarrow \infty  }{ \left( { \alpha  }_{ n }b+{ \beta  }_{ n }a+{ \alpha  }_{ n }{ \beta  }_{ n } \right) =0 } $$
so $$\lim _{ n\rightarrow \infty  }{ { a }_{ n }{ b }_{ n }=ab=\lim _{ n\rightarrow \infty  }{ { a }_{ n }\lim _{ n\rightarrow \infty  }{ { b }_{ n } }  }  } $$
A: Here's a solution which mostly consists of proving several special cases, which I saw in my first analysis textbook (don't remember the title anymore):


*

*Special case where $a = b = 0$.

*Special case where $a_n = A$ is a constant sequence and $b = 0$.

*(Possibly already done) Prove if $a_n \rightarrow a$ and $b_n \rightarrow b$ then $a_n + b_n \rightarrow a + b$ as $n \to \infty$.

*(Possibly already done) Prove if $a_n = A$ is a constant sequence then $a_n \to A$ as $n \to \infty$.


Then, to put all these together to prove the general case: use
\begin{equation}
a_n b_n = (a_n - a) (b_n - b) + a (b_n - b) + b (a_n - a) + ab.
\end{equation}
