Verify proof that $(a_n) \rightarrow l \in \mathbb{R}$ and $(b_n) \rightarrow 0$ implies $(a_nb_n) \rightarrow 0$?

Question

Let $(a_n)$,$(b_n)$ be sequences s.t. $(a_n)\rightarrow l \in \mathbb{R}$ and $(b_n)\rightarrow 0$. Show that $a_nb_n \rightarrow 0$ without using the Limit Laws.

Here's my attempt:

Fix $\epsilon > 0$.

Since $(a_n) \rightarrow l$, $\exists N_1 \in \mathbb{N}$ s.t. $\forall n \geq N_1$, $|a_n-l|< |l|$

Likewise since $(b_n) \rightarrow 0$, $\exists N_2 \in \mathbb{N}$ s.t. $\forall n \geq N_2$, $|b_n|<\frac{\epsilon}{2|l|}$

But $|a_nb_n|=|(a_n-l)b_n + lb_n| \leq |(_−)_|+|lb_n| = |a_n-l||b_n|+|l||b_n| < |l| \cdot \frac{\epsilon}{2|l|} +|l| \cdot \frac{\epsilon}{2|l|} = \epsilon$

Since $\epsilon$ was arbitrary, we've shown what's required.

Answer 1

If $l=0$ then we cannot divide by it and the argument does not make sense.

Instead, $a_n$ converges implies it is bounded, say by $M$. Then use $\frac{\varepsilon}{M}$ argument to conclude the result!