Dgdrxrt

Let $(a_n)$ and $(b_n)$ be sequences. If $(a_n) to Aneq 0$ and $(a_n b_n)to AB$ then $(b_n)to B$

I know that we need to show this for both $A > 0$ and $A < 0$. But I am having problems using the assumptions to deduce that

$$|b_n - B| < epsilon$$

I know that since $<a_n>to A$ then there exists an $N_1in mathbbN$ such that for all $n > N_1$

$$|a_n - A| < epsilon$$

Similarly, since $<a_n b_n>to AB$ then there exists an $N_2inmathbbN$ such that for all $n > N_2$

$$|a_n b_n - AB| < epsilon$$

Any help would be appreciated.

Background information:

Theorem 8.7 - If the sequence $(a_n)$ converges to $A$ and the sequence $(b_n)$ converges to $B$ then the sequence $(a_n b _n)$ converges to $AB$

Lemma 8.1 - If $a_nneq 0$ for all $ninmathbbN$ and if $(a_n)$ converges to $Aneq 0$ then the sequence $(1/a_n)$ converges to $1/A$

Attempted proof - Using these two theorems above we can write $(a_n b_n/a_n) = (b_n)$ which converges to $B$.

I don't think this is complete can someone help me add details?

Your proof is correct but I'd recommend justifying that $a_n neq 0$ for all $n$ sufficiently large. Note that this is necessary to make sense of the expression
$$
b_n = left( fraca_nb_na_n right).
$$
To justify that $a_n neq 0$ for all $n$ sufficiently large, we will use that $A = lima_n neq 0$. Because $a_n to A$, there exists $N in mathbbN$ such that
$$
|a_n - A| < epsilon = frac2
$$
for all $n geq N$. Therefore, for all such $n$,
beginalign*
|A| - |a_n| leq leftvert a_n - A rightvert< frac2
endalign*
whence $$
0 < frac2 < |a_n|, quad forall ngeq N.
$$
This means that the $N$-tail of $(a_n)$ is never zero and consequently the sequence
$$
fracb_na_n
$$
is well defined for all $n geq N$ (and we only really care about the tail of a sequence when discussing limits). Then, by invoking the theorem and lemma you've cited, it follows that
beginalign*
limb_n = limleft(fraca_n b_na_n right)
= fraclim(a_nb_n)lima_n
= fracABA = B.
endalign*

Note that if $a_n to a$ then the sequence is bounded, that is, there is some $L$ such that $|a_n| le L$.

Suppose $b_n to b$ then $|a_nb_n-ab| le |a_n b_n -a_n b| + |a_n b - ab| le L|b_n-b|+|b||a_n-a|$. Hence $a_nb_n to ab$.

Suppose $a neq 0$. Since $a_n to a$, we see that $|a_n| ge 1over 2 |a|$ for $n$ sufficiently large (and hence non zero). Then $|1 over a - 1over a_n| = le ^2 |a_n-a|$ and hence $1 over a_n to 1 over a$.

Hence if $a_nb_n to ab$ and $a_n to a neq 0$ then
$a_n b_n 1 over a_n = b_n to ab 1 over a = b$.