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

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

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

• Almost there, but you divided by $\lvert l\rvert$ without first dealing with the case $l=0$. Easy fix is to make $\lvert b_n\rvert<\epsilon/(2\lvert l\rvert+1)$ instead, and similarly changing the bound on $a_n-l$. – user10354138 Jun 5 at 5:24
• And you forgot to take/mention $n=\max\{N_1,N_2\}$ when you showed $a_nb_n\to0$ – Shubham Johri Jun 5 at 5:26
• Also your statement $|a_n-l| < |l|$ will not hold if $l=0$. – Anurag A Jun 5 at 5:26
• @user10354138, In the case of $|l|=0$ we just make both sequences $\sqrt{\epsilon}$ close to $0$ and take $max\{N_1,N_2\}$ (as I should have mentioned here also), right? – alwaysiamcaesar Jun 5 at 5:30

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!