proof $\frac{a_n}{b_n} \to 1$ $\land$ $a_n \to 0$ then $b_n \to 0$ tried to proof this statement directly but without any success.
so I tried the contradiction path but again get stucked, as follows:
let's assume $b_n \not\to 0$, then either $b_n \to K$ for any real K, or $b_n \to \pm\infty$, or $b_n$ doesn't converge.
if $b_n \to K$ then by the 'arithmetic of limits' $\frac{a_n}{b_n} \to 0$ in contradiction.
if $b_n \to \infty$ then for every M>0 there is N that for every n>N, $b_N$>M. thus  $|\frac{a_n}{b_n}-1| = \frac{|a_n-b_n|}{|b_n|} \le \frac{|a_n|+|b_n|}{M} \le \frac{\epsilon+|b_n|}{M}$ and here I get stucked. the same goes for $b_n \to -\infty$
edit: can I tell from $\frac{\epsilon+|b_n|}{M}$ that it tends to $\infty$ because $b_n \to \infty$ and thus $|\frac{a_n}{b_n}-1|$ can't be less then some $\epsilon$?
would appreciate your help
 A: You're method is perhaps too "complicated" for this level of proof. Indeed note that
$$ \lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{b_n}{a_n}a_n = \lim_{n \to \infty} \frac{b_n}{a_n} \lim_{n \to \infty}a_n = 1 \cdot 0 = 0$$
since all the limits exist the above operations are valid.
A: Hint: since $a_n \to 0$ then for sufficiently large $n$ $$\left|\frac{a_n}{b_n}\right|<\frac{1}{|b_n|}$$ Now take the limit ($n \to +\infty$) on both sides, assuming that $|b_n| \to +\infty$ and you will obtain a contradiction.
A: $$b_n = \frac{b_n}{a_n} a_n.$$
Providing that the limits exist, we know that
$$\lim_{n \to \infty} A_n B_n = \lim_{n \to \infty} A_n \lim_{n \to \infty} B_n.$$
Also, we know that, provided the limit exists and is non-zero,
$$\lim_{n \to \infty} C_n^{-1} = \left(\lim_{n \to \infty} C_n \right)^{-1}.$$
From these and my first equation, you should hopefully be able to show the result. (I don't want to give you all the details, as it's best if you determine them yourself!) :)
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If you don't know the results about convergence of pointwise products and pointwise reciprocals, then I'd recommend having a go at proving them. They're not too hard to prove, but getting the details right is certainly non-trivial; it would be a good exercise. If you can't prove it, then I'm sure you can find a reference, if not another SE question, proving them.
