TITLE: If $ \lim a_{n}b_{n}=\gamma $ and $ \lim b_{n}=1 $ then $ \lim a_{n}=\gamma $
QUESTION [0 upvotes]: let $a_n,b_n $ be sequences such that
$ \lim a_{n}b_{n}=\gamma $ (maybe infinity or minus infinity)
and $ \lim b_{n}=1 $
Prove or disprove that $ \lim a_{n}=\gamma $
I tried to prove it using epsilon-delta definition, but I couldnt finish the proof.
Any ideas would help. Thanks in advacne

REPLY [2 votes]: If $\gamma$ is not finite, we may assume that it's $+\infty$ (the proof for $-\infty$ would be similar; you can do it as an excercise). Let $x$ be given. We need to show that there exists an $N$ so that $a_n>x$ for all $n>N$. But since $\lim b_n=1$, we can find an $N_1$ so that $1/2<b_n<3/2$ for all $n>N_1$. Similarly, we can find an $N_2$ so that $a_nb_n>x$ for all $n>N_2$. Rearranging it:
$$a_n >\frac{x}{b_n}>x \frac{2}{3}$$
So we can pick an $N>\text{max}\{N_1,N_2\}$.
The proof for the $\gamma \in \mathbb{R}$ is similar: the idea is that if $b_n$ is close to $1$ and $a_nb_n$ is close to $\gamma$ then $a_n=\frac{a_nb_n}{b_n}$ must be close to $\frac{\gamma}{1}=\gamma$.
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